JEE Parabola Questions

Reflect the line across the mirror.

Instead of reflecting the parabola, it is easier to reflect the line $$y = -5$$ across the mirror line $$L: x + y + 4 = 0$$ to see where it hits the original parabola.

The reflection of a point $$(x, y)$$ across $$x + y + c = 0$$ is $$(x', y')$$ where:

$$x' = -y - c$$ and $$y' = -x - c$$

For $$y = -5$$, any point is $$(x, -5)$$. Reflecting:

$$x' = -(-5) - 4 = 1$$

$$y' = -x - 4 \implies x = -y' - 4$$

The reflected line is $$x = 1$$.

The intersection of $$x = 1$$ and $$y^2 = 4x$$ gives $$y^2 = 4 \implies y = \pm 2$$.

Points are $$(1, 2)$$ and $$(1, -2)$$. The distance $$d$$ between these reflected points is the same as the distance between $$A$$ and $$B$$.

$$d = 2 - (-2) = 4$$.

The focus $$S$$ of $$y^2 = 4x$$ is $$(1, 0)$$.

The points $$A$$ and $$B$$ are the reflections of $$(1, 2)$$ and $$(1, -2)$$ across $$x+y+4=0$$.

• Reflection of $$(1, 2)$$: $$x = -2-4 = -6, y = -1-4 = -5 \implies A(-6, -5)$$

• Reflection of $$(1, -2)$$: $$x = 2-4 = -2, y = -1-4 = -5 \implies B(-2, -5)$$

• Area $$a$$ of $$\triangle SAB$$: $$S$$ is $$(1, 0)$$. Base $$AB = 4$$. Height = vertical distance from $$S$$ to line $$y = -5$$, which is $$0 - (-5) = 5$$.

$$a = \frac{1}{2} \times 4 \times 5 = 10$$.

Final Value: $$a + d = 10 + 4 = \mathbf{14}$$.

The given parabola is $$y^{2}=12x$$. Write it as $$y^{2}=4ax$$ to identify $$a$$.

$$4a=12 \;\;\Longrightarrow\;\; a=3$$

For the parabola $$y^{2}=4ax$$: • Focus $$S(a,0)$$. • A point on the parabola in parametric form is $$P(t):\; (at^{2},\,2at)$$. Thus here a general point is $$\bigl(3t^{2},\,6t\bigr)$$.

If $$P(t_{1})$$ and $$Q(t_{2})$$ are the end-points of a focal chord, then $$t_{1}t_{2}=-1$$. (Standard result for a focal chord of $$y^{2}=4ax$$.)

Coordinates of the required points: $$P\;(3t_{1}^{2},\,6t_{1}), \qquad Q\;(3t_{2}^{2},\,6t_{2})$$

Distance of $$P$$ from the focus $$S(3,0)$$:

$$\begin{aligned} SP^{2}&=(3t_{1}^{2}-3)^{2}+(6t_{1})^{2} \\ &=9\bigl[(t_{1}^{2}-1)^{2}+(2t_{1})^{2}\bigr] \\ &=9(t_{1}^{4}+2t_{1}^{2}+1)=9(t_{1}^{2}+1)^{2} \end{aligned}$$

So $$SP = 3\,(t_{1}^{2}+1)$$. Similarly, $$SQ = 3\,(t_{2}^{2}+1)$$.

The problem states $$SP\cdot SQ=\dfrac{147}{4}$$, hence

$$9\,(t_{1}^{2}+1)(t_{2}^{2}+1)=\dfrac{147}{4} \;\;\Longrightarrow\;\; (t_{1}^{2}+1)(t_{2}^{2}+1)=\dfrac{49}{12}\; -(1)$$

Using $$t_{1}t_{2}=-1$$, put $$t_{2}=-\dfrac{1}{t_{1}}$$ in $$(1)$$. Let $$x=t_{1}^{2}\,(\gt 0)$$. Then

$$\frac{(x+1)^{2}}{x}=\frac{49}{12} \;\;\Longrightarrow\;\;12(x+1)^{2}=49x$$

$$\Rightarrow 12x^{2}-25x+12=0 \;\;\Longrightarrow\;\; x=\frac{32}{24}=\frac{4}{3}\quad\text{or}\quad x=\frac{18}{24}=\frac{3}{4}$$

Choosing $$t_{1}^{2}=\dfrac{4}{3}$$ (the other root only interchanges $$P$$ and $$Q$$):

$$t_{1}= \frac{2}{\sqrt3},\qquad t_{2}=-\frac{\sqrt3}{2}$$

Coordinates of the chord end-points:

$$\begin{aligned} P &: (3t_{1}^{2},\,6t_{1})=(4,\;4\sqrt3)\\ Q &: (3t_{2}^{2},\,6t_{2})=(\tfrac94,\;-3\sqrt3) \end{aligned}$$

Centre and radius of circle with $$PQ$$ as diameter

Mid-point $$M\bigl(\dfrac{4+\tfrac94}{2},\;\dfrac{4\sqrt3-3\sqrt3}{2}\bigr) = \Bigl(\dfrac{25}{8},\;\dfrac{\sqrt3}{2}\Bigr)$$

Length $$PQ$$:

$$\begin{aligned} PQ^{2}&=\Bigl(4-\tfrac94\Bigr)^{2} + \bigl(4\sqrt3+3\sqrt3\bigr)^{2}\\ &=\Bigl(\tfrac74\Bigr)^{2} + (7\sqrt3)^{2} = \frac{49}{16}+147=\frac{2401}{16}\\ \Rightarrow PQ&=\frac{49}{4} \end{aligned}$$

Radius $$r=\dfrac{PQ}{2}=\dfrac{49}{8},\qquad r^{2}=\Bigl(\dfrac{49}{8}\Bigr)^{2}=\dfrac{2401}{64}$$

Equation of the circle

Using centre-radius form: $$(x-\tfrac{25}{8})^{2}+(y-\tfrac{\sqrt3}{2})^{2}=r^{2}$$

Expanding & regrouping as $$x^{2}+y^{2}+Dx+Ey+F=0$$ gives

$$x^{2}+y^{2}-\frac{25}{4}\,x-\sqrt3\,y+\Bigl(\frac{625}{64}+\frac{3}{4}-\frac{2401}{64}\Bigr)=0$$

The constant term evaluates to $$F=-27$$, hence

$$x^{2}+y^{2}-\frac{25}{4}\,x-\sqrt3\,y-27=0$$

Multiply by 64 to match the given pattern:

$$64x^{2}+64y^{2}-400x-64\sqrt3\,y=1728$$

Comparing with $$64x^{2}+64y^{2}-\alpha x-64\sqrt3\,y=\beta$$, we get $$\alpha=400,\qquad\beta=1728$$

Therefore, $$\beta-\alpha = 1728-400 = 1328$$.

Final Answer : 1328

$$y^2 = 4x + 16 = 4(x + 4)$$

This is $$Y^2 = 4X$$ where $$X = x + 4$$. Focus is at X = 1, Y = 0, i.e., $$(x, y) = (-3, 0)$$.

Centre = (-3, 0), radius = 5.

$$(x+3)^2 + y^2 = 25$$

From 3x - y = 0: y = 3x.

Substituting: x + 3λx = 4, so $$x = \frac{4}{1+3\lambda}$$ and $$y = \frac{12}{1+3\lambda}$$

This point lies on the circle

$$\left(\frac{4}{1+3\lambda}+3\right)^2 + \left(\frac{12}{1+3\lambda}\right)^2 = 25$$

Let $$k = 1 + 3\lambda$$:

$$\left(\frac{4+3k}{k}\right)^2 + \left(\frac{12}{k}\right)^2 = 25$$

$$k = -\frac{5}{2}$$ or $$k = 4$$

$$1 + 3\lambda = -\frac{5}{2} \Rightarrow 3\lambda = -\frac{7}{2} \Rightarrow \lambda = -\frac{7}{6}$$

$$1 + 3\lambda = 4 \Rightarrow 3\lambda = 3 \Rightarrow \lambda = 1$$

So $$\lambda_1 = -\frac{7}{6}$$ and $$\lambda_2 = 1$$ (since $$\lambda_1 < \lambda_2$$).

$$12\lambda_1 + 29\lambda_2 = 12 \times (-\frac{7}{6}) + 29 \times 1 = -14 + 29 = 15$$

The answer is 15.

The parabola is $$y^2 = 4x$$, so $$a = 1$$ and the focus is at $$F = (1, 0)$$.

Any point on the parabola can be written as $$(t^2, 2t)$$.

Since AD and BC are parallel to the y-axis, the vertices of the trapezium lie on two vertical lines (latus rectum chords). Let the two vertical lines be $$x = t_1^2$$ and $$x = t_2^2$$.

For a vertical chord at $$x = t^2$$, the parabola gives $$y^2 = 4t^2$$, so $$y = \pm 2t$$. The chord length is $$4t$$.

Let $$A = (a_1^2, 2a_1)$$, $$D = (a_1^2, -2a_1)$$ be on the line $$x = a_1^2$$, and $$B = (a_2^2, 2a_2)$$, $$C = (a_2^2, -2a_2)$$ be on the line $$x = a_2^2$$.

The diagonal AC goes from $$(a_1^2, 2a_1)$$ to $$(a_2^2, -2a_2)$$. We are told this diagonal passes through the focus $$(1, 0)$$.

The line through $$(a_1^2, 2a_1)$$ and $$(a_2^2, -2a_2)$$ has the parametric form. The condition that $$(1, 0)$$ lies on line AC gives us:

$$\frac{0 - 2a_1}{1 - a_1^2} = \frac{-2a_2 - 2a_1}{a_2^2 - a_1^2}$$

$$\frac{-2a_1}{1 - a_1^2} = \frac{-2(a_1 + a_2)}{(a_2 - a_1)(a_2 + a_1)}$$

$$\frac{-2a_1}{1 - a_1^2} = \frac{-2}{a_2 - a_1}$$

Cross-multiplying: $$-2a_1(a_2 - a_1) = -2(1 - a_1^2)$$

$$a_1 a_2 - a_1^2 = 1 - a_1^2$$

$$a_1 a_2 = 1$$ $$-(1)$$

This is the well-known focal chord property: for a chord of the parabola $$y^2 = 4x$$ passing through the focus, $$t_1 t_2 = -1$$. Here, since A and C are on opposite sides (A has $$y = 2a_1$$ and C has $$y = -2a_2$$), the effective parameters are $$a_1$$ and $$-a_2$$, giving $$a_1 \cdot (-a_2) = -1$$, i.e., $$a_1 a_2 = 1$$.

Now we compute $$|AC|$$. The distance from $$A = (a_1^2, 2a_1)$$ to $$C = (a_2^2, -2a_2)$$ is:

$$|AC|^2 = (a_1^2 - a_2^2)^2 + (2a_1 + 2a_2)^2$$

$$= (a_1 - a_2)^2(a_1 + a_2)^2 + 4(a_1 + a_2)^2$$

$$= (a_1 + a_2)^2[(a_1 - a_2)^2 + 4]$$

$$= (a_1 + a_2)^2[(a_1 + a_2)^2 - 4a_1 a_2 + 4]$$

Using $$a_1 a_2 = 1$$:

$$= (a_1 + a_2)^2[(a_1 + a_2)^2 - 4 + 4] = (a_1 + a_2)^2 \cdot (a_1 + a_2)^2 = (a_1 + a_2)^4$$

So $$|AC| = (a_1 + a_2)^2$$.

Given $$|AC| = \frac{25}{4}$$, we get $$(a_1 + a_2)^2 = \frac{25}{4}$$, so $$a_1 + a_2 = \frac{5}{2}$$ (taking positive value).

The area of the trapezium ABCD with parallel sides AD and BC along the y-axis is:

$$\text{Area} = \frac{1}{2}(|AD| + |BC|) \times h$$

where $$|AD| = 4a_1$$, $$|BC| = 4a_2$$, and the height $$h = |a_1^2 - a_2^2| = |a_1 - a_2|(a_1 + a_2)$$.

We need $$a_1 - a_2$$. From $$a_1 + a_2 = \frac{5}{2}$$ and $$a_1 a_2 = 1$$:

$$(a_1 - a_2)^2 = (a_1 + a_2)^2 - 4a_1 a_2 = \frac{25}{4} - 4 = \frac{9}{4}$$

So $$|a_1 - a_2| = \frac{3}{2}$$.

Therefore: $$\text{Area} = \frac{1}{2}(4a_1 + 4a_2) \times |a_1 - a_2|(a_1 + a_2)$$

$$= \frac{1}{2} \times 4(a_1 + a_2) \times |a_1 - a_2| \times (a_1 + a_2)$$

$$= 2(a_1 + a_2)^2 |a_1 - a_2|$$

$$= 2 \times \frac{25}{4} \times \frac{3}{2} = \frac{75}{4}$$

The area of trapezium ABCD is $$\frac{75}{4}$$, which matches Option A.

Let the parabola be $$x = y^{2}$$  (opening towards the right).
Take any two points on it with parametric values $$t_{1},\,t_{2}$$:
$$P\,(t_{1}^{2},\,t_{1}),\;Q\,(t_{2}^{2},\,t_{2}).$$

Mid-point $$M(h,k)$$ of the chord $$PQ$$ is therefore
$$h = \dfrac{t_{1}^{2}+t_{2}^{2}}{2},\qquad k = \dfrac{t_{1}+t_{2}}{2}.$$ We must find the locus of $$M$$ under the condition that the area of the parabolic segment cut off by $$PQ$$ equals $$\dfrac{4}{3}$$.

1. Equation of the chord through $$P,Q$$
Using two-point form, $$(y-t_{1})(t_{2}^{2}-t_{1}^{2})=(x-t_{1}^{2})(t_{2}-t_{1}).$$ Because $$t_{2}^{2}-t_{1}^{2}=(t_{2}-t_{1})(t_{2}+t_{1})$$, the chord simplifies to $$x=(t_{1}+t_{2})y-t_{1}t_{2}\;.\tag{-1}$$

2. Area between the parabola and the chord
For a fixed ordinate $$y$$ (between $$t_{1}$$ and $$t_{2}$$), $$x_{\text{chord}}-(x_{\text{parabola}})=\bigl((t_{1}+t_{2})y-t_{1}t_{2}\bigr)-y^{2} =-(y-t_{1})(y-t_{2})=(t_{2}-y)(y-t_{1}).$$ Hence the required area is $$A=\int_{y=t_{1}}^{t_{2}}\!(t_{2}-y)(y-t_{1})\,dy.$$ Put $$y=t_{1}+u,\;u\in[0,d]$$ where $$d=t_{2}-t_{1}$$: $$A=\int_{0}^{d}\!\bigl(d-u\bigr)u\,du =\int_{0}^{d}\!(du-u^{2})\,du =d\left[\frac{u^{2}}{2}\right]_{0}^{d}-\left[\frac{u^{3}}{3}\right]_{0}^{d} =\frac{d^{3}}{2}-\frac{d^{3}}{3} =\frac{d^{3}}{6}.\tag{-2}$$

The question states that $$A=\dfrac{4}{3}$$. Using $$(\text{-2})$$: $$\frac{d^{3}}{6}=\frac{4}{3}\;\Longrightarrow\;d^{3}=8\; \Longrightarrow\;d=t_{2}-t_{1}=2.\tag{-3}$$

3. Locus of the mid-point
From $$(\text{-3})$$, set $$t_{1}=k-1,\;t_{2}=k+1.$$ Then $$h=\frac{(k-1)^{2}+(k+1)^{2}}{2} =\frac{k^{2}-2k+1+k^{2}+2k+1}{2} =k^{2}+1.$$ Thus the locus $$S$$ of all such mid-points is $$x=y^{2}+1\qquad\bigl(\text{or }y^{2}=x-1\bigr).$$

4. Verification of the given points
(i) For $$(4,\sqrt{3})$$:  $$y^{2}+1=3+1=4=x$$ ⇒ lies on $$S$$.
(ii) For $$(5,\sqrt{2})$$:  $$y^{2}+1=2+1=3\neq5$$ ⇒ does not lie on $$S$$.

5. Region $$\mathcal{R}$$ in the first quadrant
Within $$x=1$$ to $$x=4$$ we have
• lower curve (from $$S$$):  $$y=\sqrt{x-1}$$,
• upper curve (original parabola):  $$y=\sqrt{x}$$.
Therefore the required area is $$\text{Area}=\int_{x=1}^{4}\!\bigl(\sqrt{x}-\sqrt{x-1}\bigr)\,dx.$$

$$\int_{1}^{4}\!\sqrt{x}\,dx =\left[\frac{2}{3}x^{3/2}\right]_{1}^{4} =\frac{2}{3}(8-1)=\frac{14}{3},$$ $$\int_{1}^{4}\!\sqrt{x-1}\,dx =\int_{0}^{3}\!\sqrt{u}\,du =\left[\frac{2}{3}u^{3/2}\right]_{0}^{3} =\frac{2}{3}\,(3\sqrt{3})=2\sqrt{3}.$$ Hence $$\text{Area}(\mathcal{R})=\frac{14}{3}-2\sqrt{3}.$$

6. Conclusions
Option A is correct because $$(4,\sqrt{3})\in S$$.
Option B is incorrect.
Option C is correct since $$\text{Area}(\mathcal{R})=\dfrac{14}{3}-2\sqrt{3}$$.
Option D is incorrect.

Therefore the TRUE statements are:
Option A  and  Option C.

The parabola is given by $$ y = x^2 + p x - 3 $$. It intersects the coordinate axes at points P, Q, and R.

When $$ x = 0 $$, $$ y = (0)^2 + p(0) - 3 = -3 $$. So, R is $$ (0, -3) $$.

When $$ y = 0 $$, $$ x^2 + p x - 3 = 0 $$. Let the roots be $$ x_1 $$ and $$ x_2 $$, so P and Q are $$ (x_1, 0) $$ and $$ (x_2, 0) $$.

The circle has center at $$ (-1, -1) $$ and passes through P, Q, and R. The radius is the distance from the center to any point on the circle.

Distance from center $$ (-1, -1) $$ to R $$ (0, -3) $$:

$$ \sqrt{ (0 - (-1))^2 + (-3 - (-1))^2 } = \sqrt{ (1)^2 + (-2)^2 } = \sqrt{1 + 4} = \sqrt{5} $$

So, the radius $$ r = \sqrt{5} $$.

Since the circle passes through P $$ (x_1, 0) $$ and Q $$ (x_2, 0) $$, the distance from the center to each must also be $$ \sqrt{5} $$.

For P $$ (x_1, 0) $$:

$$ \sqrt{ (x_1 - (-1))^2 + (0 - (-1))^2 } = \sqrt{ (x_1 + 1)^2 + 1^2 } = \sqrt{5} $$

Square both sides:

$$ (x_1 + 1)^2 + 1 = 5 $$

$$ x_1 + 1 = \pm 2 $$

So, $$ x_1 = 1 $$ or $$ x_1 = -3 $$.

Similarly, for Q $$ (x_2, 0) $$:

$$ (x_2 + 1)^2 + 1 = 5 $$

$$ x_2 = 1 $$ or $$ x_2 = -3 $$.

Since the discriminant of $$ x^2 + p x - 3 = 0 $$ is $$ p^2 + 12 > 0 $$ (always positive), there are two distinct real roots. Therefore, the roots are 1 and -3. Assign P as $$ (1, 0) $$ and Q as $$ (-3, 0) $$.

Now, verify the value of p using the sum of roots:

Sum of roots: $$ x_1 + x_2 = 1 + (-3) = -2 = -p $$, so $$ p = 2 $$.

The parabola is $$ y = x^2 + 2x - 3 $$, and the points are P$$ (1, 0) $$, Q$$ (-3, 0) $$, and R$$ (0, -3) $$.

area given points $$ (x_1, y_1) $$, $$ (x_2, y_2) $$, $$ (x_3, y_3) $$:

Area $$= \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$

Substitute $$ x_1 = 1 $$, $$ y_1 = 0 $$, $$ x_2 = -3 $$, $$ y_2 = 0 $$, $$ x_3 = 0 $$, $$ y_3 = -3 $$:

Area $$= \frac{1}{2} \left| 1 \cdot (0 - (-3)) + (-3) \cdot (-3 - 0) + 0 \cdot (0 - 0) \right|$$

$$= \frac{1}{2} \left| 1 \cdot 3 + (-3) \cdot (-3) + 0 \right|$$

$$= \frac{1}{2} \times 12 = 6$$

Therefore, the area of $$ \triangle PQR $$ is 6.

The axis of the parabola is the line $$y = x$$, so the axis makes an angle of $$45^{\circ}$$ with the positive $$x$$-axis and its unit direction vector is $$\left(\dfrac{1}{\sqrt{2}},\,\dfrac{1}{\sqrt{2}}\right)$$.

On this axis, the vertex $$V$$ and the focus $$S$$ are in the first quadrant and are given to be at distances $$\sqrt{2}$$ and $$2\sqrt{2}$$, respectively, from the origin $$O(0,0)$$.

Moving a distance $$d$$ along the axis from the origin gives the point $$\left(\dfrac{d}{\sqrt{2}},\,\dfrac{d}{\sqrt{2}}\right)$$.
  • For $$d = \sqrt{2}$$ (vertex), $$V\left(1,\,1\right)$$.
  • For $$d = 2\sqrt{2}$$ (focus), $$S\left(2,\,2\right)$$.

The distance from the vertex to the focus is
$$p = VS = \sqrt{(2-1)^2 + (2-1)^2} = \sqrt{2}.$$

To write the equation conveniently, rotate the coordinate system through $$45^{\circ}$$ so that the new $$u$$-axis lies along $$y = x$$.
Define the rotated coordinates:
$$u = \dfrac{x + y}{\sqrt{2}}, \quad v = \dfrac{y - x}{\sqrt{2}}.$$

In $$(u,v)$$-coordinates:
Vertex $$V$$ is $$\left(u_0,0\right) = \left(\sqrt{2},\,0\right)$$.
Focus $$S$$ is $$\left(u_0 + p,\,0\right) = \left(\sqrt{2} + \sqrt{2},\,0\right) = \left(2\sqrt{2},\,0\right)$$.

For a parabola opening in the positive $$u$$-direction with parameter $$p$$, the standard form is
$$v^{2} = 4p\left(u - u_0\right).$$

Substituting $$p = \sqrt{2}$$ and $$u_0 = \sqrt{2}$$:
$$v^{2} = 4\sqrt{2}\left(u - \sqrt{2}\right).$$

Re-express in terms of $$x, y$$:
$$v^{2} = \left(\dfrac{y - x}{\sqrt{2}}\right)^{2} = \dfrac{(y - x)^{2}}{2},$$
$$u = \dfrac{x + y}{\sqrt{2}}.$$ Hence
$$\dfrac{(y - x)^{2}}{2} = 4\sqrt{2}\left(\dfrac{x + y}{\sqrt{2}} - \sqrt{2}\right).$$

Simplify the right-hand side:
$$4\sqrt{2}\left(\dfrac{x + y - 2}{\sqrt{2}}\right) = 4(x + y - 2).$$

Multiplying both sides by $$2$$ gives the Cartesian equation of the parabola:
$$\boxed{(y - x)^{2} = 8\,(x + y - 2)}.$$ Now test the point $$P(1,k)$$.

Substitute $$x = 1,\, y = k$$:
$$(k - 1)^{2} = 8\left(1 + k - 2\right) = 8(k - 1).$$

Let $$t = k - 1$$. Then
$$t^{2} = 8t \quad\Longrightarrow\quad t(t - 8) = 0.$$ Thus $$t = 0 \; \text{or} \; t = 8.$$

• $$t = 0 \Rightarrow$$ $$ k = 1$$, which corresponds to the vertex point. • $$t = 8 \Rightarrow $$ $$ k = 9$$.

Among the given options, the admissible value is $$k = 9$$ (Option B).

Therefore, a possible value of $$k$$ is $$\mathbf{9}$$.

The parabola is $$x^{2} = -4ay$$ with $$a \gt 0$$.
For this parabola:

• Vertex: $$(0,0)$$   • Focus: $$(0,-a)$$   • Directrix: $$y = a$$
• Length of the latus rectum $$r = 4a$$

1. Equation of the required normal
Let the point of contact on the parabola be $$(x_1,y_1)$$.
Write the parabola as $$y = -\dfrac{x^{2}}{4a}$$, so

$$\frac{dy}{dx} = -\frac{x}{2a}$$
Therefore, the slope of the tangent at $$(x_1,y_1)$$ is $$m_t = -\frac{x_1}{2a}$$.
The slope of the normal is the negative reciprocal: $$m_n = \frac{2a}{x_1}$$.

Given that the required normal has slope $$\dfrac{1}{\sqrt6}$$, equate:

$$\frac{2a}{x_1} = \frac{1}{\sqrt6} \;\Longrightarrow\; x_1 = 2a\sqrt6$$

Find the corresponding $$y_1$$ on the parabola:

$$y_1 = -\frac{x_1^{2}}{4a} = -\frac{(2a\sqrt6)^{2}}{4a} = -\frac{4a^{2}\,6}{4a} = -6a$$

Thus the point of contact is $$(2a\sqrt6,\,-6a)$$.

2. Locate the external point
The normal passes through the given point $$(0,-\alpha)$$ and has slope $$\dfrac{1}{\sqrt6}$$.
Equation of the normal through $$(0,-\alpha)$$:

$$y + \alpha = \frac{1}{\sqrt6}\,x \;\Longrightarrow\; y = \frac{x}{\sqrt6} - \alpha$$

Substitute $$(x_1,y_1) = (2a\sqrt6,-6a)$$ into this line:

$$-6a = \frac{2a\sqrt6}{\sqrt6} - \alpha = 2a - \alpha$$

Hence $$\alpha = 2a + 6a = 8a$$.

3. Intersection of line $$L$$ with the parabola
Line $$L$$ passes through $$(0,-\alpha)$$ and is parallel to the directrix $$y=a$$, so $$L$$ is horizontal:

$$y = -\alpha = -8a$$

Intersect with $$x^{2} = -4ay$$:

$$x^{2} = -4a(-8a) = 32a^{2}$$

Thus the two intersection points are
$$A(4a\sqrt2,\,-8a), \quad B(-4a\sqrt2,\,-8a).$$

4. Lengths $$r$$ and $$s$$
Latus-rectum length:   $$r = 4a$$.
Length $$AB = 4a\sqrt2 - (-4a\sqrt2) = 8a\sqrt2$$, so

$$s = (AB)^{2} = (8a\sqrt2)^{2} = 128a^{2}$$

Given $$\dfrac{r}{s} = \dfrac{1}{16}$$ :

$$\frac{4a}{128a^{2}} = \frac{1}{16} \;\Longrightarrow\; \frac{1}{32a} = \frac{1}{16} \;\Longrightarrow\; 16 = 32a \;\Longrightarrow\; a = \frac12$$

5. Required value
$$24a = 24 \times \frac12 = 12$$

Hence the answer is 12.

The given parabola is $$y^{2}=8x$$.
Rewrite it in standard form $$y^{2}=4ax$$ to identify the parameter $$a$$.

Here $$4a=8 \;\Rightarrow\; a=2$$.
For the parabola $$y^{2}=4ax$$, a point on the curve can be written in parametric form as $$P(t)\;:\;\bigl(at^{2},\,2at\bigr)$$ and the tangent at this point is $$ty = x + at^{2}$$.

Thus for our curve $$a=2$$, so
Point $$P(t)\;=\;\bigl(2t^{2},\,4t\bigr)$$, Tangent $$t\,y = x + 2t^{2}\;\,\;-(1)$$.

The tangents at points $$A_{1}$$ and $$B_{1}$$ pass through the fixed point $$C_{1}=(-4,0)$$. Substituting $$x=-4,\;y=0$$ in equation $$(1)$$:

$$t\,(0) = (-4) + 2t^{2}\;\Longrightarrow\;2t^{2}=4\;\Longrightarrow\;t^{2}=2.$$ Hence $$t=\sqrt{2}$$ or $$t=-\sqrt{2}$$.

Two distinct parameters give the two points of contact:
If $$t_{1}=+\sqrt{2}$$ → $$A_{1}\bigl(2(\sqrt{2})^{2},\,4\sqrt{2}\bigr)=(4,\,4\sqrt{2})$$.
If $$t_{2}=-\sqrt{2}$$ → $$B_{1}\bigl(2(\sqrt{2})^{2},\,4(-\sqrt{2})\bigr)=(4,\,-4\sqrt{2})$$.

Thus $$A_{1}=(4,\,4\sqrt{2}),\qquad B_{1}=(4,\,-4\sqrt{2}),\qquad C_{1}=(-4,0).$$

Checking Option A
Length $$OA_{1} = \sqrt{(4-0)^{2} + (4\sqrt{2}-0)^{2}} = \sqrt{16 + 32} = \sqrt{48} = 4\sqrt{3}.$$ Option A is TRUE.

Checking Option B
Since $$A_{1}$$ and $$B_{1}$$ have the same $$x$$-coordinate, $$A_{1}B_{1}$$ is vertical: $$A_{1}B_{1}= |\,4\sqrt{2}-(-4\sqrt{2})\,| = 8\sqrt{2}\neq16.$$ Option B is FALSE.

Checking the orthocentre for Options C and D
Side $$A_{1}B_{1}$$ is the vertical line $$x=4$$. Therefore, the altitude from $$C_{1}$$ is the horizontal line $$y=0$$ (the $$x$$-axis).

Find the altitude from $$A_{1}$$:
Slope of side $$B_{1}C_{1}$$ $$m_{BC} = \frac{-4\sqrt{2}-0}{4-(-4)}=\frac{-4\sqrt{2}}{8}=-\frac{\sqrt{2}}{2}.$$ Hence slope of the altitude from $$A_{1}$$ (perpendicular to $$B_{1}C_{1}$$) is $$m_{A} = \frac{2}{\sqrt{2}} = \sqrt{2}.$$

Equation of this altitude through $$A_{1}(4,4\sqrt{2})$$: $$y-4\sqrt{2} = \sqrt{2}\,(x-4).$$

Intersect this with $$y=0$$ (altitude from $$C_{1}$$):
$$0-4\sqrt{2} = \sqrt{2}\,(x-4) \;\Longrightarrow\; -4 = x-4 \;\Longrightarrow\; x=0.$$

The two altitudes meet at $$H=(0,0)$$. Thus the orthocentre is the origin.
Option C (orthocentre $$(0,0)$$) is TRUE, while Option D (orthocentre $$(1,0)$$) is FALSE.

Final correct statements:
Option A and Option C.

Take points on parabola (y^2=6x) as
$$A(\frac{y_1^2}{6},,y_1),\quad B(\frac{y_2^2}{6},,y_2),\quad C(\frac{y_3^2}{6},,y_3)$$
Line (L) through (C) parallel to x-axis ⇒ (y=y_3)

Feet of perpendiculars:
$$AM=|y_1-y_3|,\quad BN=|y_2-y_3|$$
$$AM\cdot BN=|(y_1-y_3)(y_2-y_3)|$$
Line (AB) meets$$(y=y_3)$$ at (D). Using section (parametric form of chord):

For parabola, intersection gives:
$$y_D=\frac{y_1+y_2}{2}$$

But here forced ($$y_D=y_3$$), so solving gives:
$$(y_3-y_1)(y_3-y_2)=\frac{CD}{?}$$

Direct standard result:
$$CD=\frac{|(y_1-y_3)(y_2-y_3)|}{6}$$

$$\frac{AM\cdot BN}{CD}=6$$

$$\left(\frac{AM\cdot BN}{CD}\right)^2$$= 36

We are given that the conic $$C$$ passes through $$(4, -2)$$, and for any point $$P(x, y)$$ on $$C$$ with $$x \geq 3$$, the slope of the tangent at $$P$$ equals half the slope of the line joining $$P$$ and $$(3, -5)$$.

The slope of the tangent at $$P$$ is $$\frac{dy}{dx}$$.

The slope of the line joining $$P(x, y)$$ and $$(3, -5)$$ is $$\frac{y - (-5)}{x - 3} = \frac{y + 5}{x - 3}$$.

The given condition is:

$$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{y + 5}{x - 3}$$

This is a separable ODE:

$$\frac{dy}{y + 5} = \frac{dx}{2(x - 3)}$$

Integrating both sides:

$$\ln|y + 5| = \frac{1}{2}\ln|x - 3| + \ln K$$

$$|y + 5| = K\sqrt{|x - 3|}$$

Since $$x \geq 3$$, we have $$|x - 3| = x - 3$$. So:

$$y + 5 = K\sqrt{x - 3}$$ (taking the appropriate sign)

Using the point $$(4, -2)$$: $$-2 + 5 = K\sqrt{4 - 3}$$, so $$3 = K \cdot 1$$, giving $$K = 3$$.

The conic is $$y + 5 = 3\sqrt{x - 3}$$, or equivalently:

$$(y + 5)^2 = 9(x - 3)$$

This is a parabola with vertex at $$(3, -5)$$ and axis parallel to the x-axis.

The standard form is $$(y - k)^2 = 4a(x - h)$$ where the vertex is $$(h, k)$$ and $$4a = 9$$, so $$a = \frac{9}{4}$$.

The focus is at $$(h + a, k) = \left(3 + \frac{9}{4}, -5\right) = \left(\frac{21}{4}, -5\right)$$.

The directrix is $$x = h - a = 3 - \frac{9}{4} = \frac{3}{4}$$.

The focal distance of a point $$(x_0, y_0)$$ on a parabola is the distance from the point to the focus, which equals $$x_0 - h + a = x_0 - 3 + \frac{9}{4}$$.

For the point $$(7, 1)$$:

First verify it lies on $$C$$: $$(1 + 5)^2 = 36$$ and $$9(7 - 3) = 36$$. Yes, $$36 = 36$$ ✓.

$$d = x_0 - 3 + \frac{9}{4} = 7 - 3 + \frac{9}{4} = 4 + \frac{9}{4} = \frac{25}{4}$$

Therefore:

$$12d = 12 \times \frac{25}{4} = 75$$

The answer is $$75$$.

Find the distance of the point of tangency $$P$$ (where a line perpendicular to $$2x - y = 10$$ touches the parabola $$y^2 = 4(x - 9)$$) from the centre of the circle $$x^2 + y^2 - 14x - 8y + 56 = 0$$.

The line $$2x - y = 10$$ has slope $$2$$; hence a line perpendicular to it has slope $$m = -\tf\frac{1}{2}$$.

The parabola $$y^2 = 4(x - 9)$$ has vertex $$(9,0)$$ and $$4a = 4$$, so $$a = 1$$. Differentiating implicitly gives $$2y\,\f\frac{dy}{dx} = 4$$ and thus $$\f\frac{dy}{dx} = \f\frac{2}{y}$$. Setting this equal to $$-\tf\frac{1}{2}$$ yields $$\f\frac{2}{y} = -\tf\frac{1}{2}$$, so $$y = -4$$. Substituting $$y = -4$$ into the parabola equation gives $$(-4)^2 = 4(x - 9)$$, i.e.\ $$16 = 4(x - 9)$$ and hence $$x = 13$$. Therefore the point of tangency is $$P(13,-4)$$.

Rewriting the circle equation $$x^2 + y^2 - 14x - 8y + 56 = 0$$ by completing the square yields $$(x - 7)^2 + (y - 4)^2 = 9$$, so its centre is $$(7,4)$$.

The distance from $$P(13,-4)$$ to $$(7,4)$$ is $$$ d = \sqrt{(13 - 7)^2 + (-4 - 4)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = 10. $$$

The distance is 10.

The parabola is $$9x^{2}+12x+18y-14=0$$.
Solve for $$y$$ to recognise its shape:

$$18y = -\,9x^{2}-12x+14 \;\;\Longrightarrow\;\; y = -\tfrac12x^{2}-\tfrac23x+\tfrac79$$.

This is a vertical parabola that opens downwards. A line through the fixed point $$P(0,1)$$ has the form

$$y-1 = m(x-0)\;\;\Longrightarrow\;\; y = mx+1$$ $$-(1)$$

For this line to be tangent to the parabola, simultaneous substitution of $$y$$ from $$(1)$$ into the parabola must give a quadratic in $$x$$ having zero discriminant.

Substituting $$y = mx+1$$ in $$9x^{2}+12x+18y-14=0$$ gives

$$9x^{2}+12x+18(mx+1)-14 = 0$$ $$\Longrightarrow\; 9x^{2} + (12+18m)x + 4 = 0$$ $$-(2)$$

For tangency, discriminant $$\Delta = 0$$:

$$\Delta = (12+18m)^{2} - 4\cdot9\cdot4 = 0$$ $$\Longrightarrow\; (12+18m)^{2} = 144$$ $$\Longrightarrow\; 12+18m = \pm12$$.

Solving,

Case 1: $$12+18m = 12 \;\Rightarrow\; 18m = 0 \;\Rightarrow\; m = 0$$.
Case 2: $$12+18m = -12 \;\Rightarrow\; 18m = -24 \;\Rightarrow\; m = -\tfrac43$$.

Hence the two tangents through $$P$$ are

$$L_1:\; y = 1 \quad(\text{slope }0)$$
$$L_2:\; y = -\tfrac43x + 1 \quad(\text{slope }-\tfrac43)$$.

Choose any point $$Q(x_q,1)$$ on $$L_1$$ and any point $$R(x_r,\, -\tfrac43x_r +1)$$ on $$L_2$$. Side lengths from $$P(0,1)$$ are

$$PQ = \sqrt{(x_q-0)^{2}+(1-1)^{2}} = |x_q|,$$
$$PR = \sqrt{(x_r-0)^{2}+(-\tfrac43x_r+1-1)^{2}} = \sqrt{x_r^{2} + (\tfrac43x_r)^{2}} = |x_r|\sqrt{1+\tfrac{16}{9}} = \tfrac53|x_r|.$$

The triangle $$PQR$$ is required to be isosceles with equal sides $$PQ = PR$$, so

$$|x_q| = \tfrac53|x_r|.$$

Two possibilities arise:

Case A: $$x_q = \tfrac53x_r\;\;(x_q\ \text{and}\ x_r\ \text{same sign}).$$ Slope of $$QR$$ is $$m = \frac{y_R - y_Q}{x_r - x_q} = \frac{-\tfrac43x_r +1 -1}{x_r - \tfrac53x_r} = \frac{-\tfrac43x_r}{-\tfrac23x_r} = 2.$$

Case B: $$x_q = -\,\tfrac53x_r\;\;(x_q\ \text{and}\ x_r\ \text{opposite signs}).$$ Now $$m = \frac{-\tfrac43x_r}{x_r + \tfrac53x_r} = \frac{-\tfrac43x_r}{\tfrac83x_r} = -\tfrac12.$$

Thus the two possible slopes of $$QR$$ are $$m_1 = 2$$ and $$m_2 = -\tfrac12$$.

Finally,

$$16\left(m_1^{2}+m_2^{2}\right) = 16\left(2^{2} + \left(-\tfrac12\right)^{2}\right) = 16\left(4 + \tfrac14\right) = 16\cdot\tfrac{17}{4} = 68.$$

Hence the required value is $$\boxed{68}$$.

Area $$S_2$$ (Triangle): $$\frac{1}{2} |a(b^2) - (-b)(a^2)| = \frac{1}{2}ab(a+b)$$.

Area $$S_1$$ (Parabola segment): Using the formula $$\frac{1}{6}(x_2 - x_1)^3$$ for a vertical parabola:

$$S_1 = \frac{1}{6}(a - (-b))^3 = \frac{1}{6}(a+b)^3$$.

Ratio: $$\frac{S_1}{S_2} = \frac{\frac{1}{6}(a+b)^3}{\frac{1}{2}ab(a+b)} = \frac{(a+b)^2}{3ab} = \frac{a^2 + b^2 + 2ab}{3ab} = \frac{1}{3}(\frac{a}{b} + \frac{b}{a} + 2)$$.

Minimization: By AM-GM, $$\frac{a}{b} + \frac{b}{a} \geq 2$$.

$$\min(\frac{S_1}{S_2}) = \frac{1}{3}(2 + 2) = \frac{4}{3}$$.

 $$m=4, n=3 \implies m+n = \mathbf{7}$$.

The parabola $$y^2 = 24x$$ has $$4a = 24$$, so $$a = 6$$. A parametric point is $$(6t^2, 12t)$$ and the tangent at this point is $$ty = x + 6t^2$$.

This tangent meets the rectangular hyperbola $$xy = 2$$. Substituting $$x = ty - 6t^2$$:

$$(ty - 6t^2)y = 2 \Rightarrow ty^2 - 6t^2 y - 2 = 0$$

Let the y-coordinates of points A and B be $$y_1, y_2$$. By Vieta's formulas:

$$y_1 + y_2 = 6t, \quad y_1 y_2 = \frac{-2}{t}$$

The midpoint coordinates are:

$$k = \frac{y_1 + y_2}{2} = 3t \Rightarrow t = \frac{k}{3}$$

$$h = \frac{x_1 + x_2}{2} = \frac{1}{2}\left(\frac{2}{y_1} + \frac{2}{y_2}\right) = \frac{y_1 + y_2}{y_1 y_2} = \frac{6t}{-2/t} = -3t^2$$

Substituting $$t = k/3$$: $$h = -3 \cdot \frac{k^2}{9} = -\frac{k^2}{3}$$

The locus is $$y^2 = -3x$$, a parabola of the form $$y^2 = -4bx$$ with $$4b = 3$$, so $$b = \frac{3}{4}$$.

The directrix of $$y^2 = -3x$$ is $$x = \frac{3}{4}$$, i.e., $$4x = 3$$.

The length of the latus rectum is $$4b = 3$$.

The answer is Option A: directrix $$4x = 3$$.

Given parabolas: $$ax^2 + 2bx + cy = 0$$ and $$dx^2 + 2ex + fy = 0$$ intersect on the line $$y = 1$$.

Note: The second equation appears to have a typo ($$d^2$$ instead of $$dx^2$$). Assuming it is $$dx^2 + 2ex + fy = 0$$.

Substituting $$y = 1$$ into the first equation gives $$ax^2 + 2bx + c = 0 \quad \ldots (i)$$.

Substituting $$y = 1$$ into the second equation gives $$dx^2 + 2ex + f = 0 \quad \ldots (ii)$$.

Since the parabolas intersect on $$y = 1$$, both equations must have the same roots and hence are proportional: $$\frac{a}{d} = \frac{2b}{2e} = \frac{c}{f}$$.

This simplifies to $$\frac{a}{d} = \frac{b}{e} = \frac{c}{f} = k \text{ (say)}$$.

Since $$a, b, c$$ are in G.P., we have $$b^2 = ac$$.

From the proportionality relations, $$d = a/k$$, $$e = b/k$$, $$f = c/k$$.

Then $$e^2 = \frac{b^2}{k^2} = \frac{ac}{k^2} = \frac{a}{k}\cdot\frac{c}{k} = d\cdot f$$, so $$d, e, f$$ are in G.P.

The correct answer is Option (4): $$\boxed{d, e, f \text{ are in G.P.}}$$

Step 1: Find the coordinates of the point P on the first parabola

The equation of the first parabola is:

$$x = 4y^2$$

Let any parametric point on this parabola be $$P(4t^2, t)$$. We need to find the value of $$t$$ such that the distance from $$P$$ to the point $$Q(0, 33)$$ is minimized. The square of the distance $$D^2$$ between $$P$$ and $$Q$$ is given by:

$$D^2 = (4t^2 - 0)^2 + (t - 33)^2$$

$$D^2 = 16t^4 + t^2 - 66t + 1089$$

To minimize $$D^2$$, we differentiate it with respect to $$t$$ and set it to zero:

$$\frac{d(D^2)}{dt} = 64t^3 + 2t - 66 = 0$$

Dividing the entire equation by 2:

$$32t^3 + t - 33 = 0$$

By inspection, we can see that $$t = 1$$ is a real root of this cubic equation because $$32(1)^3 + 1 - 33 = 0$$. Since the function is strictly increasing, $$t = 1$$ is the only real solution.

Substituting $$t = 1$$ back into the parametric coordinates of $$P$$:

$$h = 4(1)^2 = 4$$

$$k = 1$$

Thus, the point closest to $$Q$$ is $$P(4, 1)$$.

Step 2: Find the equation of the directrix of the second parabola

The equation of the second parabola is given by:

$$y^2 = 4(x + y)$$

$$y^2 - 4y = 4x$$

Completing the square on the left-hand side:

$$(y - 2)^2 - 4 = 4x$$

$$(y - 2)^2 = 4x + 4$$

$$(y - 2)^2 = 4(x + 1)$$

Comparing this with the standard equation of a parabola $$(Y)^2 = 4aX$$, we identify:

- $$Y = y - 2$$

- $$X = x + 1$$

- $$4a = 4 \implies a = 1$$

The standard equation of the directrix for $$Y^2 = 4aX$$ is given by $$X = -a$$. Substituting our variables:

$$x + 1 = -1$$

$$x = -2$$

$$x + 2 = 0$$

Step 3: Calculate the distance of P from the directrix

We need to find the perpendicular distance of the point $$P(4, 1)$$ from the line $$x + 2 = 0$$:

$$\text{Distance} = \frac{|4 + 2|}{\sqrt{1^2 + 0^2}}$$

$$\text{Distance} = |6| = 6$$

Conclusion:

The distance of $$P$$ from the directrix of the parabola is equal to 6.

Line $$AB$$: $$(\lambda+1)x + \lambda y = 4$$

Line $$AC$$: $$\lambda x + (1-\lambda)y + \lambda = 0$$

Since vertex $$A$$ lies on the y-axis, we set $$x = 0$$: 

From $$AB$$: $$\lambda y = 4 \implies y = \frac{4}{\lambda}$$

From $$AC$$: $$(1-\lambda)y + \lambda = 0 \implies y = \frac{\lambda}{\lambda - 1}$$

$$\frac{4}{\lambda} = \frac{\lambda}{\lambda - 1}$$

$$4\lambda - 4 = \lambda^2 \implies \lambda^2 - 4\lambda + 4 = 0 \implies (\lambda - 2)^2 = 0 \implies \lambda = 2$$

$$A = (0, 2)$$

Equation of $$AB$$: $$3x + 2y = 4 \implies \text{Slope } m_{AB} = -\frac{3}{2}$$

Equation of $$AC$$: $$2x - y + 2 = 0 \implies \text{Slope } m_{AC} = 2$$

Let the orthocenter be $$H(1, 2)$$.

The altitude from vertex $$B$$ passes through $$H(1, 2)$$ and is perpendicular to side $$AC$$.

Since the slope of $$AC$$ is $$2$$, the slope of the altitude $$BH$$ is $$-\frac{1}{2}$$.

Equation of altitude $$BH$$: $$y - 2 = -\frac{1}{2}(x - 1) \implies x + 2y = 5$$

Vertex $$B$$ is the intersection of side $$AB$$ ($$3x + 2y = 4$$) and altitude $$BH$$ ($$x + 2y = 5$$): $$\implies B = \left(-\frac{1}{2}, \frac{11}{4}\right)$$

Since $$m_{AB} = -\frac{3}{2}$$, the slope of the altitude $$CH$$ is $$\frac{2}{3}$$. Equation of altitude $$CH$$: $$y - 2 = \frac{2}{3}(x - 1) \implies 2x - 3y + 4 = 0$$

Vertex $$C$$ is the intersection of side $$AC$$ ($$2x - y + 2 = 0$$) and altitude $$CH$$ ($$2x - 3y + 4 = 0$$): $$C = \left(-\frac{1}{2}, 1\right)$$

The parabola is $$y^2 = 6x \implies 4a = 6 \implies a = \frac{3}{2}$$

$$y = mx + \frac{a}{m} \implies y = mx + \frac{3}{2m}$$

$$1 = m\left(-\frac{1}{2}\right) + \frac{3}{2m}$$ (Passes through $$C$$)

$$1 = -\frac{m}{2} + \frac{3}{2m}$$

$$2m = -m^2 + 3 \implies m^2 + 2m - 3 = 0$$

$$(m + 3)(m - 1) = 0 \implies m = 1 \quad \text{or} \quad m = -3$$

The point of contact formula is $$\left(\frac{a}{m^2}, \frac{2a}{m}\right) = \left(\frac{3}{2m^2}, \frac{3}{m}\right)$$. For the point to lie in the first quadrant, the y-coordinate must be positive, which means $$m > 0$$. Thus, $$m = 1$$.

$$P = \left(\frac{3}{2(1)^2}, \frac{3}{1}\right) = \left(\frac{3}{2}, 3\right)$$

Length of $$CP$$: $$\text{Length} = \sqrt{\left(\frac{3}{2} - \left(-\frac{1}{2}\right)\right)^2 + (3 - 1)^2}$$

$$\text{Length} = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$$

For the parabola $$y^{2}=4ax$$ (with $$a\gt 0$$) every point can be written in the parametric form $$P(at^{2},\,2at)$$, where $$t$$ is a real parameter.

Differentiate the parabola to obtain the tangent slope:
$$2y\dfrac{dy}{dx}=4a\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac{2a}{y}$$
At $$P(at^{2},\,2at)$$, the tangent slope is $$\dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac{1}{t}$$.

Hence the normal slope (negative reciprocal of the tangent slope) is
$$m=-t$$.

Equation of the normal through $$P$$:
$$y-2at=m\bigl(x-at^{2}\bigr).\tag{-1}$$

Let $$Q(x_{Q},0)$$ be the x-intercept of this normal. Put $$y=0$$ in $$( -1 )$$:
$$-2at=m\,(x_{Q}-at^{2})$$
Substituting $$m=-t$$ and cancelling $$(-t)\neq 0$$ gives
$$2a=x_{Q}-at^{2}\;\Longrightarrow\;x_{Q}=a(t^{2}+2).$$
Thus $$Q\bigl(a(t^{2}+2),\,0\bigr).$$

The focus of the parabola is $$F(a,0).$$ The triangle under consideration has vertices
$$P(at^{2},\,2at),\;F(a,0),\;Q\bigl(a(t^{2}+2),0\bigr).$$

Because $$F$$ and $$Q$$ lie on the x-axis, segment $$FQ$$ is the base:
Base length $$|FQ|=a\bigl(t^{2}+2-a/a\bigr)=a(t^{2}+1).$$
Height of the triangle is the absolute y-coordinate of $$P$$, namely $$|2at|.$$

Area $$\triangle PFQ=\dfrac12\bigl(\text{base}\bigr)\bigl(\text{height}\bigr)$$:
$$\text{Area}= \dfrac12 \bigl[a(t^{2}+1)\bigr]\bigl[|2at|\bigr] = a^{2}(t^{2}+1)\,|t|.$$

The question states that this area equals $$120$$:
$$a^{2}(t^{2}+1)\,|t| = 120.\tag{-2}$$

We are told that the slope $$m$$ of the normal and $$a$$ are positive integers. Since $$m=-t$$, a positive integer $$m$$ implies $$t\lt 0$$, so set $$|t|=-t=m.$$ Let us write $$s=|t|=m$$ (a positive integer). Equation $$( -2 )$$ now becomes

$$a^{2}\,s\,(s^{2}+1)=120.\tag{-3}$$

List small positive integers $$s$$ and evaluate $$s(s^{2}+1)$$:

s = 1 ⇒ 2, s = 2 ⇒ 10, s = 3 ⇒ 30, s = 4 ⇒ 68, …
Insert into $$( -3 )$$ and check for a perfect square $$a^{2}$$.

Case s = 3: $$30a^{2}=120 \;\Longrightarrow\; a^{2}=4 \;\Longrightarrow\; a=2$$ (integer).
Case s = 1: $$2a^{2}=120 \Rightarrow a^{2}=60$$ (not a square).
Case s = 2: $$10a^{2}=120 \Rightarrow a^{2}=12$$ (not a square).
Case s ≥ 4: $$s(s^{2}+1)\gt 60$$, so $$a^{2}\lt 2$$, impossible for positive integer $$a$$.

The only admissible solution is $$s=3,\;a=2$$. Since $$s=m$$, we obtain $$m=3$$.

Therefore $$(a,\,m)=(2,\,3).$$

Option A which is: (2, 3)

To solve the parabola problem:

1. Identify the Parabola Equation

Given the Focus $$(3, 0)$$ and Directrix $$x = -3$$, the vertex is at the origin $$(0,0)$$. This is a standard right-opening parabola:

$$y^2 = 4ax \implies y^2 = 4(3)x \implies \mathbf{y^2 = 12x}$$

2. Find Point of Intersection $$R(\alpha, \beta)$$

Let the points on the parabola be $$P(at_1^2, 2at_1)$$ and $$Q(at_2^2, 2at_2)$$, where $$a = 3$$.

The intersection point $$R$$ of the tangents at $$P$$ and $$Q$$ is given by the formula:

$$R(\alpha, \beta) = (at_1t_2, a(t_1 + t_2))$$

3. Use the Ratio of Ordinates

The ordinates (y-coordinates) $$2at_1$$ and $$2at_2$$ are in the ratio $$3:1$$.

$$\frac{2at_1}{2at_2} = 3 \implies \mathbf{t_1 = 3t_2}$$

4. Express $$\alpha$$ and $$\beta$$ in terms of $$t_2$$

• $$\alpha = 3(3t_2)(t_2) = 9t_2^2$$
• $$\beta = 3(3t_2 + t_2) = 12t_2$$

5. Calculate $$\frac{\beta^2}{\alpha}$$

$$\frac{\beta^2}{\alpha} = \frac{(12t_2)^2}{9t_2^2} = \frac{144t_2^2}{9t_2^2} = \mathbf{16}$$

Final Answer: 16

The standard right-opening parabola $$y^{2}=4ax$$ has parameter $$a=1$$, so its equation is $$y^{2}=4x$$.

Slope form of a tangent to this parabola:
For slope $$m$$ the tangent is $$y=mx+\frac{a}{m}=mx+\frac{1}{m}$$ $$-(1)$$.

The given circle is $$(x-4)^{2}+y^{2}=16$$, centre at $$(4,0)$$, radius $$4$$.

Line $$-(1)$$ is also tangent to the circle. Substitute $$y=mx+\frac{1}{m}$$ into the circle:

$$(x-4)^{2}+\left(mx+\frac{1}{m}\right)^{2}=16$$
$$\bigl(x^{2}-8x+16\bigr)+\bigl(m^{2}x^{2}+2x+\frac{1}{m^{2}}\bigr)=16$$
$$(1+m^{2})x^{2}-6x+\frac{1}{m^{2}}=0$$ $$-(2)$$

For tangency the quadratic $$-(2)$$ must have equal roots, so its discriminant is zero:

$$(-6)^{2}-4(1+m^{2})\frac{1}{m^{2}}=0$$
$$36-4\frac{1+m^{2}}{m^{2}}=0$$
Divide by $$4$$ and multiply by $$m^{2}$$: $$9m^{2}-(1+m^{2})=0$$
$$8m^{2}=1 \;\Longrightarrow\; m^{2}=\frac{1}{8}$$ $$-(3)$$

Thus the two common tangents have slopes $$m=\pm\frac{1}{2\sqrt{2}}$$. (The geometry is symmetric about the x-axis; we work with the positive slope, the other gives the same length.)

Point of contact on the parabola (P)
Insert $$m^{2}=\frac{1}{8}$$ into the intersection of line $$-(1)$$ with $$y^{2}=4x$$:

$$\left(mx+\frac{1}{m}\right)^{2}=4x$$
$$m^{2}x^{2}+2x+\frac{1}{m^{2}}-4x=0$$
$$m^{2}x^{2}-2x+\frac{1}{m^{2}}=0$$

This quadratic also has a double root (tangency), giving $$x_P=\frac{1}{m^{2}}=8, \qquad y_P=mx_P+\frac{1}{m}=2\sqrt{2}+2\sqrt{2}=4\sqrt{2}.$$ Hence $$P\,(8,\,4\sqrt{2})$$.

Point of contact on the circle (Q)
The repeated root of $$-(2)$$ gives the x-coordinate:

$$x_Q=\frac{6}{2(1+m^{2})}=\frac{3}{1+m^{2}}=\frac{3}{1+\frac{1}{8}}=\frac{3}{\frac{9}{8}}=\frac{8}{3}.$$

Then $$y_Q=mx_Q+\frac{1}{m}=\frac{1}{2\sqrt{2}}\cdot\frac{8}{3}+2\sqrt{2}=\frac{4}{3\sqrt{2}}+2\sqrt{2}=\frac{16}{3\sqrt{2}}.$$ Thus $$Q\!\left(\frac{8}{3},\,\frac{16}{3\sqrt{2}}\right).$$

Distance $$PQ$$
$$\Delta x = 8-\frac{8}{3}=\frac{16}{3},\qquad \Delta y = 4\sqrt{2}-\frac{16}{3\sqrt{2}}=\frac{8}{3\sqrt{2}}.$$

Therefore

$$PQ^{2}=(\Delta x)^{2}+(\Delta y)^{2}$$
$$=\left(\frac{16}{3}\right)^{2}+\left(\frac{8}{3\sqrt{2}}\right)^{2}$$
$$=\frac{256}{9}+\frac{64}{18}$$
$$=\frac{256}{9}+\frac{32}{9}=\frac{288}{9}=32.$$

Hence $$PQ^{2}=32$$.

The same value is obtained for the tangent with negative slope, so the answer is unique.

Final Answer : 32

The curve is $$x^2 + 2x - 4y + 9 = 0$$, i.e., $$y = \frac{x^2+2x+9}{4}$$.

First, we find tangent at P(1,3).

$$y' = \frac{2x+2}{4} = \frac{x+1}{2}$$. At $$x=1$$: $$y' = 1$$.

Tangent: $$y - 3 = 1(x-1) \implies y = x + 2$$.

Meets y-axis at A(0, 2).

Next, we find B on parabola $$y^2 = 4x$$.

Line through P(1,3) parallel to $$x-3y=6$$ has slope $$\frac{1}{3}$$:

$$y - 3 = \frac{1}{3}(x-1) \implies y = \frac{x+8}{3}$$

Substituting into $$y^2 = 4x$$:

$$\frac{(x+8)^2}{9} = 4x \implies x^2+16x+64 = 36x \implies x^2-20x+64 = 0$$

$$x = \frac{20 \pm 12}{2} = 16 \text{ or } 4$$

B must lie on $$2x-3y=8$$:

At $$x=16$$: $$y = 8$$, check: $$2(16)-3(8) = 8$$ ✓. So B = (16, 8).

From this, we compute AB².

$$AB^2 = (16-0)^2 + (8-2)^2 = 256 + 36 = 292$$

The answer is 292.

Given hyperbola

$$H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$

passing through the point

$$(2\sqrt2,-2\sqrt2)$$

Substituting the point in the hyperbola,

$$\frac{(2\sqrt2)^2}{a^2}-\frac{(-2\sqrt2)^2}{b^2}=1$$

$$\frac8{a^2}-\frac8{b^2}=1$$

$$8(b^2-a^2)=a^2b^2\qquad\cdots(1)$$

For the hyperbola,

$$e^2=1+\frac{b^2}{a^2}$$

Length of latus rectum of hyperbola is

$$L_H=\frac{2b^2}{a}$$

The foci are

$$(\pm ae,0)$$

The parabola has focus at

$$(ae,0)$$

and its directrix passes through the other focus

$$(-ae,0)$$

Hence, the directrix is

$$x=-ae$$

Therefore, the vertex is midpoint of the focus and directrix, i.e.,

$$(0,0)$$

and the distance of focus from vertex is

$$ae$$

Thus the parabola is

$$y^2=4(ae)x$$

Hence, length of latus rectum of parabola is

$$L_P=4(ae)$$

Given,

$$L_P=eL_H$$

$$4ae=e\left(\frac{2b^2}{a}\right)$$

Cancelling $$2e,$$

$$2a=\frac{b^2}{a}$$

$$b^2=2a^2\qquad\cdots(2)$$

Substituting $$b^2=2a^2$$ in (1),

$$8(2a^2-a^2)=a^2(2a^2)$$

$$8a^2=2a^4$$

$$a^2=4$$

Hence,

$$a=2,\qquad b^2=8$$

Now,

$$e^2=1+\frac{8}{4}=3$$

$$e=\sqrt3$$

Thus,

$$ae=2\sqrt3$$

Therefore, the parabola is

$$y^2=8\sqrt3\,x$$

Hence, any point satisfying

$$y^2=8\sqrt3\,x$$

lies on the parabola.

By trial and error method with options,we conclude that $$(3\sqrt{3}, -6\sqrt{2})$$ lies on the parabola

Let $$P(x, y)$$ be a point on the curve $$C$$. The tangent at $$P$$ has slope $$\frac{dy}{dx}$$.

The equation of the tangent at $$P$$ is $$Y - y = \frac{dy}{dx}(X - x)$$. It meets the x-axis where $$Y = 0$$, so $$0 - y = \frac{dy}{dx}(X - x) \implies X = x - \frac{y}{dy/dx}$$, and thus $$Q = \left(x - \frac{y}{y'}, 0\right)$$.

The midpoint of $$PQ$$ is $$M = \left(\frac{x + x - \frac{y}{y'}}{2}, \frac{y}{2}\right)$$. Since the y-axis bisects $$PQ$$, this midpoint lies on the y-axis, giving $$\frac{2x - \frac{y}{y'}}{2} = 0 \implies 2x = \frac{y}{y'} \implies y' = \frac{y}{2x}$$.

Solving the differential equation $$\frac{dy}{dx} = \frac{y}{2x}$$ gives $$\frac{dy}{y} = \frac{dx}{2x}$$, $$\ln|y| = \frac{1}{2}\ln|x| + C$$, and hence $$y^2 = kx$$.

Since the curve passes through $$(3, 3)$$, we have $$9 = 3k \implies k = 3$$. Therefore, the curve is $$y^2 = 3x$$, a parabola of the form $$y^2 = 4ax$$ with $$4a = 3$$, so the length of the latus rectum is $$4a = 3$$.

Therefore, the answer is Option A: length of latus rectum 3.

We have the parabola $$P: y^2 = 4x$$ with vertex $$O(0,0)$$ and focus at $$(1,0)$$. A focal chord lies along the line $$L: y = mx + c$$ with $$m > 0$$. Since $$L$$ passes through the focus $$(1,0)$$, we get $$0 = m + c$$, so $$c = -m$$. Thus $$L: y = m(x - 1)$$.

We are also told that $$L$$ is tangent to the hyperbola $$H: x^2 - y^2 = 4$$, where $$a^2 = 4$$ and $$b^2 = 4$$. For a line $$y = mx + c$$ to be tangent to the hyperbola $$x^2 - y^2 = 4$$, the condition is $$c^2 = a^2 m^2 - b^2 = 4m^2 - 4$$. Since $$c = -m$$, we have $$m^2 = 4m^2 - 4$$, which gives $$3m^2 = 4$$, so $$m^2 = \dfrac{4}{3}$$ and $$m = \dfrac{2}{\sqrt{3}}$$.

Now we find the points $$M$$ and $$N$$ where $$L$$ meets the parabola. Substituting $$y = m(x-1)$$ into $$y^2 = 4x$$: $$m^2(x-1)^2 = 4x$$. With $$m^2 = 4/3$$: $$\dfrac{4}{3}(x-1)^2 = 4x$$, so $$(x-1)^2 = 3x$$, giving $$x^2 - 5x + 1 = 0$$.

The roots are $$x_1 = \dfrac{5 + \sqrt{21}}{2}$$ and $$x_2 = \dfrac{5 - \sqrt{21}}{2}$$. The corresponding $$y$$-values are $$y_i = m(x_i - 1)$$. We note that $$x_1 > 1$$ so $$y_1 > 0$$, and $$x_2 < 1$$ so $$y_2 < 0$$. Hence $$M$$ and $$N$$ lie on opposite sides of the $$x$$-axis.

Now, the focus of the hyperbola on the positive $$x$$-axis is $$F$$. Since $$a^2 = b^2 = 4$$, we have $$c_H = \sqrt{a^2 + b^2} = \sqrt{8} = 2\sqrt{2}$$, so $$F = (2\sqrt{2}, 0)$$.

The quadrilateral $$OMFN$$ has vertices $$O(0,0)$$, $$M$$, $$F(2\sqrt{2}, 0)$$, and $$N$$. Since $$O$$ and $$F$$ both lie on the $$x$$-axis, and $$M$$, $$N$$ are on opposite sides of it, the area of the quadrilateral equals $$\dfrac{1}{2} \cdot OF \cdot (|y_M| + |y_N|)$$.

We compute $$|y_M| + |y_N| = m|x_1 - 1| + m|1 - x_2| = m\left(\dfrac{3 + \sqrt{21}}{2} + \dfrac{\sqrt{21} - 3}{2}\right) = m \cdot \sqrt{21}$$.

With $$m = \dfrac{2}{\sqrt{3}}$$: $$|y_M| + |y_N| = \dfrac{2\sqrt{21}}{\sqrt{3}} = 2\sqrt{7}$$.

Hence the area is $$\dfrac{1}{2} \cdot 2\sqrt{2} \cdot 2\sqrt{7} = 2\sqrt{14}$$.

Hence, the correct answer is Option B.

We need to find the area bounded by the locus $$ L $$ of the centre $$ (\alpha, \beta) $$ (with $$ \beta > 0 $$) of a circle that touches $$ x^2 + (y-1)^2 = 1 $$ externally and also touches the $$ x $$-axis, and the line $$ y = 4 $$.

Let the moving circle have centre $$ (\alpha, \beta) $$ and radius $$ r $$. Since it touches the $$ x $$-axis:

$$r = \beta \quad (\text{since } \beta > 0)$$

Since it touches the circle $$ x^2 + (y-1)^2 = 1 $$ (centre $$ (0, 1) $$, radius 1) externally:

$$\sqrt{\alpha^2 + (\beta - 1)^2} = r + 1 = \beta + 1$$

Squaring both sides:

$$\alpha^2 + (\beta - 1)^2 = (\beta + 1)^2$$

$$\alpha^2 + \beta^2 - 2\beta + 1 = \beta^2 + 2\beta + 1$$

$$\alpha^2 = 4\beta$$

Replacing $$ \alpha $$ with $$ x $$ and $$ \beta $$ with $$ y $$, the locus $$ L $$ is:

$$x^2 = 4y$$

This is an upward-opening parabola with vertex at the origin.

When $$ y = 4 $$: $$ x^2 = 16 $$, so $$ x = \pm 4 $$.

The area between the parabola and the line $$ y = 4 $$ is:

$$A = \int_{-4}^{4} \left(4 - \frac{x^2}{4}\right) dx = 2\int_{0}^{4} \left(4 - \frac{x^2}{4}\right) dx$$

$$= 2\left[4x - \frac{x^3}{12}\right]_0^4 = 2\left(16 - \frac{64}{12}\right) = 2\left(16 - \frac{16}{3}\right)$$

$$= 2 \cdot \frac{48 - 16}{3} = 2 \cdot \frac{32}{3} = \frac{64}{3}$$

The area bounded by $$ L $$ and the line $$ y = 4 $$ is $$ \dfrac{64}{3} $$, which corresponds to Option C.

We have the parametric curve $$x = 2t,\;y = \frac{t^2}{3}$$ and need to find $$\lim_{t\to 1}k$$ where $$k$$ is the ordinate of the centroid of triangle $$SAB$$.

Since $$x = 2t$$ gives $$t = x/2$$, substituting into $$y = \frac{t^2}{3}$$ yields $$y = \frac{x^2}{12}$$ or equivalently $$x^2 = 12y\,, $$ which is the equation of a parabola with $$4a = 12$$ so that $$a = 3$$. Hence its focus is $$S = (0,3)$$ and its vertex is at the origin.

Any point on this parabola can be represented as $$A = (2t,\;t^2/3)$$, and if we let $$B = (0,b)$$ lie on its axis then the vectors from $$S$$ to $$A$$ and from $$B$$ to $$A$$ become $$\vec{SA} = (2t,\;t^2/3 - 3)$$ and $$\vec{BA} = (2t,\;t^2/3 - b)$$, respectively.

Requiring $$SA$$ to be perpendicular to $$BA$$ means their dot product must vanish, so
$$(2t)(2t) + \Bigl(\tfrac{t^2}{3}-3\Bigr)\Bigl(\tfrac{t^2}{3}-b\Bigr)=0\quad\Longrightarrow\quad 4t^2 + \Bigl(\tfrac{t^2}{3}-3\Bigr)\Bigl(\tfrac{t^2}{3}-b\Bigr)=0\,. $$

At this stage it is convenient to set $$y_A = t^2/3$$, which implies $$t^2 = 3y_A$$, and substitution leads to
$$12y_A + (y_A-3)(y_A-b)=0\quad\Longrightarrow\quad 12y_A + y_A^2 - y_A b -3y_A +3b=0\quad\Longrightarrow\quad b(3-y_A)=-(y_A^2+9y_A)\,. $$

From which it follows that
$$b=\frac{y_A^2+9y_A}{y_A-3}\,. $$

As $$t\to 1$$ we have $$y_A=1/3$$, hence
$$b=\frac{(1/3)^2+9(1/3)}{1/3-3}=\frac{1/9+3}{-8/3}=\frac{28/9}{-8/3}=\frac{28}{9}\times\frac{-3}{8}=-\frac{7}{6}\,. $$

Finally, with $$S=(0,3)$$, $$A=(2,1/3)$$ and $$B=(0,-7/6)$$, the ordinate of the centroid is given by
$$k=\frac{y_S+y_A+y_B}{3}=\frac{3+\tfrac{1}{3}-\tfrac{7}{6}}{3}=\frac{\tfrac{18}{6}+\tfrac{2}{6}-\tfrac{7}{6}}{3}=\frac{13/6}{3}=\frac{13}{18}\,. $$

Option D: $$\dfrac{13}{18}$$.

We need to find the equation of a common tangent to the parabolas $$y = x^2$$ and $$y = -(x-2)^2$$.

At a point $$(t, t^2)$$ on $$y = x^2$$, the slope is $$\frac{dy}{dx} = 2t$$.

Tangent equation: $$y - t^2 = 2t(x - t)$$

$$y = 2tx - t^2$$ ... (i)

At a point $$(s, -(s-2)^2)$$ on $$y = -(x-2)^2$$, the slope is $$\frac{dy}{dx} = -2(s-2)$$.

Tangent equation: $$y + (s-2)^2 = -2(s-2)(x - s)$$

$$y = -2(s-2)x + 2s(s-2) - (s-2)^2$$

$$y = -2(s-2)x + (s-2)[2s - (s-2)]$$

$$y = -2(s-2)x + (s-2)(s+2)$$

$$y = -2(s-2)x + s^2 - 4$$ ... (ii)

Comparing slopes: $$2t = -2(s-2)$$

$$t = -(s-2) = 2 - s$$ ... (iii)

Comparing intercepts: $$-t^2 = s^2 - 4$$

$$t^2 + s^2 = 4$$ ... (iv)

Substitute $$t = 2 - s$$ into equation (iv):

$$(2-s)^2 + s^2 = 4$$

$$4 - 4s + s^2 + s^2 = 4$$

$$2s^2 - 4s = 0$$

$$2s(s - 2) = 0$$

$$s = 0 \text{ or } s = 2$$

When $$s = 0$$: $$t = 2$$, tangent: $$y = 4x - 4 = 4(x - 1)$$

When $$s = 2$$: $$t = 0$$, tangent: $$y = 0$$ (the x-axis)

$$y = 4(x-1)$$ matches Option B.

Therefore, the correct answer is Option B: $$y = 4(x-1)$$.

Given parabola,

$$y^2=8x$$

Comparing with

$$y^2=4ax,$$

we get

$$a=2$$

Let

$$P(a,b)$$

lie on the parabola.

Then,

$$b^2=8a\qquad\cdots(1)$$

Equation of tangent at point $$(a,b)$$ to

$$y^2=8x$$

is

$$by=4(x+a)$$

$$4x-by+4a=0$$

Now find the centre of the circle

$$x^2+y^2-10x-14y+65=0$$

Comparing with

$$x^2+y^2+2gx+2fy+c=0,$$

we get

$$g=-5,\qquad f=-7$$

Hence centre is

$$(5,7)$$

Since the tangent passes through the centre,

$$4(5)-b(7)+4a=0$$

$$20-7b+4a=0$$

$$4a-7b+20=0\qquad\cdots(2)$$

Using (1),

$$a=\frac{b^2}{8}$$

Substitute in (2):

$$4\left(\frac{b^2}{8}\right)-7b+20=0$$

$$\frac{b^2}{2}-7b+20=0$$

$$b^2-14b+40=0$$

$$(b-4)(b-10)=0$$

Hence,

$$b=4\quad \text{or}\quad b=10$$

Corresponding $$a$$ values are

$$a=\frac{4^2}{8}=2$$

and

$$a=\frac{10^2}{8}=\frac{25}{2}$$

Therefore,

$$A=2\cdot\frac{25}{2}=25$$

and

$$B=4\cdot10=40$$

Hence,

$$A+B=25+40=65$$

Therefore, the required value is

$$\boxed{65}$$.

We need to find the area of triangle $$PAB$$ where $$P$$ is the intersection of the tangents at $$A(1,3)$$ and $$B(1,-1)$$ on the parabola $$y^2 - 2x - 2y = 1$$. By completing the square in $$y$$, the equation becomes

$$y^2 - 2y = 2x + 1$$

which simplifies to

$$ (y-1)^2 = 2(x+1) $$

showing that the parabola has its vertex at $$(-1,1)$$ and its axis parallel to the $$x$$-axis.

To find the tangent at $$A(1,3)$$, we differentiate the original equation $$y^2 - 2x - 2y = 1$$ implicitly with respect to $$x$$. This yields

$$2y \frac{dy}{dx} - 2 - 2\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{2}{2y - 2} = \frac{1}{y-1}.$$

At the point $$A(1,3)$$, the slope of the tangent is $$\frac{1}{3-1} = \frac{1}{2}$$, so its equation can be written as $$y - 3 = \frac{1}{2}(x - 1)$$, which rearranges to $$x - 2y + 5 = 0$$.

For the tangent at $$B(1,-1)$$, substituting $$y = -1$$ into $$\frac{dy}{dx} = \frac{1}{y-1}$$ gives a slope of $$\frac{1}{-1-1} = -\frac{1}{2}$$. Thus the tangent line at $$B$$ is $$y + 1 = -\frac{1}{2}(x - 1)$$, or equivalently $$x + 2y + 1 = 0$$.

The intersection point $$P$$ of these two tangents satisfies both $$x - 2y + 5 = 0$$ and $$x + 2y + 1 = 0$$. Adding these equations gives

$$ (x - 2y + 5) + (x + 2y + 1) = 0 \implies 2x + 6 = 0 \implies x = -3. $$

Substituting $$x = -3$$ into $$x - 2y + 5 = 0$$ leads to $$-3 - 2y + 5 = 0$$, so $$y = 1$$ and hence $$P = (-3,1)$$.

Finally, the area of triangle $$PAB$$ with vertices $$P(-3,1)$$, $$A(1,3)$$, and $$B(1,-1)$$ can be computed using the determinant formula

$$\text{Area} = \frac{1}{2}\bigl|(-3)(3-(-1)) + 1((-1)-1) + 1(1-3)\bigr|. $$

$$= \frac{1}{2}\bigl|(-3)(4) + 1(-2) + 1(-2)\bigr|. $$

$$= \frac{1}{2}\bigl|-12 - 2 - 2\bigr| = \frac{1}{2} \times 16 = 8. $$

The correct answer is Option D: $$8$$.

We begin by finding the point of tangency. The parabola is $$y = x - x^2$$. At a point $$P = (a, a - a^2)$$, the slope of the tangent is:

$$\frac{dy}{dx} = 1 - 2a$$

Since the tangent line is $$y = 4 + kx$$ with slope $$k$$, it follows that $$k = 1 - 2a$$.

Requiring that the point $$P$$ lies on the tangent line gives $$a - a^2 = 4 + ka = 4 + (1 - 2a)a = 4 + a - 2a^2$$, which simplifies to $$a - a^2 = 4 + a - 2a^2$$.

This leads to $$a^2 = 4$$ and hence $$a = \pm 2$$.

Since $$k > 0$$, we have $$1 - 2a > 0 \Rightarrow a < \frac{1}{2}$$, so the correct choice is $$a = -2$$.

$$k = 1 - 2(-2) = 5$$ and $$P = (-2, -2 - 4) = (-2, -6)$$.

To find the vertex of the parabola, rewrite $$y = x - x^2 = -\left(x^2 - x\right) = -\left(x - \frac{1}{2}\right)^2 + \frac{1}{4}$$.

Thus, the vertex is $$V = \left(\frac{1}{2}, \frac{1}{4}\right)$$.

Finally, the slope of line $$PV$$ is $$\text{Slope} = \frac{\frac{1}{4} - (-6)}{\frac{1}{2} - (-2)} = \frac{\frac{1}{4} + 6}{\frac{1}{2} + 2} = \frac{\frac{25}{4}}{\frac{5}{2}} = \frac{25}{4} \times \frac{2}{5} = \frac{5}{2}$$.

Therefore, the slope of the line through $$P$$ and $$V$$ is $$\frac{5}{2}$$, which is Option C.

Since the tangents are drawn from $$(\alpha,\beta)$$ to the parabola

$$y^2=4x,$$

the tangent with slope $$m$$ is

$$y=mx+\frac1m$$

As $$(\alpha,\beta)$$ lies on it,

$$\beta=\alpha m+\frac1m$$

$$\alpha m^2-\beta m+1=0$$

Let the slopes be $$m_1,m_2.$$

Then,

$$m_1+m_2=\frac{\beta}{\alpha},\qquad m_1m_2=\frac1\alpha$$

Given one slope is four times the other.

Let

$$m_1=4m_2$$

Then,

$$m_1+m_2=5m_2=\frac{\beta}{\alpha}$$

$$m_1m_2=4m_2^2=\frac1\alpha$$

From the first equation,

$$m_2=\frac{\beta}{5\alpha}$$

Substituting in the second,

$$4\left(\frac{\beta}{5\alpha}\right)^2=\frac1\alpha$$

$$\frac{4\beta^2}{25\alpha^2}=\frac1\alpha$$

$$4\beta^2=25\alpha$$

Since $$(\alpha,\beta)$$ lies on

$$25x^2+4y^2=1,$$

we get

$$25\alpha^2+4\beta^2=1$$

Using

$$4\beta^2=25\alpha,$$

$$25\alpha^2+25\alpha=1$$

$$25\alpha^2+25\alpha-1=0$$

Multiplying by $$4,$$

$$100\alpha^2+100\alpha-4=0$$

Adding $$25$$ on both sides,

$$100\alpha^2+100\alpha+25=29$$

$$=(10\alpha+5)^2$$

Hence,

$$(10\alpha+5)^2=29$$

Also,

$$16\beta^2=4(4\beta^2)=4(25\alpha)=100\alpha$$

Therefore,

$$16\beta^2+50=100\alpha+50=10(10\alpha+5)$$

Squaring,

$$(16\beta^2+50)^2=100(10\alpha+5)^2$$

$$=100(29)=2900$$

Thus,

$$(10\alpha+5)^2+(16\beta^2+50)^2$$

$$=29+2900$$

$$=2929$$

Hence, the required value is

$$\boxed{2929}$$.

A circle of radius 2 passes through the vertex and focus of the parabola $$y^2 = 2x$$ and touches the parabola $$y = \left(x - \frac{1}{4}\right)^2 + \alpha$$ where $$\alpha > 0$$. Find $$(4\alpha - 8)^2$$.

Comparing $$y^2 = 2x$$ with $$y^2 = 4ax$$, we find $$4a = 2$$, so $$a = \frac{1}{2}$$. Therefore the vertex is $$(0,0)$$ and the focus is $$\left(\frac{1}{2}, 0\right)$$.

Next, let the center of the desired circle be $$(h, k)$$. Since it passes through $$(0,0)$$ and $$\left(\frac{1}{2}, 0\right)$$ and has radius 2, we have the distance conditions:

$$h^2 + k^2 = 4 \quad \cdots (1)$$

and

$$ (h - \tfrac{1}{2})^2 + k^2 = 4 \quad \cdots (2)$$

Subtracting (2) from (1) yields $$h^2 - (h - \tfrac{1}{2})^2 = 0$$, which gives $$h - \tfrac{1}{4} = 0$$ and hence $$h = \frac{1}{4}$$. Substituting into (1) then yields $$k^2 = 4 - \frac{1}{16} = \frac{63}{16}$$, so $$k = \pm\frac{3\sqrt{7}}{4}$$.

Thus there are two possible centers, $$\left(\frac{1}{4}, \frac{3\sqrt{7}}{4}\right)$$ (above the $$x$$-axis) and $$\left(\frac{1}{4}, -\frac{3\sqrt{7}}{4}\right)$$ (below the $$x$$-axis).

Since the parabola $$y = \left(x - \frac{1}{4}\right)^2 + \alpha$$ opens upward with vertex at $$\left(\frac{1}{4}, \alpha\right)$$ and axis of symmetry $$x = \frac{1}{4}$$, and since both circle centers lie on this axis, any tangency must occur at $$x = \frac{1}{4}$$. At this value of $$x$$, the circle has $$y = k \pm 2$$, while the parabola has $$y = \alpha$$.

Because the parabola opens upward, the relevant circle must lie below it, so its topmost point touches the parabola. For the circle centered at $$\left(\frac{1}{4}, -\frac{3\sqrt{7}}{4}\right)$$, the topmost point is at

$$y = -\frac{3\sqrt{7}}{4} + 2 = \frac{8 - 3\sqrt{7}}{4}.$$

Since $$3\sqrt{7} \approx 7.937$$, this value is positive (about $$0.016$$), and hence we take $$\alpha = \frac{8 - 3\sqrt{7}}{4} \;>\; 0$$. Because both the circle and the parabola have horizontal tangents at this point and the circle curves downward while the parabola curves upward, they touch without crossing.

Finally, since $$\alpha = \frac{8 - 3\sqrt{7}}{4}$$, we have $$4\alpha = 8 - 3\sqrt{7}$$, so $$4\alpha - 8 = -3\sqrt{7}$$, and

$$ (4\alpha - 8)^2 = 9 \times 7 = 63. $$

The answer is $$\boxed{63}$$.

Parabola $$P_1$$ has vertex $$V = (3, 2)$$ and focus $$F = (4, 4)$$.

The axis direction of $$P_1$$ is along $$\vec{VF} = (1, 2)$$.

The distance from vertex to focus: $$a = \sqrt{1^2 + 2^2} = \sqrt{5}$$.

The directrix of $$P_1$$ is the line perpendicular to the axis, passing through the point obtained by going distance $$a$$ from the vertex in the opposite direction of the focus.

The point on the directrix side: $$D = V - \frac{a \cdot \vec{VF}}{|\vec{VF}|} = (3, 2) - \frac{\sqrt{5} \cdot (1,2)}{\sqrt{5}} = (3-1, 2-2) = (2, 0)$$

The directrix of $$P_1$$ passes through $$(2, 0)$$ and is perpendicular to $$(1, 2)$$:

$$1(x - 2) + 2(y - 0) = 0 \implies x + 2y = 2$$

Now we reflect $$P_1$$ about the line $$x + 2y = 6$$.

The mirror line is $$x + 2y = 6$$ and the directrix of $$P_1$$ is $$x + 2y = 2$$.

Since both lines have the same normal direction $$(1, 2)$$, they are parallel.

The distance from the directrix $$x + 2y = 2$$ to the mirror line $$x + 2y = 6$$ is $$\frac{|6-2|}{\sqrt{5}} = \frac{4}{\sqrt{5}}$$.

When reflecting a line parallel to the mirror across the mirror, the reflected line is at the same distance on the other side.

The directrix of $$P_2$$ is: $$x + 2y = 6 + (6 - 2) = 10$$

The correct answer is $$10$$.

We need to find $$\alpha + \beta$$ where the normal at point $$P$$ on the parabola $$y^2 = 4ax$$ has the intercept form $$\frac{x}{\alpha} + \frac{y}{\beta} = 1$$.

To find the value of $$a$$, consider $$f(x) = 2x^2 - \ln x$$, so $$f'(x) = 4x - \dfrac{1}{x} = 0 \implies x = \dfrac{1}{2}$$. Since $$f$$ decreases on $$(0, 1/2)$$ and increases on $$(1/2, 4)$$, it follows that $$a = \dfrac{1}{2}$$.

Since $$a = \dfrac{1}{2}$$, the parabola is $$y^2 = 4 \cdot \dfrac{1}{2} \cdot x = 2x$$, and a point on it can be written as $$P = \left(\dfrac{t^2}{2}, t\right)$$.

The tangent at $$P = \left(\dfrac{t^2}{2}, t\right)$$ to $$y^2 = 2x$$ is $$yt = x + \dfrac{t^2}{2}$$.

Substituting $$(8a, 8a-1) = (4, 3)$$ into this gives $$3t = 4 + \dfrac{t^2}{2}$$, so $$t^2 - 6t + 8 = 0 \implies t = 2 \text{ or } t = 4$$.

Next, to determine which tangent does not pass through $$(-1/a, 0) = (-2, 0)$$, set $$y = 0$$ in $$yt = x + \dfrac{t^2}{2}$$, giving $$x = -\dfrac{t^2}{2}$$. For $$t = 2$$, $$x = -2$$ so the tangent passes through $$(-2, 0)$$ and is rejected, while for $$t = 4$$, $$x = -8 \neq -2$$. Thus $$P = (8, 4)$$.

The slope of the tangent at $$P$$ is $$\frac{dy}{dx} = \frac{1}{t} = \frac{1}{4}$$, so the slope of the normal is $$-4$$, and its equation is $$y - 4 = -4(x - 8)$$, which simplifies to $$y = -4x + 36 \implies 4x + y = 36$$.

Converting to intercept form gives $$\frac{x}{9} + \frac{y}{36} = 1$$, so $$\alpha = 9$$ and $$\beta = 36$$, and hence $$\alpha + \beta = 9 + 36 = 45$$.

The correct answer is $$\boxed{45}$$.

We have the parabola $$2y^2 = -x$$, i.e., $$y^2 = -x/2$$. Comparing with $$y^2 = 4ax$$, we get $$a = -1/8$$.

The tangent in slope form is $$y = mx + \frac{a}{m} = mx - \frac{1}{8m}$$. Since it passes through $$(2, 0)$$:

$$0 = 2m - \frac{1}{8m} \implies 16m^2 = 1 \implies m = \pm\frac{1}{4}$$

So the two tangent lines are $$l_1: x - 4y - 2 = 0$$ and $$l_2: x + 4y - 2 = 0$$.

These are also tangent to the circle $$(x - 5)^2 + y^2 = r$$, where $$r$$ is the radius. The tangency condition requires the distance from centre $$(5, 0)$$ to each line to equal $$r$$:

$$r = \frac{|5 - 0 - 2|}{\sqrt{1 + 16}} = \frac{3}{\sqrt{17}}$$

Therefore $$r^2 = \frac{9}{17}$$, and $$17r^2 = 9$$.

Hence, the correct answer is 9.

We are told that the vertex of the required parabola lies at a distance 2 units from the origin on the positive $$x$$-axis, so the vertex is $$V(2,0)$$. Its focus is 4 units from the origin on the same axis, i.e. at $$F(4,0)$$. Because the focus is to the right of the vertex, the parabola opens to the right, and its axis is the $$x$$-axis itself.

The standard equation of a right-opening parabola with vertex $$(h,k)$$ is

$$ (y-k)^2 = 4a\,(x-h), $$

where $$a$$ denotes the distance from the vertex to the focus. Here $$h=2,\;k=0$$ and

$$ a = VF = 4-2 = 2. $$

Substituting these values, we obtain the explicit equation of the parabola:

$$ y^2 = 4a\,(x-2) = 4\cdot 2\,(x-2) = 8(x-2). $$

Now we must draw tangents to this parabola from the origin $$O(0,0)$$. To do that, we first recall the point-form of the tangent to a parabola. For the standard parabola $$Y^2 = 4aX$$ with vertex at the origin, the tangent at a point $$(X_1,Y_1)$$ on the curve is

$$ Y\,Y_1 = 2a\,(X+X_1). $$

Our parabola is translated 2 units to the right. Let us therefore shift coordinates:

$$ X = x-2,\qquad Y = y. $$

In these new coordinates the equation becomes $$Y^2 = 8X,$$ still with $$a=2$$. Hence, at $$(X_1,Y_1)$$ (equivalently $$(x_1,y_1)$$), the tangent is

$$ Y\,Y_1 = 2a\,(X+X_1)\; \Longrightarrow\; y\,y_1 \;=\; 2\bigl[(x-2)+(x_1-2)\bigr]. $$

Simplifying the right-hand side, this gives

$$ y\,y_1 = 2\,(x + x_1 - 4). \quad -(1) $$

Because we want this tangent to pass through the origin $$O(0,0)$$, we substitute $$(x,y)=(0,0)$$ into (1):

$$ 0\cdot y_1 = 2\,(0 + x_1 - 4)\;\; \Longrightarrow\;\; 2\,(x_1 - 4)=0. $$

Therefore

$$ x_1 = 4. $$

To find the corresponding $$y_1$$-coordinates of the points of contact, we insert $$x_1=4$$ into the parabola’s equation:

$$ y_1^2 = 8\,(4-2)=8\cdot 2 = 16 \;\Longrightarrow\; y_1 = \pm 4. $$

Hence the two points of tangency are

$$ S(4,4)\quad\text{and}\quad R(4,-4). $$

We now possess the three vertices of the required triangle: $$O(0,0),\;S(4,4),\;R(4,-4).$$ Since one vertex is the origin, the area formula simplifies. For a triangle with vertices $$O(0,0),\;B(x_1,y_1),\;C(x_2,y_2)$$ the area is

$$ \text{Area} = \dfrac12\,\bigl|\,x_1y_2 - x_2y_1\,\bigr|. $$

Substituting $$B=S(4,4)$$ and $$C=R(4,-4)$$, we get

$$ \text{Area} = \frac12\,\Bigl|\,4\cdot(-4)\;-\;4\cdot4\,\Bigr| = \frac12\,\bigl|\, -16 - 16 \bigr| = \frac12\,(32) = 16. $$

Hence, the required area is $$16$$ square units.

Hence, the correct answer is Option B.

The parabola is $$y^2 = 4x$$. The mirror image of this curve with respect to the line $$y = x$$ is obtained by interchanging $$x$$ and $$y$$, giving $$x^2 = 4y$$. This is the curve $$C$$.

To find the tangent to $$C: x^2 = 4y$$ at $$P(2, 1)$$, we differentiate implicitly: $$2x = 4\frac{dy}{dx}$$, so $$\frac{dy}{dx} = \frac{x}{2}$$. At $$(2, 1)$$, the slope is $$\frac{2}{2} = 1$$.

The equation of the tangent is $$y - 1 = 1(x - 2)$$, which simplifies to $$x - y = 1$$.

Let $$P = (t,\, 4t^2 + 1)$$ be a variable point on the parabola $$y = 4x^2 + 1$$.

The foot of the perpendicular from a point $$(a, b)$$ to the line $$y = x$$ is $$Q = \left(\frac{a+b}{2},\, \frac{a+b}{2}\right)$$. Here, $$Q = \left(\frac{t + 4t^2 + 1}{2},\, \frac{t + 4t^2 + 1}{2}\right).$$

Let $$M = (h, k)$$ be the midpoint of $$P$$ and $$Q$$: $$h = \frac{t + \frac{t+4t^2+1}{2}}{2} = \frac{4t^2 + 3t + 1}{4}, \quad k = \frac{(4t^2+1) + \frac{t+4t^2+1}{2}}{2} = \frac{12t^2 + t + 3}{4}.$$

From the expression for $$k$$: $$4k = 12t^2 + t + 3$$. Using $$4h = 4t^2 + 3t + 1$$, we substitute $$12t^2 = 3(4t^2) = 3(4h - 3t - 1)$$: $$4k = 3(4h - 3t - 1) + t + 3 = 12h - 8t,$$ which gives $$t = \frac{3h - k}{2}$$.

Substituting back into $$4h = 4t^2 + 3t + 1$$: $$4h = (3h-k)^2 + \frac{3(3h-k)}{2} + 1.$$ Multiplying by 2: $$8h = 2(3h-k)^2 + 3(3h-k) + 2 = 2(3h-k)^2 + 9h - 3k + 2.$$

Rearranging: $$2(3h-k)^2 + h - 3k + 2 = 0.$$

Replacing $$h \to x$$ and $$k \to y$$, the locus is $$2(3x - y)^2 + (x - 3y) + 2 = 0.$$

The parabola is $$y^2 = 2x$$. Differentiating: $$2y\frac{dy}{dx} = 2$$, so the slope at $$P(2,2)$$ is $$\frac{dy}{dx} = \frac{1}{y} = \frac{1}{2}$$.

Equation of the tangent at $$P(2,2)$$: $$y - 2 = \frac{1}{2}(x-2)$$, which simplifies to $$x - 2y + 2 = 0$$. Setting $$y = 0$$ gives $$x = -2$$, so $$Q = (-2, 0)$$.

The slope of the normal at $$P$$ is $$-2$$ (negative reciprocal of the tangent slope). Equation of the normal: $$y - 2 = -2(x - 2)$$, i.e., $$y = -2x + 6$$.

To find where the normal meets the parabola again, substitute $$y = -2x+6$$ into $$y^2 = 2x$$: $$(-2x+6)^2 = 2x \implies 4x^2 - 24x + 36 = 2x \implies 4x^2 - 26x + 36 = 0 \implies 2x^2 - 13x + 18 = 0.$$

The roots are $$x = \frac{13 \pm \sqrt{169 - 144}}{4} = \frac{13 \pm 5}{4}$$, giving $$x = 2$$ (point $$P$$) and $$x = \frac{9}{2}$$. At $$x = \frac{9}{2}$$: $$y = -2\cdot\frac{9}{2} + 6 = -3$$. So $$R = \left(\frac{9}{2}, -3\right)$$.

Area of triangle $$PQR$$ with $$P(2,2)$$, $$Q(-2,0)$$, $$R\!\left(\tfrac{9}{2},-3\right)$$: Area $$= \frac{1}{2}\left|x_P(y_Q - y_R) + x_Q(y_R - y_P) + x_R(y_P - y_Q)\right|$$ $$= \frac{1}{2}\left|2(0-(-3)) + (-2)((-3)-2) + \frac{9}{2}(2-0)\right|$$ $$= \frac{1}{2}\left|6 + 10 + 9\right| = \frac{25}{2}.$$

The area of triangle $$PQR$$ is $$\dfrac{25}{2}$$ square units.

The parabola is $$y^2 = 4ax$$, whose focus is at $$(a, 0)$$. Let a moving point on the parabola be $$(at^2, 2at)$$.

The mid-point of the line segment joining the focus $$(a, 0)$$ and $$(at^2, 2at)$$ is $$(h, k)$$, where $$h = \frac{a + at^2}{2}$$ and $$k = \frac{0 + 2at}{2} = at$$.

From $$k = at$$, we get $$t = \frac{k}{a}$$.

Substituting in the expression for $$h$$: $$h = \frac{a + a \cdot \frac{k^2}{a^2}}{2} = \frac{a^2 + k^2}{2a}$$.

So $$2ah = a^2 + k^2$$, which gives $$k^2 = 2ah - a^2 = 2a\left(h - \frac{a}{2}\right)$$.

Replacing $$h$$ with $$x$$ and $$k$$ with $$y$$, the locus is $$y^2 = 2a\left(x - \frac{a}{2}\right)$$.

This is a parabola with vertex at $$\left(\frac{a}{2}, 0\right)$$ and $$4A = 2a$$, so $$A = \frac{a}{2}$$.

The directrix of this parabola is $$x = \frac{a}{2} - \frac{a}{2} = 0$$.

Hence, the correct answer is Option B.

We have the parabola $$y^{2}=8x$$.

First we compare it with the standard form $$y^{2}=4ax$$. Thus $$4a=8$$ and we get $$a=2$$. The directrix of a parabola $$y^{2}=4ax$$ is always $$x=-a$$, so here the directrix is the vertical line $$x=-2$$.

The point given on the curve is $$P(2,-4)$$. Because $$( -4)^{2}=16=8\cdot 2$$, the point indeed lies on the parabola.

For a parabola in the form $$y^{2}=4ax$$, the tangent at a point $$(x_{1},y_{1})$$ on the curve is given by the standard formula

$$yy_{1}=2a\,(x+x_{1}).$$

Substituting $$x_{1}=2,\;y_{1}=-4,\;a=2$$ gives

$$y(-4)=2\cdot 2\,(x+2).$$

So $$-4y=4(x+2)\;\Longrightarrow\;-4y=4x+8.$$

Dividing by $$-4$$ we obtain the equation of the tangent:

$$x+y+2=0.$$

To find its intersection with the directrix $$x=-2$$, we put $$x=-2$$ in this equation:

$$-2+y+2=0\;\Longrightarrow\;y=0.$$

Hence the tangent meets the directrix at $$A(-2,0).$$

Now we need the normal at $$P(2,-4)$$. First we compute the slope of the tangent. Differentiating $$y^{2}=8x$$ gives $$2y\dfrac{dy}{dx}=8,$$ hence

$$\dfrac{dy}{dx}=\frac{4}{y}.$$

At $$P(2,-4)$$ we have $$\dfrac{dy}{dx}=\dfrac{4}{-4}=-1.$$ Therefore the slope of the tangent is $$-1,$$ while the slope of the normal is the negative reciprocal, namely $$1.$$

The normal passing through $$P(2,-4)$$ with slope $$1$$ has equation

$$y+4=1\,(x-2),$$

which simplifies to

$$x-y-6=0.$$

This normal meets the directrix $$x=-2$$ at the point obtained by substituting $$x=-2$$:

$$-2-y-6=0\;\Longrightarrow\;-y-8=0\;\Longrightarrow\;y=-8.$$

Hence the normal meets the directrix at $$B(-2,-8).$$

At this stage we know three vertices of the desired square: $$A(-2,0),\;P(2,-4),\;B(-2,-8).$$

Let us check some distances. The segment $$AB$$ lies entirely on the directrix and its length is

$$AB=\sqrt{(-2+2)^{2}+(0+8)^{2}}=\sqrt{0^{2}+8^{2}}=8.$$

The segments $$PA$$ and $$PB$$ have lengths

$$PA=\sqrt{(2+2)^{2}+(-4-0)^{2}}=\sqrt{4^{2}+(-4)^{2}}=\sqrt{16+16}=4\sqrt{2},$$

$$PB=\sqrt{(2+2)^{2}+(-4+8)^{2}}=\sqrt{4^{2}+4^{2}}=\sqrt{16+16}=4\sqrt{2}.$$

Thus $$PA=PB,$$ and $$AB=8=4\sqrt{2}\times\sqrt{2}.$$ Therefore $$AB$$ is a diagonal of a square whose side is $$4\sqrt{2}.$$ The square is $$AQBP$$ in that cyclic order, so indeed $$A$$ and $$B$$ are opposite vertices, while $$P$$ and the unknown $$Q(a,b)$$ are the other opposite vertices.

For any square, the two diagonals are perpendicular and bisect each other. Hence their common midpoint is the centre of the square. The midpoint of the diagonal $$AB$$ is therefore the centre $$O$$ of the square:

$$O\left(\frac{-2+(-2)}{2},\;\frac{0+(-8)}{2}\right)=(-2,-4).$$

Because $$OP$$ is half of one diagonal and $$OA$$ is half of the other, we calculate $$\overrightarrow{OP}=(2-(-2),\; -4-(-4))=(4,0),$$ $$\overrightarrow{OA}=(-2-(-2),\; 0-(-4))=(0,4).$$ Indeed $$\overrightarrow{OP}$$ is perpendicular to $$\overrightarrow{OA},$$ as required.

The second vertex on the same diagonal as $$P$$ is obtained by reversing this vector through the centre, i.e. by taking the negative of $$\overrightarrow{OP}$$. Hence

$$\overrightarrow{OQ}=-\overrightarrow{OP}=(-4,0).$$

Adding this to the coordinates of $$O(-2,-4)$$ gives the coordinates of $$Q$$:

$$Q(-2-4,\;-4+0)=(-6,-4).$$

Thus $$a=-6,\;b=-4,$$ and we require $$2a+b$$:

$$2a+b=2(-6)+(-4)=-12-4=-16.$$

Hence, the correct answer is Option C.

The parabola $$y^2 = -64x$$ has $$4a = -64$$, so $$a = -16$$ and the focus is at $$(-16, 0)$$.

A focal chord passes through the focus, so substituting $$(-16, 0)$$ into $$y = mx + c$$ gives $$0 = -16m + c$$, hence $$c = 16m$$.

For the line $$y = mx + c$$ (i.e., $$mx - y + c = 0$$) to be tangent to the circle $$(x + 10)^2 + y^2 = 4$$ with centre $$(-10, 0)$$ and radius 2, the distance from the centre to the line must equal 2:

$$\frac{|m(-10) + c|}{\sqrt{m^2 + 1}} = 2$$

Substituting $$c = 16m$$: $$\frac{|-10m + 16m|}{\sqrt{m^2+1}} = \frac{6|m|}{\sqrt{m^2+1}} = 2$$

Squaring: $$36m^2 = 4(m^2 + 1) \implies 32m^2 = 4 \implies m^2 = \frac{1}{8} \implies m = \frac{1}{2\sqrt{2}}$$ (since $$m > 0$$).

Then $$c = 16m = \frac{16}{2\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2}$$.

$$4\sqrt{2}(m + c) = 4\sqrt{2}\left(\frac{1}{2\sqrt{2}} + 4\sqrt{2}\right) = 4\sqrt{2} \cdot \frac{1}{2\sqrt{2}} + 4\sqrt{2} \cdot 4\sqrt{2} = 2 + 32 = 34$$

We begin with the parabola whose equation is $$y^{2}=8x$$. In the standard form $$y^{2}=4ax$$ we identify $$4a=8$$, so we get $$a=2$$ for the parameter of the parabola.

For a parabola of the type $$y^{2}=4ax$$, the equation of the tangent at a point $$(x_{1},y_{1})$$ on the curve is given by the standard tangent formula

$$yy_{1}=2a\left(x+x_{1}\right).$$

Here the given point is $$(2,-4)$$, so we have $$x_{1}=2$$ and $$y_{1}=-4$$. Substituting these values along with $$a=2$$ into the tangent formula we obtain

$$y(-4)=2\cdot2\,(x+2).$$

Simplifying the right-hand side first, $$2\cdot2=4$$, so

$$-4y=4(x+2).$$

Next we divide every term by $$-4$$ to get the equation in a neater form:

$$y=-x-2.$$

It is often convenient to write a straight-line equation in the form $$Ax+By+C=0$$. Adding $$x$$ and $$2$$ to both sides gives

$$x+y+2=0.$$

Thus the tangent line $$L$$ has the Cartesian equation $$x+y+2=0.$$ The same line is also tangent to the circle whose equation is $$x^{2}+y^{2}=a.$$ The centre of this circle is clearly the origin $$(0,0)$$ and its radius is $$\sqrt{a}.$$

For a line $$Ax+By+C=0$$, the perpendicular distance of a point $$(x_{0},y_{0})$$ from the line is given by the distance formula

$$\text{Distance}=\dfrac{|Ax_{0}+By_{0}+C|}{\sqrt{A^{2}+B^{2}}}.$$

To enforce tangency, the distance from the centre of the circle to the line must be equal to the radius. Taking $$(x_{0},y_{0})=(0,0)$$ and the coefficients $$A=1,\;B=1,\;C=2,$$ we compute

$$\text{Distance}=\dfrac{|1\cdot0+1\cdot0+2|}{\sqrt{1^{2}+1^{2}}} =\dfrac{|2|}{\sqrt{2}} =\dfrac{2}{\sqrt{2}} =\sqrt{2}.$$

This distance must equal the radius $$\sqrt{a},$$ so we set

$$\sqrt{a}=\sqrt{2}.$$

Squaring both sides gives

$$a=2.$$

So, the answer is $$2$$.

We parametrize the parabola $$y^2 = 6x$$ using the standard form with $$4a = 6$$, giving $$a = \frac{3}{2}$$. A general point on the parabola is $$\left(\frac{3t^2}{2},\, 3t\right)$$.

The square of the distance from this point to $$\left(3, \frac{3}{2}\right)$$ is $$D(t) = \left(\frac{3t^2}{2} - 3\right)^2 + \left(3t - \frac{3}{2}\right)^2$$. Differentiating and setting $$D'(t) = 0$$ yields $$9t^3 - 9 = 0$$, so $$t^3 = 1$$, giving the only real solution $$t = 1$$.

At $$t = 1$$, the point on the parabola is $$\alpha = \frac{3(1)^2}{2} = \frac{3}{2}$$ and $$\beta = 3(1) = 3$$. We can verify: $$\beta^2 = 9 = 6 \cdot \frac{3}{2} = 6\alpha$$, confirming the point lies on the parabola.

Therefore $$2(\alpha + \beta) = 2\!\left(\frac{3}{2} + 3\right) = 2 \cdot \frac{9}{2} = \boxed{9}$$.

1. Finding the Common Tangent

The equation of a tangent of slope $$m$$

to the parabola  $$y^2 = 4ax$$ is given by:

$$y = mx + \frac{a}{m}$$

For the given parabola $$y^2 = 4x$$, we have $$a = 1$$

. Therefore, the tangent is:

$$y = mx + \frac{1}{m}$$

The equation of a tangent of slope $$m$$

to the parabola $$x^2 = 4ay$$ is given by:

$$y = mx - am^2$$

For the given parabola $$x^2 = 4y$$ , we have $$a = 1$$.

 Therefore, the tangent is: $$y = mx - m^2$$

For a line to be a common tangent to both parabolas, the equations must be identical. This means their y-intercepts must be equal:

$$\frac{1}{m} = -m^2$$

$$m^3 = -1$$

$$m = -1$$

Substitute

$$m = -1$$

back into the tangent equation to get the equation of the common tangent:

$$y = (-1)x + \frac{1}{-1}$$

$$y = -x - 1$$

$$x + y + 1 = 0$$

2. Condition for Tangency to the Circle

We are given that the common tangent $$x + y + 1 = 0$$

also touches the circle: $$x^2 + y^2 = c^2$$

For a line to be tangent to a circle, the perpendicular distance from the center of the circle to the line must be equal to the radius of the circle.

The center of the circle

$$x^2 + y^2 = c^2$$

is at $$(0, 0)$$ and its radius is $$c$$.

Using the perpendicular distance formula from a point

$$(x_1, y_1)$$ to a line $$Ax + By + C = 0$$:

$$\text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

Substitute the center

$$(0, 0)$$

and the line

$$x + y + 1 = 0$$

into the formula, and set it equal to the radius

$$c$$

:

$$c = \frac{|1(0) + 1(0) + 1|}{\sqrt{1^2 + 1^2}}$$

$$c = \frac{|0 + 0 + 1|}{\sqrt{1 + 1}}$$

$$c = \frac{1}{\sqrt{2}}$$

Final Answer

The correct value of

$$c$$

is:

$$c = \frac{1}{\sqrt{2}}$$

This corresponds to Option B.

Let us denote the fixed point by $$A(0,-1)$$ and take an arbitrary point on the parabola $$x^{2}=4y$$ as $$B(X,Y)$$, where the capital letters $$X$$ and $$Y$$ will later be related by the equation of the parabola.

The required point $$P(x,y)$$ divides the segment $$AB$$ internally in the ratio $$1:2$$, that is, $$AP:PB = 1:2$$.

For internal division, we first recall the section-formula: if a point $$P$$ divides the segment joining $$A(x_{1},y_{1})$$ and $$B(x_{2},y_{2})$$ in the ratio $$m:n$$ (with $$m$$ attaching to the farther end $$B$$ and $$n$$ to $$A$$), then

$$P\bigl(\,\dfrac{mx_{2}+nx_{1}}{m+n},\; \dfrac{my_{2}+ny_{1}}{m+n}\bigr).$$

Here $$A(0,-1)$$, $$B(X,Y)$$, $$m=1$$, $$n=2$$. Substituting each symbol, we obtain the coordinates of $$P$$:

$$x \;=\; \dfrac{1\cdot X + 2\cdot 0}{1+2} \;=\; \dfrac{X}{3},$$

$$y \;=\; \dfrac{1\cdot Y + 2\cdot (-1)}{1+2} \;=\; \dfrac{Y-2}{3}.$$

Next we make use of the fact that $$B(X,Y)$$ lies on the parabola. From $$x^{2}=4y$$ we have for point $$B$$

$$X^{2}=4Y \;\;\Longrightarrow\;\; Y=\dfrac{X^{2}}{4}.$$

Our aim is to eliminate $$X$$ and $$Y$$ in favour of the point $$P(x,y)$$. From the relations for $$x$$ and $$y$$ found above we solve for $$X$$ and $$Y$$:

From $$x = \dfrac{X}{3} \; \Longrightarrow\; X = 3x.$$

From $$y = \dfrac{Y-2}{3} \; \Longrightarrow\; Y = 3y + 2.$$

Substituting these expressions into $$X^{2}=4Y$$ gives

$$\bigl(3x\bigr)^{2} \;=\; 4\bigl(3y+2\bigr).$$

Carrying out the algebra step by step,

$$9x^{2} \;=\; 12y + 8.$$

Re-arranging, we bring all terms to one side:

$$9x^{2} - 12y = 8.$$

Thus the locus of the point $$P(x,y)$$ is represented by the equation

$$9x^{2} - 12y = 8.$$

Comparing with the given options, this matches option A.

Hence, the correct answer is Option A.

The equation $$y^{2}=12x$$ can be compared with the standard parabola $$y^{2}=4ax$$. Equating the coefficients we obtain $$4a=12$$, so $$a=3$$.

For a parabola of the form $$y^{2}=4ax$$ the parametric coordinates of any point are given by the well-known formula $$P(at^{2},\,2at)$$. Substituting $$a=3$$ we write the point $$P$$ on the given parabola as $$P\;(3t^{2},\,6t).$$

The axis of this parabola is the $$x$$-axis, i.e. the line $$y=0$$. The foot of the perpendicular drawn from $$P$$ to the axis therefore has the same $$x$$-coordinate as $$P$$ and a $$y$$-coordinate of zero. Hence $$N\;(3t^{2},\,0).$$

The length of the segment $$PN$$ is simply the absolute difference of their $$y$$-coordinates because the segment is vertical: $$PN=\bigl|6t-0\bigr|=6|t|.$$

Now we need the mid-point of $$PN$$. Using the mid-point formula $$M\left(\frac{x_{P}+x_{N}}{2},\,\frac{y_{P}+y_{N}}{2}\right),$$ we substitute $$P(3t^{2},6t)$$ and $$N(3t^{2},0)$$ to get $$M\left(\frac{3t^{2}+3t^{2}}{2},\,\frac{6t+0}{2}\right) =\bigl(3t^{2},\,3t\bigr).$$

Through $$M$$ we draw a line parallel to the axis of the parabola. Because the axis is the $$x$$-axis, a parallel line is a horizontal line. Thus the required line is simply $$y=3t.$$

This horizontal line meets the parabola again at a point $$Q$$. To find the coordinates of $$Q$$ we substitute $$y=3t$$ in the parabola’s equation $$y^{2}=12x$$:

$$\bigl(3t\bigr)^{2}=12x \;\Longrightarrow\;9t^{2}=12x \;\Longrightarrow\;x=\frac{9}{12}t^{2}=\frac{3}{4}t^{2}.$$

Hence $$Q\left(\frac{3}{4}t^{2},\,3t\right).$$

Observe that $$M$$ and $$Q$$ have the same $$y$$-coordinate. The distance $$MQ$$ is therefore purely horizontal: $$MQ=\bigl|x_{M}-x_{Q}\bigr| =\Bigl|3t^{2}-\frac{3}{4}t^{2}\Bigr| =\left(3-\frac{3}{4}\right)t^{2} =\frac{9}{4}\,t^{2}.$$

Next we use the information about the line $$NQ$$. First we find its slope. Using the two points $$N(3t^{2},\,0),\qquad Q\!\left(\frac{3}{4}t^{2},\,3t\right),$$ the slope is

$$m=\frac{3t-0}{\dfrac{3}{4}t^{2}-3t^{2}} =\frac{3t}{\left(\dfrac{3}{4}-3\right)t^{2}} =\frac{3t}{\left(\dfrac{3}{4}-\dfrac{12}{4}\right)t^{2}} =\frac{3t}{-\dfrac{9}{4}t^{2}} =-\frac{4}{3}\,\frac{1}{t}.$$

Using point-slope form, the equation of $$NQ$$ is $$y-0=m\bigl(x-3t^{2}\bigr) \;\Longrightarrow\;y=-\frac{4}{3t}\,\bigl(x-3t^{2}\bigr).$$

The $$y$$-intercept is obtained by putting $$x=0$$:

$$y=-\frac{4}{3t}\,(0-3t^{2}) =-\frac{4}{3t}\,(-3t^{2}) =\frac{12t^{2}}{3t} =4t.$$

We are told that this intercept equals $$\dfrac{4}{3}$$, so

$$4t=\frac{4}{3}\;\Longrightarrow\;t=\frac{1}{3}.$$

With $$t=\dfrac{1}{3}$$ we now evaluate $$MQ$$:

$$MQ=\frac{9}{4}\,t^{2} =\frac{9}{4}\left(\frac{1}{3}\right)^{2} =\frac{9}{4}\cdot\frac{1}{9} =\frac{1}{4}.$$

Thus $$MQ=\dfrac{1}{4}$$.

Looking at the given options, this corresponds to Option C.

Hence, the correct answer is Option C.

We need two tangents, one to each parabola, which finally cut each other at a right angle. The first parabola is $$y^{2}=4(x+1)$$. For the standard parabola $$y^{2}=4ax$$ the tangent having slope $$m$$ is given by the well-known slope form

$$y=mx+\dfrac{a}{m}\,.$$

Here, after shifting the origin, $$a=1$$ and $$x$$ is replaced by $$x+1$$. So the tangent to the first parabola with slope $$m$$ is

$$y=m(x+1)+\dfrac{1}{m}\;. \quad -(1)$$

The second parabola is $$y^{2}=8(x+2)=4\cdot2\,(x+2)$$, hence $$a=2$$. Again using the slope form with the horizontal shift $$x\mapsto x+2$$, the tangent having slope $$n$$ is

$$y=n(x+2)+\dfrac{2}{n}\;. \quad -(2)$$

The two tangents meet at right angles, therefore the product of their slopes equals $$-1$$, i.e.

$$mn=-1\quad\Longrightarrow\quad n=-\dfrac1m\;. \quad -(3)$$

Let the two tangents intersect at the point $$P(h,k)$$. By definition, $$P$$ must satisfy both (1) and (2).

Substituting $$x=h,\;y=k$$ in (1) we get

$$k=m(h+1)+\dfrac1m\;. \quad -(4)$$

Likewise, substituting $$x=h,\;y=k$$ in (2) gives

$$k=n(h+2)+\dfrac{2}{n}\;. \quad -(5)$$

Because both right-hand sides equal $$k$$, we equate them:

$$m(h+1)+\dfrac1m=n(h+2)+\dfrac{2}{n}\;. \quad -(6)$$

Now we use relation (3), namely $$n=-1/m$$, on the right-hand side of (6).

Replacing $$n$$ by $$-1/m$$ gives

$$m(h+1)+\dfrac1m=-\dfrac{h+2}{m}-2m\;. \quad -(7)$$

To clear the denominators we multiply every term in (7) by $$m$$:

$$m^{2}(h+1)+1=-(h+2)-2m^{2}\;. \quad -(8)$$

Now we bring every term to the left side:

$$m^{2}(h+1)+1+(h+2)+2m^{2}=0\;. \quad -(9)$$

Combining the $$m^{2}$$ terms first, we notice

$$m^{2}(h+1)+2m^{2}=m^{2}\bigl((h+1)+2\bigr)=m^{2}(h+3)\;.$$

Similarly, collecting the constant terms, we have

$$1+(h+2)=h+3\;.$$

Hence equation (9) simplifies neatly to

$$(h+3)\bigl(m^{2}+1\bigr)=0\;. \quad -(10)$$

Because $$m^{2}+1$$ is always positive (it can never vanish), the only possibility for (10) to hold is

$$h+3=0\quad\Longrightarrow\quad h=-3\;. \quad -(11)$$

Thus the $$x$$-coordinate of the intersection point $$P$$ is fixed at $$-3$$, whereas the $$y$$-coordinate $$k$$ remains unrestricted. Consequently, every such pair of perpendicular tangents meets somewhere on the vertical line

$$x=-3\quad\Longleftrightarrow\quad x+3=0\;.$$

Hence, the correct answer is Option A.

We consider the parabola $$y^2 = 8x$$ whose vertex is at the origin. Let us place one vertex of the desired equilateral triangle at this point, calling it $$A(0,0).$$ Because the curve is symmetric about the $$x$$-axis, it is natural to look for the other two vertices as a pair of points that are mirror images with respect to this axis. Hence we write

$$B(x_0,\;y_0)\quad\text{and}\quad C(x_0,\;-y_0)$$

with $$y_0 > 0$$ and $$B,\,C$$ lying on the parabola. So we have

$$y_0^2 = 8x_0. \quad -(1)$$

Now an equilateral triangle requires all three sides to be equal, so we equate the squared lengths:

Length $$AB^2 = (x_0-0)^2 + (y_0-0)^2 = x_0^2 + y_0^2,$$

Length $$BC^2 = (x_0 - x_0)^2 + (y_0 + y_0)^2 = 0^2 + (2y_0)^2 = 4y_0^2.$$

Setting $$AB = BC$$ gives

$$x_0^2 + y_0^2 = 4y_0^2.$$

Simplifying,

$$x_0^2 = 3y_0^2. \quad -(2)$$

From (2) we obtain $$x_0 = \sqrt{3}\,y_0.$$ Substituting this value of $$x_0$$ into the parabola condition (1) yields

$$y_0^2 = 8(\sqrt{3}\,y_0).$$

Dividing by $$y_0$$ (which is positive) we get

$$y_0 = 8\sqrt{3}.$$

Putting this back into $$x_0 = \sqrt{3}\,y_0$$ gives

$$x_0 = \sqrt{3}\,(8\sqrt{3}) = 8\cdot3 = 24.$$

Hence the coordinates of the three vertices are

$$A(0,0),\; B(24,\,8\sqrt{3}),\; C(24,\,-8\sqrt{3}).$$

The side length can now be found. Using $$AB$$,

$$AB^2 = 24^2 + (8\sqrt{3})^2 = 576 + 64\cdot3 = 576 + 192 = 768,$$

so

$$AB = \sqrt{768} = \sqrt{256\cdot3} = 16\sqrt{3}.$$

The standard formula for the area of an equilateral triangle with side $$s$$ is

$$\text{Area} = \frac{\sqrt{3}}{4}\,s^2.$$

Substituting $$s = 16\sqrt{3}$$ we have

$$\text{Area} = \frac{\sqrt{3}}{4}\,(16\sqrt{3})^2 = \frac{\sqrt{3}}{4}\,(256\cdot3) = \frac{768\sqrt{3}}{4} = 192\sqrt{3}\ \text{square units}.$$

Hence, the correct answer is Option C.

We have a straight line $$y = mx$$ with the given condition $$m \gt 0$$. This line meets the parabola $$y^2 = x$$ at two points. One of them is obviously the origin $$O(0,0)$$. To obtain the other point, we substitute $$x = \dfrac{y}{m}$$ (from $$y = mx$$) into the parabola’s equation:

$$y^2 \;=\; x \;=\; \dfrac{y}{m}.$$

Re-arranging,

$$y^2 - \dfrac{y}{m} = 0 \;\;\Longrightarrow\;\; y\Bigl(y - \dfrac1m\Bigr) = 0.$$

This gives two possible values of $$y$$:

$$y = 0 \quad\text{or}\quad y = \dfrac1m.$$

The value $$y = 0$$ corresponds to the origin, so the non-trivial point of intersection is obtained from

$$y_P = \dfrac1m.$$

For this $$y_P$$, the corresponding $$x_P$$ is found from $$x = \dfrac{y}{m}$$:

$$x_P = \dfrac{1/m}{m} = \dfrac1{m^2}.$$

Thus, the coordinates of the required point $$P$$ are

$$P\!\Bigl(\dfrac1{m^2},\; \dfrac1m\Bigr).$$

Next we need the tangent to the parabola $$y^2 = x$$ at the point $$P$$. First we find the slope of the tangent. Differentiating $$y^2 = x$$ implicitly with respect to $$x$$:

$$2y\dfrac{dy}{dx} = 1 \;\;\Longrightarrow\;\; \dfrac{dy}{dx} = \dfrac1{2y}.$$

At $$P$$, where $$y = \dfrac1m$$, the slope of the tangent is

$$m_{\text{tangent}} = \dfrac1{2\bigl(1/m\bigr)} = \dfrac1{2/m} = \dfrac{m}{2}.$$

The equation of the tangent at $$P\bigl(x_P,y_P\bigr)$$ is, therefore,

$$y - y_P = m_{\text{tangent}}\bigl(x - x_P\bigr),$$

that is,

$$y - \dfrac1m = \dfrac{m}{2}\!\left( x - \dfrac1{m^2} \right).$$

This tangent meets the $$x$$-axis where $$y = 0$$. Putting $$y = 0$$ in the above equation gives

$$0 - \dfrac1m = \dfrac{m}{2}\!\left( x_Q - \dfrac1{m^2} \right).$$

We solve for $$x_Q$$ step by step:

$$-\dfrac1m = \dfrac{m}{2}\,x_Q - \dfrac{m}{2}\cdot\dfrac1{m^2}$$ $$-\dfrac1m = \dfrac{m}{2}\,x_Q - \dfrac1{2m}$$

Now bring the second term on the right to the left:

$$-\dfrac1m + \dfrac1{2m} = \dfrac{m}{2}\,x_Q$$ $$-\dfrac1{2m} = \dfrac{m}{2}\,x_Q$$

Multiplying both sides by $$\dfrac2m$$:

$$x_Q = -\,\dfrac2{m^2}\cdot\dfrac12 = -\dfrac1{m^2}.$$

Thus, the tangent meets the $$x$$-axis at

$$Q\!\Bigl(-\dfrac1{m^2},\,0\Bigr).$$

Now we have the three vertices of $$\triangle OPQ$$:

$$O(0,0), \quad P\!\Bigl(\dfrac1{m^2},\,\dfrac1m\Bigr), \quad Q\!\Bigl(-\dfrac1{m^2},\,0\Bigr).$$

The area of a triangle with vertices $$(x_1,y_1),(x_2,y_2),(x_3,y_3)$$ is given by the determinant formula

$$\text{Area} \;=\; \dfrac12\Bigl|\,x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\Bigr|.$$

Substituting $$(x_1,y_1) = (0,0)$$, $$(x_2,y_2) = \bigl(\tfrac1{m^2}, \tfrac1m\bigr)$$ and $$(x_3,y_3) = \bigl(-\tfrac1{m^2},0\bigr)$$, we get

$$\text{Area} = \dfrac12\Bigl|\; 0\bigl(\tfrac1m - 0\bigr) \;+\; \dfrac1{m^2}\bigl(0 - 0\bigr) \;+\; \Bigl(-\dfrac1{m^2}\Bigr)\bigl(0 - \tfrac1m\bigr) \Bigr|.$$

The first two terms vanish, so

$$\text{Area} = \dfrac12\Bigl|\, \Bigl(-\dfrac1{m^2}\Bigr)\Bigl(-\dfrac1m\Bigr) \Bigr| = \dfrac12\Bigl(\dfrac1{m^3}\Bigr) = \dfrac1{2m^3}.$$

According to the question, this area equals $$4$$ square units:

$$\dfrac1{2m^3} = 4.$$

Multiplying both sides by $$2m^3$$:

$$1 = 8m^3 \;\;\Longrightarrow\;\; m^3 = \dfrac18.$$

Taking the positive cube root (because $$m \gt 0$$), we obtain

$$m = \dfrac12 = 0.5.$$

So, the answer is $$m = 0.5.$$

We have the parabola $$y^{2}+4(x-a^{2})=0.$$

First we rewrite it in the standard horizontal‐opening form. Moving the term $$4(x-a^{2})$$ to the other side gives

$$y^{2}=-4(x-a^{2}).$$

So we may write

$$y^{2}=-4x+4a^{2}\quad\Longrightarrow\quad x=a^{2}-\frac{y^{2}}{4}.$$

Comparing with the standard equation $$y^{2}=4p(x-h)$$, we see that the vertex is at $$\bigl(a^{2},\,0\bigr).$$

Now we find the points where the parabola meets the y-axis. On the y-axis we have $$x=0,$$ so we substitute $$x=0$$ in the original equation:

$$y^{2}+4(0-a^{2})=0\quad\Longrightarrow\quad y^{2}-4a^{2}=0.$$

This factors as

$$y^{2}=4a^{2}\quad\Longrightarrow\quad y=\pm2a.$$

Thus the two intersection points are $$\bigl(0,\,2a\bigr)$$ and $$\bigl(0,\,-2a\bigr).$$

Hence the three vertices of the required triangle are

$$V\bigl(a^{2},0\bigr),\;P_{1}\bigl(0,2a\bigr),\;P_{2}\bigl(0,-2a\bigr).$$

To obtain the area, we use the shoelace (determinant) formula for a triangle whose vertices are $$\bigl(x_{1},y_{1}\bigr),\bigl(x_{2},y_{2}\bigr),\bigl(x_{3},y_{3}\bigr):$$

$$\text{Area}=\frac12\,\Bigl|\,x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\,\Bigr|.$$

Assigning the coordinates, we take

$$\bigl(x_{1},y_{1}\bigr)=\bigl(a^{2},0\bigr),\quad \bigl(x_{2},y_{2}\bigr)=\bigl(0,2a\bigr),\quad \bigl(x_{3},y_{3}\bigr)=\bigl(0,-2a\bigr).$$

Now we compute each term:

$$x_{1}(y_{2}-y_{3})=a^{2}\bigl(2a-(-2a)\bigr)=a^{2}\cdot4a=4a^{3},$$

$$x_{2}(y_{3}-y_{1})=0\cdot\bigl(-2a-0\bigr)=0,$$

$$x_{3}(y_{1}-y_{2})=0\cdot\bigl(0-2a\bigr)=0.$$

Adding these gives

$$x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})=4a^{3}.$$

Therefore the area is

$$\text{Area}=\frac12\,\bigl|\,4a^{3}\bigr|=2\,|a^{3}|.$$

We are told that this area equals $$250$$ square units, so

$$2\,|a^{3}|=250\quad\Longrightarrow\quad|a^{3}|=125.$$

Taking the real cube root (remembering that $$|a^{3}|=(|a|)^{3}$$), we get

$$|a|=125^{1/3}=5.$$

Thus the parameter $$a$$ can be $$5$$ or $$-5.$$ Among the given options, only the positive value appears, so we select

$$a=5.$$

Hence, the correct answer is Option D.

We are given the parabola $$x^{2}=4y$$ and the line $$x-\sqrt{2}\,y+4\sqrt{2}=0$$. The points where this line meets the parabola will form the required chord; so we first find their coordinates by solving the two equations simultaneously.

From the line we can express $$y$$ in terms of $$x$$. Rearranging we get

$$x-\sqrt{2}\,y+4\sqrt{2}=0 \quad\Longrightarrow\quad \sqrt{2}\,y=x+4\sqrt{2}\quad\Longrightarrow\quad y=\dfrac{x+4\sqrt{2}}{\sqrt{2}}\;.$$

Now we substitute this value of $$y$$ in the parabola $$x^{2}=4y$$. Thus we have

$$x^{2}=4\left(\dfrac{x+4\sqrt{2}}{\sqrt{2}}\right).$$

Next we simplify the right-hand side. Since $$4/\sqrt{2}=\dfrac{4\sqrt{2}}{2}=2\sqrt{2}$$, the equation becomes

$$x^{2}=2\sqrt{2}\,x+16.$$

Bringing every term to the left, we obtain the quadratic equation

$$x^{2}-2\sqrt{2}\,x-16=0.$$

To solve it we use the quadratic‐formula $$x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$, where here $$a=1,\,b=-2\sqrt{2}$$ and $$c=-16$$. Hence

$$x=\dfrac{2\sqrt{2}\pm\sqrt{(2\sqrt{2})^{2}-4(1)(-16)}}{2}.$$

Calculating inside the square root,

$$(2\sqrt{2})^{2}=4\cdot2=8,\qquad -4ac=-4(1)(-16)=64,$$

so

$$b^{2}-4ac=8+64=72\quad\Longrightarrow\quad\sqrt{72}=6\sqrt{2}.$$

Thus

$$x=\dfrac{2\sqrt{2}\pm6\sqrt{2}}{2}.$$

Separating the two possibilities,

$$x_{1}=\dfrac{2\sqrt{2}+6\sqrt{2}}{2}=\dfrac{8\sqrt{2}}{2}=4\sqrt{2},$$

$$x_{2}=\dfrac{2\sqrt{2}-6\sqrt{2}}{2}=\dfrac{-4\sqrt{2}}{2}=-2\sqrt{2}.$$

Corresponding $$y$$‐coordinates are obtained from $$y=\dfrac{x+4\sqrt{2}}{\sqrt{2}}$$.

For $$x_{1}=4\sqrt{2}$$ we get

$$y_{1}=\dfrac{4\sqrt{2}+4\sqrt{2}}{\sqrt{2}}=\dfrac{8\sqrt{2}}{\sqrt{2}}=8.$$

For $$x_{2}=-2\sqrt{2}$$ we get

$$y_{2}=\dfrac{-2\sqrt{2}+4\sqrt{2}}{\sqrt{2}}=\dfrac{2\sqrt{2}}{\sqrt{2}}=2.$$

Therefore the end‐points of the chord are $$P_{1}(4\sqrt{2},\,8)$$ and $$P_{2}(-2\sqrt{2},\,2).$$

Now we find the distance between these points. The distance formula is

$$\text{Distance}=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}.$$

We compute the differences:

$$x_{1}-x_{2}=4\sqrt{2}-(-2\sqrt{2})=6\sqrt{2},$$

$$y_{1}-y_{2}=8-2=6.$$

Substituting,

$$\text{Distance}=\sqrt{(6\sqrt{2})^{2}+6^{2}}=\sqrt{36\cdot2+36}=\sqrt{72+36}=\sqrt{108}.$$

Since $$108=36\cdot3$$, we have $$\sqrt{108}=6\sqrt{3}.$$

Hence the length of the chord is $$6\sqrt{3}$$ units.

Hence, the correct answer is Option A.

We begin with the two curves that have been given.

The parabola is $$y^{2}=4x$$ and the circle is $$x^{2}+y^{2}=5.$$

We want the common point of these two curves that lies in the first quadrant, that is, with $$x>0$$ and $$y>0.$$

For a parabola of the form $$y^{2}=4ax,$$ a convenient parametric representation is

$$x = at^{2},\qquad y = 2at.$$

Here the parameter is $$t$$ and the constant $$a=1$$ (because the right side is $$4x = 4\cdot1\cdot x$$). Hence for our parabola we set

$$x = t^{2},\qquad y = 2t.$$

Now we substitute these expressions for $$x$$ and $$y$$ in the equation of the circle $$x^{2}+y^{2}=5.$$ So we have

$$\left(t^{2}\right)^{2} + \left(2t\right)^{2} = 5.$$

Simplifying each term gives

$$t^{4} + 4t^{2} = 5.$$

To solve this quartic, let us set $$u = t^{2}.$$ Because $$t^{2} \ge 0,$$ we will look only for non-negative solutions of $$u.$$ Rewriting, we get

$$u^{2} + 4u - 5 = 0.$$

This is a quadratic in $$u.$$ Using the quadratic formula $$u=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a},$$ with $$a=1,\; b=4,\; c=-5,$$ we find

$$u = \dfrac{-4 \pm \sqrt{16 + 20}}{2} = \dfrac{-4 \pm \sqrt{36}}{2} = \dfrac{-4 \pm 6}{2}.$$

That yields two numerical values:

$$u = \dfrac{-4 + 6}{2} = 1, \qquad u = \dfrac{-4 - 6}{2} = -5.$$

Because $$u = t^{2} \ge 0,$$ we discard the negative value $$u=-5$$ and keep $$u=1.$$

Thus $$t^{2}=1 \implies t=\pm1.$$ We need the point in the first quadrant where $$y=2t$$ is positive, so we choose $$t=+1.$$

With $$t=1$$ we obtain the coordinates of the point of intersection: $$x = t^{2} = 1,\qquad y = 2t = 2.$$ Therefore the required point on both the parabola and the circle in the first quadrant is $$P(1,2).$$

Next, we find the equation of the tangent to the parabola at the point $$P.$$

First, we recall that for any curve given implicitly as $$F(x,y)=0,$$ the slope of the tangent is obtained via implicit differentiation: $$\frac{dy}{dx} = -\frac{F_{x}}{F_{y}}.$$

For the parabola $$y^{2}=4x,$$ let us differentiate both sides with respect to $$x.$$ We have

$$\frac{d}{dx}\bigl(y^{2}\bigr) = \frac{d}{dx}\bigl(4x\bigr).$$

Using the power rule and the chain rule,

$$2y\frac{dy}{dx} = 4.$$

Hence

$$\frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y}.$$

At the point $$P(1,2),$$ we substitute $$y=2$$ to get the slope of the tangent:

$$m = \frac{dy}{dx}\Big|_{(1,2)} = \frac{2}{2} = 1.$$

Now we use the point-slope form of a straight line:

$$y - y_{1} = m(x - x_{1}).$$

With $$m = 1,\; (x_{1},y_{1}) = (1,2),$$ this becomes

$$y - 2 = 1\,(x - 1).$$

Simplifying gives

$$y - 2 = x - 1 \;\;\Longrightarrow\;\; y = x + 1.$$

This is the equation of the tangent line to the parabola at the required point.

Finally, we check which of the four option points lies on the line $$y = x + 1.$$ For a point $$\bigl(x_{0},y_{0}\bigr)$$ to be on this line we must have $$y_{0}=x_{0}+1.$$

Let us test each option:

Option A $$\left(\frac14,\frac34\right):$$ $$\frac34 \neq \frac14 + 1 = \frac54,$$ so it is not on the line.

Option B $$\left(-\frac13,\frac43\right):$$ $$\frac43 \neq -\frac13 + 1 = \frac23,$$ so it fails as well.

Option C $$\left(-\frac14,\frac12\right):$$ $$\frac12 \neq -\frac14 + 1 = \frac34,$$ so it is not on the line.

Option D $$\left(\frac34,\frac74\right):$$ Here $$\frac74 = \frac34 + 1,$$ which is exactly satisfied.

Therefore the tangent line $$y = x + 1$$ passes through the point $$\left(\frac34,\frac74\right),$$ which corresponds to Option 4.

Hence, the correct answer is Option 4.

First we read the geometrical data. The axis of the parabola is along the $$x$$-axis, its vertex is 2 units away from the origin on the positive $$x$$-axis, and its focus is 4 units away from the origin on the same axis.

Because the distance is measured from the origin along the positive $$x$$-axis, we immediately obtain the co-ordinates

$$\text{Vertex } V(2,0)$$   and   $$\text{Focus } F(4,0).$$

The focus lies to the right of the vertex, so the parabola opens rightwards. Let $$h$$ and $$k$$ be the co-ordinates of the vertex. Here $$h=2$$ and $$k=0.$$ The distance between the vertex and the focus is denoted by $$p$$, so

$$p = 4-2 = 2.$$

For a parabola whose axis is horizontal (parallel to the $$x$$-axis) and which opens to the right, the standard equation is stated first:

$$ (y-k)^2 = 4p\,(x-h). $$

Now we substitute the values $$k=0,\, h=2,\, p=2$$ obtained above. This gives

$$ (y-0)^2 = 4\,(2)\,(x-2). $$

Simplifying the constants on the right-hand side, we have

$$ y^2 = 8\,(x-2). $$

This is the explicit Cartesian equation of the required parabola. Any point $$P(x,y)$$ will lie on the parabola if and only if its co-ordinates satisfy $$y^2 = 8(x-2).$$ We now test each option one by one, substituting the given values of $$x$$ and $$y$$ and checking equality.

Option A: $$P(6,\,4\sqrt2).$$ We compute the left-hand side (LHS) and the right-hand side (RHS) of the parabola’s equation.

$$\text{LHS}=y^2 = (4\sqrt2)^2 = 16\cdot2 = 32,$$

$$\text{RHS}=8(x-2)=8(6-2)=8\cdot4=32.$$

LHS = RHS, so Option A satisfies the equation and the point lies on the parabola.

Option B: $$P(5,\,2\sqrt6).$$

$$\text{LHS}=y^2 = (2\sqrt6)^2 = 4\cdot6 = 24,$$

$$\text{RHS}=8(x-2)=8(5-2)=8\cdot3=24.$$

LHS = RHS once again, so Option B also lies on the parabola.

Option C: $$P(8,\,6).$$

$$\text{LHS}=y^2 = 6^2 = 36,$$

$$\text{RHS}=8(x-2)=8(8-2)=8\cdot6=48.$$

Here $$36 \neq 48,$$ so the equality fails; the point does not satisfy the parabola’s equation.

Option D: $$P(4,\,-4).$$

$$\text{LHS}=y^2 = (-4)^2 = 16,$$

$$\text{RHS}=8(x-2)=8(4-2)=8\cdot2=16.$$

LHS = RHS, so Option D is also a point on the parabola.

We have evaluated all four candidates and discovered that every point except Option C satisfies the defining equation of the parabola. Therefore Option C is the lone point that does not lie on the given parabola.

Hence, the correct answer is Option C.

We are given the parabola $$y^{2}=16x$$.

First we rewrite it in the standard form $$y^{2}=4ax$$ to identify the value of $$a$$.

Comparing $$y^{2}=16x$$ with $$y^{2}=4ax$$, we obtain $$4a = 16 \;\Rightarrow\; a = 4$$.

For the parabola $$y^{2}=4ax$$, the focus is known to be $$\bigl(a,0\bigr)$$. Hence here the focus is $$\bigl(4,0\bigr)$$.

Any point on the parabola can be written in the standard parametric form. The formula is:

“For $$y^{2}=4ax$$, a point corresponding to parameter $$t$$ is $$\bigl(at^{2},\,2at\bigr)$$.”

With $$a = 4$$, the coordinates become $$\bigl(4t^{2},\,8t\bigr)$$.

One end of the focal chord is given as $$(1,4)$$. Because this point lies on the parabola, there exists a parameter, say $$t_{1}$$, such that

$$\bigl(4t_{1}^{2},\,8t_{1}\bigr) = (1,4).$$

We equate the coordinates one by one:

From the $$x$$-coordinate: $$4t_{1}^{2}=1 \;\Longrightarrow\; t_{1}^{2}=\dfrac14 \;\Longrightarrow\; t_{1}=\pm\dfrac12.$$

From the $$y$$-coordinate: $$8t_{1}=4 \;\Longrightarrow\; t_{1}=\dfrac48=\dfrac12.$$

Combining these, we fix $$t_{1}=\dfrac12$$ (the positive sign satisfies both equations).

For a chord that passes through the focus—called a focal chord—there is a well-known relation between the parameters of its two endpoints.

The property is:

“If the endpoints of a chord of $$y^{2}=4ax$$ correspond to parameters $$t_{1}$$ and $$t_{2}$$, the chord passes through the focus if and only if $$t_{1}t_{2}=-1$$.”

We already have $$t_{1}=\dfrac12$$, so we obtain $$t_{2}$$ directly:

$$t_{1}t_{2}=-1 \;\Longrightarrow\; \dfrac12\,t_{2}=-1 \;\Longrightarrow\; t_{2}=-2.$$

Now we compute the coordinates of the second endpoint using $$t_{2}=-2$$:

$$x_{2}=4t_{2}^{2}=4(-2)^{2}=4\cdot4=16,$$

$$y_{2}=8t_{2}=8(-2)=-16.$$

So the second endpoint is $$Q(16,-16)$$.

The focal chord therefore has endpoints $$P(1,4)$$ and $$Q(16,-16)$$.

To find its length, we use the distance formula between two points $$\bigl(x_{1},y_{1}\bigr)$$ and $$\bigl(x_{2},y_{2}\bigr)$$:

$$\text{Distance}=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}.$$

Substituting $$P(1,4)$$ and $$Q(16,-16)$$, we have

$$$ \begin{aligned} PQ &= \sqrt{(16-1)^{2}+(-16-4)^{2}} \\ &= \sqrt{15^{2}+(-20)^{2}} \\ &= \sqrt{225+400} \\ &= \sqrt{625} \\ &= 25. \end{aligned} $$$

Thus the length of the required focal chord is $$25$$.

Hence, the correct answer is Option B.

We have the straight line

$$ax+y=c$$

which is required to touch (be tangent to) both the circle

$$x^{2}+y^{2}=1$$

and the parabola

$$y^{2}=4\sqrt{2}\,x.$$

Tangency condition with the circle. For a line written as $$Ax+By+C=0,$$ the perpendicular distance from the centre $$O(0,0)$$ to the line is

$$\dfrac{|C|}{\sqrt{A^{2}+B^{2}}}.$$

Here the given line can be rewritten in the standard form as

$$ax+y-c=0,$$

so that $$A=a,\;B=1,\;C=-c.$$ The distance of the centre from the line therefore equals

$$\dfrac{|\, -c\,|}{\sqrt{a^{2}+1}}=\dfrac{|c|}{\sqrt{a^{2}+1}}.$$

Because the line is tangent to the circle of radius $$1,$$ this distance must be $$1.$$ Hence

$$\dfrac{|c|}{\sqrt{a^{2}+1}}=1 \;\;\Longrightarrow\;\; c^{2}=a^{2}+1. \quad -(1)$$

Tangency condition with the parabola. Let us write the line in slope-intercept form:

$$y=-ax+c.$$

Its slope is therefore $$m=-a.$$ For the parabola $$y^{2}=4p\,x,$$ a tangent having slope $$m$$ is known to be

$$y=mx+\dfrac{p}{m}\quad\Bigl(\text{standard slope form}\Bigr).$$

In our problem the parabola is $$y^{2}=4\sqrt{2}\,x,$$ so that $$p=\sqrt{2}.$$ Comparing the required tangent $$y=-ax+c$$ with the standard tangent $$y=mx+\dfrac{p}{m},$$ we equate

$$m=-a\qquad\text{and}\qquad c=\dfrac{p}{m}.$$

Substituting $$m=-a$$ and $$p=\sqrt{2}$$ in the second relation gives

$$c=\dfrac{\sqrt{2}}{-a}=-\dfrac{\sqrt{2}}{a},$$

and therefore

$$ac=-\sqrt{2}. \quad -(2)$$

Combining the two conditions. From (1) we have $$a^{2}=c^{2}-1.$$ Squaring (2) gives

$$(ac)^{2}=2 \;\;\Longrightarrow\;\; a^{2}c^{2}=2. \quad -(3)$$

Now substitute $$a^{2}=c^{2}-1$$ from (1) into (3):

$$(c^{2}-1)\,c^{2}=2.$$

Expanding the left-hand side, we obtain

$$c^{4}-c^{2}-2=0.$$

This is a quadratic equation in $$c^{2}.$$ Let $$t=c^{2}.$$ Then

$$t^{2}-t-2=0.$$

Using the quadratic formula,

$$t=\dfrac{1\pm\sqrt{1+8}}{2} =\dfrac{1\pm3}{2}.$$

So

$$t=2\quad\text{or}\quad t=-1.$$

Because $$t=c^{2}$$ cannot be negative, we keep only $$t=2,$$ i.e.

$$c^{2}=2.$$

Extracting $$c.$$ Taking the (positive) square root gives

$$c=\sqrt{2}.$$

(The negative root $$c=-\sqrt{2}$$ also satisfies the algebraic conditions, but the question asks for the value of $$c$$; conventionally we report the positive one.)

Hence, the correct answer is Option B.

We begin with the given parabola $$x^{2}=8y$$.

First we compare this with the standard form $$x^{2}=4ay$$. By comparison we have $$4a=8\;\Longrightarrow\;a=2.$$

Now we recall the standard result: for the parabola $$x^{2}=4ay,$$ the equation of the tangent having slope $$m$$ is given by the formula

$$y=mx-am^{2}.$$

In the present problem the tangent is said to make an angle $$\theta$$ with the positive direction of the x-axis. The slope of such a line is therefore

$$m=\tan\theta.$$

Substituting $$m=\tan\theta$$ and $$a=2$$ into the tangent formula, we get

$$y=\bigl(\tan\theta\bigr)x-2\bigl(\tan\theta\bigr)^{2}.$$

To compare this with the options, we now isolate $$x$$. Transposing terms yields

$$x\tan\theta=y+2\tan^{2}\theta.$$

Dividing every term by $$\tan\theta$$ (which is non-zero for a valid angle of inclination), we have

$$x=y\cot\theta+2\tan\theta.$$

This equation exactly matches Option C.

Hence, the correct answer is Option C.

First, let us rewrite the equation of the circle in its centre-radius form.

We have $$x^{2}+y^{2}-6x=0 \;.$$

Completing the square in $$x$$:

$$x^{2}-6x+y^{2}=0 \;\Longrightarrow\; (x^{2}-6x+9)+y^{2}=9 \;\Longrightarrow\; (x-3)^{2}+y^{2}=3^{2}\;.$$

So the circle has centre $$C(3,0)$$ and radius $$r=3$$.

Now let us take a straight line with slope $$m$$ in its slope-intercept form

$$y=mx+c\;.$$

For this line to be tangent to the circle, the perpendicular distance from the centre to the line must be equal to the radius. The perpendicular-distance formula is

$$\text{Distance}=\frac{|Ax_{0}+By_{0}+C|}{\sqrt{A^{2}+B^{2}}}$$

for a line $$Ax+By+C=0$$ and a point $$(x_{0},y_{0})$$. Writing $$y=mx+c$$ as $$mx-y+c=0$$, we identify $$A=m,\;B=-1,\;C=c$$ and substitute $$(x_{0},y_{0})=(3,0)$$:

$$\frac{|m\cdot3+(-1)\cdot0+c|}{\sqrt{m^{2}+(-1)^{2}}}=3 \;\Longrightarrow\; \frac{|3m+c|}{\sqrt{m^{2}+1}}=3\;.$$

Hence

$$|3m+c|=3\sqrt{m^{2}+1} \;\Longrightarrow\; c=-3m\pm3\sqrt{m^{2}+1}\;. \quad -(1)$$

Next, for the parabola $$y^{2}=4x$$ we recall the standard tangent formula. For the parabola $$y^{2}=4ax$$, a line of slope $$m$$ is a tangent iff

$$y=mx+\frac{a}{m}\;.$$

Here $$4a=4\;\Longrightarrow\;a=1$$, so the condition is

$$c=\frac{1}{m}\;. \quad -(2)$$

A common tangent must satisfy both (1) and (2). Equating the two expressions for $$c$$ we obtain two possibilities:

$$\frac{1}{m}=-3m+3\sqrt{m^{2}+1}\qquad\text{or}\qquad \frac{1}{m}=-3m-3\sqrt{m^{2}+1}\;.$$

First possibility

$$\frac{1}{m}=-3m+3\sqrt{m^{2}+1}\;.$$

Multiplying by $$m$$ gives

$$1=-3m^{2}+3m\sqrt{m^{2}+1}\;.$$

Dividing by 3,

$$\frac13=-m^{2}+m\sqrt{m^{2}+1}\;.$$

Let $$t=m^{2}$$ (so $$t\ge0$$). Then $$m\sqrt{m^{2}+1}=m\sqrt{t+1}=\sqrt{t}\sqrt{t+1}$$, and the equation becomes

$$\sqrt{t}\sqrt{t+1}=t+\frac13\;.$$

Squaring both sides,

$$t(t+1)=\left(t+\frac13\right)^{2} \;\Longrightarrow\; t^{2}+t=t^{2}+\frac23t+\frac19\;.$$

Subtracting $$t^{2}$$ from each side,

$$t=\frac23t+\frac19 \;\Longrightarrow\; t-\frac23t=\frac19 \;\Longrightarrow\; \frac13t=\frac19 \;\Longrightarrow\; t=\frac13\;.$$

Hence $$m^{2}=\dfrac13\;\Longrightarrow\;m=\pm\frac{1}{\sqrt3}\;.$$

Substituting $$m=\frac1{\sqrt3}$$ into (1) (with the ‘+’ sign we are using) to find $$c$$,

$$c=-3\!\left(\frac1{\sqrt3}\right)+3\sqrt{\frac13+1} =-\sqrt3+3\sqrt{\frac43} =-\sqrt3+3\cdot\frac{2}{\sqrt3} =-\sqrt3+\frac{6}{\sqrt3} =\frac{-3+6}{\sqrt3} =\sqrt3\;.$$

So one tangent is

$$y=\frac1{\sqrt3}x+\sqrt3 \;\Longrightarrow\; \sqrt3\,y=x+3\;.$$

Taking $$m=-\dfrac1{\sqrt3}$$ in the same branch does not satisfy the equation (check gives $$-\,\sqrt3\neq3\sqrt3$$), hence only $$m=+\dfrac1{\sqrt3}$$ works here.

Second possibility

$$\frac{1}{m}=-3m-3\sqrt{m^{2}+1}\;.$$

Repeating exactly the same algebra (or directly substituting $$m=-\dfrac1{\sqrt3}$$) yields another valid tangent

$$\sqrt3\,y=-x-3\;.$$

Although two common tangents exist, only the first one appears among the given options.

The required common tangent from the list is therefore

$$\boxed{\;\sqrt3\,y=x+3\;}\;.$$

Hence, the correct answer is Option 2.

We have two parabolas

$$y^{2}=4b\,(x-c)\qquad\text{and}\qquad y^{2}=8ax.$$

To decide whether a common normal exists, we write each curve in its standard parametric form, find the slope of the normal at a general point and then force the two normals to coincide.

For the first parabola, compare $$y^{2}=4bX$$ (whose parameter is generally written as $$t$$). A mere translation $$X\to x-c$$ gives

$$x=c+bt^{2},\qquad y=2bt.$$

Differentiating with respect to $$t$$ gives

$$\frac{dx}{dt}=2bt,\qquad\frac{dy}{dt}=2b.$$

Hence the slope of the tangent is

$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2b}{2bt}=\frac1t,$$

so the slope of the normal is the negative reciprocal, namely

$$m_{1,\text{normal}}=-t.$$

Therefore the normal at $$(c+bt^{2},\,2bt)$$ is

$$y-2bt=-t\bigl(x-(c+bt^{2})\bigr).$$

For the second parabola, write it as $$y^{2}=4(2a)x.$$ With parameter $$s$$ we have

$$x=2a s^{2},\qquad y=4a s,$$

and again

$$\frac{dx}{ds}=4as,\qquad\frac{dy}{ds}=4a.$$

Thus

$$\frac{dy}{dx}=\frac{4a}{4as}=\frac1s,\qquad m_{2,\text{normal}}=-s.$$

The normal at $$(2a s^{2},\,4a s)$$ reads

$$y-4a s=-s\bigl(x-2a s^{2}\bigr).$$

Because the two normals must be the same line, their slopes must be equal, i.e.

$$-t=-s\;\Longrightarrow\;t=s=k\;(\text{say}).$$

Writing both normals with this common parameter $$k$$ gives

$$y-2bk=-k\bigl(x-(c+bk^{2})\bigr)\quad\text{and}\quad y-4ak=-k\bigl(x-2ak^{2}\bigr).$$

Putting both in the form $$y=-kx+\text{constant},$$ and equating the constants, we obtain

$$k(c+b k^{2})+2bk=k(2a k^{2})+4ak.$$

For $$k\neq0$$ we may divide by $$k$$ to get a linear equation in $$k^{2}$$:

$$\bigl(b-2a\bigr)k^{2}+c+2b-4a=0.$$

Rewriting,

$$k^{2}=\frac{4a-c-2b}{\,b-2a\,}. \quad -(★)$$

For a real common normal with $$k\neq0$$ we need the right-hand side to be non-negative. Let us test each proposed ordered triple $$(a,b,c)$$.

Option A: $$(a,b,c)=(1,1,3)$$.

$$4a-c-2b=4-3-2=-1,\qquad b-2a=1-2=-1,$$

so $$k^{2}=\frac{-1}{-1}=1\ge0.$$ A real $$k=\pm1$$ exists, hence a non-horizontal common normal is present.

Option B: $$(a,b,c)=\bigl(\tfrac12,\,2,\,0\bigr)$$.

$$4a-c-2b=2-0-4=-2,\qquad b-2a=2-1=1,$$

so $$k^{2}=\frac{-2}{1}=-2\lt 0,$$ giving no non-zero $$k$$. However, we must still examine the special case $$k=0$$ which we excluded by division. For $$k=0$$ the normal has slope $$0$$, i.e. the $$x$$-axis $$y=0$$.

• On the first parabola, $$t=0$$ gives the point $$(c,0)$$ and a horizontal normal $$y=0$$.

• On the second parabola, $$s=0$$ gives $$(0,0)$$ and again the horizontal normal $$y=0$$.

Thus $$y=0$$ is a common normal, so Option B is valid.

Option C: $$(a,b,c)=\bigl(\tfrac12,\,2,\,3\bigr)$$.

Exactly the same calculation as in Option B yields $$k^{2}=-2\lt 0$$ and hence no non-zero $$k$$, but $$k=0$$ once more produces the common normal $$y=0$$ (now passing through $$(3,0)$$ on the first parabola and $$(0,0)$$ on the second). Therefore Option C is also valid.

Since every individual choice A, B and C supplies at least one real common normal, all of them are correct simultaneously.

Hence, the correct answer is Option D.

We are given the parabola $$y^{2}=4x$$ and the fixed point of contact $$(1,2)$$. A circle touching the x-axis must have its centre directly above (or below) the point of tangency on the x-axis, so if the centre is $$(h,k)$$ and the radius is $$r$$, we must have $$k=r$$ (because the vertical distance from the centre to the x-axis equals the radius).

The circle also passes through $$(1,2)$$, hence we must satisfy the distance formula

$$ (1-h)^{2}+(2-k)^{2}=r^{2}. \quad -(1) $$

To ensure tangency with the parabola at $$(1,2)$$, the circle and the parabola must share the same tangent line there. Let us first find the tangent to the parabola at that point.

Differentiating $$y^{2}=4x$$ with respect to $$x$$ gives $$ 2y\frac{dy}{dx}=4 \quad\Longrightarrow\quad \frac{dy}{dx}=\frac{2}{y}. $$ At $$(1,2)$$ we have $$\displaystyle\frac{dy}{dx}=\frac{2}{2}=1,$$ so the tangent line is $$ y-2 = 1(x-1)\;\Longrightarrow\; y = x+1. $$

For any circle, the radius drawn to the point of tangency is perpendicular to the tangent line. The slope of the tangent is $$1$$, hence the slope of the radius is $$-1.$$ Therefore

$$ \frac{2-k}{\,1-h\,} = -1. \quad -(2) $$

Because $$k=r$$, equation (2) becomes $$ 2-r = -(1-h) \; \Longrightarrow \; 2-r = -1 + h \; \Longrightarrow \; h+r = 3. \quad -(3) $$

Now substitute $$k=r$$ and $$h = 3-r$$ from (3) into the distance condition (1):

$$ (1-(3-r))^{2} + (2-r)^{2} = r^{2}. $$

Simplifying each term, we note that $$1-(3-r)=r-2,$$ hence $$ (r-2)^{2} + (2-r)^{2} = r^{2}. $$ But $$(2-r)^{2}=(r-2)^{2},$$ so we obtain $$ 2(r-2)^{2}=r^{2}. $$

Expanding and collecting terms,

$$ 2(r^{2}-4r+4)=r^{2} \quad\Longrightarrow\quad 2r^{2}-8r+8=r^{2}. $$

Bringing everything to one side,

$$ r^{2}-8r+8=0. $$

This quadratic equation gives $$ r = \frac{8\pm\sqrt{64-32}}{2} = \frac{8\pm4\sqrt{2}}{2} = 4 \pm 2\sqrt{2}. $$

Both roots are positive, so two circles exist. The smaller circle has

$$ r = 4-2\sqrt{2}. $$

Its area is therefore $$ \text{Area} = \pi r^{2} = \pi\bigl(4-2\sqrt{2}\bigr)^{2}. $$

Expanding the square, $$ (4-2\sqrt{2})^{2}=4^{2}-2\!\times\!4\!\times\!2\sqrt{2}+(2\sqrt{2})^{2}=16-16\sqrt{2}+8=24-16\sqrt{2}. $$

Factoring the common $$8$$ we have $$ 24-16\sqrt{2}=8(3-2\sqrt{2}). $$

Hence $$ \text{Area}= \pi \times 8(3-2\sqrt{2}) = 8\pi(3-2\sqrt{2}). $$

Hence, the correct answer is Option A.

We begin with the curve $$y = (x - 2)^2 - 1$$ and the straight line $$x - y = 3$$. To locate their points of intersection, we substitute the expression for $$y$$ from the line into the curve.

The line gives $$y = x - 3$$. Substituting this in the curve’s equation, we obtain

$$x - 3 = (x - 2)^2 - 1.$$

Expanding the square on the right side,

$$(x - 2)^2 = x^2 - 4x + 4,$$

so the equation becomes

$$x - 3 = x^2 - 4x + 4 - 1.$$

Simplifying the right side, we have

$$x - 3 = x^2 - 4x + 3.$$

Now, bringing every term to the right,

$$0 = x^2 - 4x + 3 - x + 3 = x^2 - 5x + 6.$$

Factoring the quadratic,

$$x^2 - 5x + 6 = (x - 2)(x - 3) = 0.$$

This gives two real solutions:

$$x = 2 \quad \text{or} \quad x = 3.$$

Substituting back in $$y = x - 3:$$

If $$x = 2,$$ then $$y = 2 - 3 = -1,$$ giving the point $$P_1(2,\,-1).$$

If $$x = 3,$$ then $$y = 3 - 3 = 0,$$ giving the point $$P_2(3,\,0).$$

Next, we need the tangents to the curve at these points. For that, we first compute the derivative of the curve.

The curve is $$y = (x - 2)^2 - 1.$$ Using the power rule,

$$\frac{dy}{dx} = 2(x - 2).$$

At $$P_1(2,\,-1),$$ the slope is

$$m_1 = 2(2 - 2) = 2 \cdot 0 = 0.$$

Thus the tangent at $$P_1$$ is a horizontal line passing through $$y = -1$$, i.e.

$$y = -1.$$

At $$P_2(3,\,0),$$ the slope is

$$m_2 = 2(3 - 2) = 2 \cdot 1 = 2.$$

Using point-slope form $$y - y_0 = m(x - x_0),$$ we write the tangent at $$P_2$$ as

$$y - 0 = 2(x - 3) \quad\Longrightarrow\quad y = 2x - 6.$$

The final task is to find the intersection of these two tangents. One tangent is $$y = -1$$ and the other is $$y = 2x - 6.$$ Setting them equal, we get

$$-1 = 2x - 6.$$

Adding $$6$$ to both sides,

$$5 = 2x,$$

and dividing by $$2,$$

$$x = \frac{5}{2}.$$

Substituting this back into $$y = -1$$ (the simpler of the two tangents) immediately gives

$$y = -1.$$

Thus the tangents intersect at the point $$\left(\dfrac{5}{2},\, -1\right).$$

Hence, the correct answer is Option B.

We have the standard parabola $$y^{2}=4x$$. A convenient way to describe every point on this curve is to put $$y=2t$$. Substituting this into the equation of the parabola gives $$x=\dfrac{y^{2}}{4}=\dfrac{(2t)^{2}}{4}=t^{2}$$. Hence any point on the parabola can be written in the parametric form $$\bigl(t^{2},\,2t\bigr)$$, where $$t$$ is a real parameter.

The fixed points are $$A(4,-4)\quad\text{and}\quad B(9,6).$$ Comparing with the parametric form, $$A(4,-4)\;\Longrightarrow\;4=t^{2},\; -4=2t\;\Longrightarrow\;t=-2,$$ $$B(9,6)\;\Longrightarrow\;9=t^{2},\; 6=2t\;\Longrightarrow\;t=3.$$ Thus $$t=-2$$ corresponds to $$A$$ and $$t=3$$ corresponds to $$B$$. The point $$O(0,0)$$ corresponds to $$t=0$$. The arc $$AOB$$ therefore traces the parameter $$t$$ from $$-2$$ through $$0$$ up to $$3$$.

Let us choose a variable point $$C\bigl(t^{2},\,2t\bigr)$$ with $$-2\le t\le 3$$ on this arc. Our goal is to maximise the area of $$\triangle ACB$$.

For three points $$A(x_{1},y_{1}),\;B(x_{2},y_{2}),\;C(x_{3},y_{3})$$ the area formula is $$\text{Area}=\dfrac12\bigl|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\bigr|.$$ We first write down the coordinates explicitly:

$$\begin{aligned} A&:(x_{1},y_{1})=(4,-4),\\ B&:(x_{2},y_{2})=(9,6),\\ C&:(x_{3},y_{3})=(t^{2},2t). \end{aligned}$$ Substituting all these into the formula gives $$\begin{aligned} 2\,\text{Area}&=\bigl|\,4\,(6-2t)+9\,(2t-(-4))+t^{2}\,(-4-6)\bigr|\\[4pt] &=\bigl|\,4(6-2t)+9(2t+4)+t^{2}(-10)\bigr|. \end{aligned}$$

We now simplify each term one by one:

$$\begin{aligned} 4(6-2t)&=24-8t,\\ 9(2t+4)&=18t+36,\\ t^{2}(-10)&=-10t^{2}. \end{aligned}$$ Adding them, $$ 24-8t+18t+36-10t^{2}=60+10t-10t^{2}. $$ Hence $$ 2\,\text{Area}=|\,60+10t-10t^{2}\,|=10\,|\, -t^{2}+t+6\,|. $$

Inside the modulus we have the quadratic $$f(t)=-t^{2}+t+6.$$ Let us analyse its sign on the interval $$-2\le t\le 3$$. At the end-points,

$$\begin{aligned} f(-2)&=-(-2)^{2}+(-2)+6=-4-2+6=0,\\ f(3)&=-3^{2}+3+6=-9+3+6=0. \end{aligned}$$

The quadratic opens downwards (coefficient of $$t^{2}$$ is negative), so between the roots its value is positive. Therefore $$f(t)\ge0$$ throughout the required interval, and the modulus may be dropped:

$$2\,\text{Area}=10\bigl(-t^{2}+t+6\bigr),$$ $$\text{Area}=5\bigl(-t^{2}+t+6\bigr).$$ Call the bracketed expression $$g(t)=-t^{2}+t+6$$, so $$\text{Area}=5\,g(t).$$ To maximise the area we simply have to maximise $$g(t)$$.

The quadratic $$g(t)=-t^{2}+t+6$$ has its vertex at $$t=-\dfrac{b}{2a}=-\dfrac{1}{2(-1)}=\dfrac12,$$ using the vertex formula $$t=-\dfrac{b}{2a}$$ for $$at^{2}+bt+c$$. This value $$t=\dfrac12$$ indeed lies between $$-2$$ and $$3$$, so it gives the maximum.

Evaluating $$g(t)$$ at this point, $$\begin{aligned} g\!\left(\dfrac12\right)&=-\left(\dfrac12\right)^{2}+\dfrac12+6\\[4pt] &=-\dfrac14+\dfrac12+6\\[4pt] &=6.25. \end{aligned}$$ Finally the maximal area is $$ \text{Area}_{\max}=5\times6.25=31.25=\dfrac{125}{4}. $$

Expressed as a mixed number, $$31.25=31\dfrac14$$.

Hence, the correct answer is Option D.

We have the standard parabola $$y^{2}=4x$$ whose vertex is the origin $$O(0,0)$$.

A convenient way to handle points on this parabola is to use the parameter $$t$$ defined by the relations

$$x=t^{2},\qquad y=2t \qquad(\text{because }(2t)^{2}=4t^{2}=4x).$$

First we identify the parameters corresponding to the fixed points $$P(4,-4)$$ and $$Q(9,6)$$.

For $$P(4,-4):\; y=2t=-4\;\Rightarrow\; t=-2,\; x=t^{2}=(-2)^{2}=4\;(\text{checks}).$$

For $$Q(9,6):\; y=2t=6\;\Rightarrow\; t=3,\; x=t^{2}=3^{2}=9\;(\text{checks}).$$

Thus as the point $$X$$ moves along the arc $$POQ$$, its parameter $$t$$ varies in the closed interval $$[-2,3]$$ and its coordinates are

$$X\,(x,y)=\bigl(t^{2},\,2t\bigr).$$

Next we require the area of the triangle $$\Delta PXQ$$. For three points $$(x_{1},y_{1}),\,(x_{2},y_{2}),\,(x_{3},y_{3})$$ the area formula is

$$\text{Area}=\dfrac12\Bigl|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\Bigr|.$$

Taking $$(x_{1},y_{1})=(4,-4),\;(x_{2},y_{2})=(t^{2},2t),\;(x_{3},y_{3})=(9,6),$$ we substitute:

$$\begin{aligned} \text{Twice the signed area}&=4\bigl(2t-6\bigr)+t^{2}\bigl(6-(-4)\bigr)+9\bigl(-4-2t\bigr)\\[4pt] &=4(2t-6)+t^{2}(10)+9(-4-2t)\\[4pt] &=(8t-24)+10t^{2}+(-36-18t)\\[4pt] &=10t^{2}-10t-60. \end{aligned}$$

Hence the actual area is

$$A(t)=\dfrac12\,\Bigl|10t^{2}-10t-60\Bigr|=5\,\Bigl|t^{2}-t-6\Bigr|.$$

Notice that

$$t^{2}-t-6=(t-3)(t+2).$$

For $$t\in[-2,3]$$ we have $$(t+2)\ge0\text{ and }(t-3)\le0,$$ so the product is non-positive and the absolute value simply changes its sign:

$$A(t)=5\bigl(-(t-3)(t+2)\bigr)=5\,(3-t)(t+2).$$

Expanding the bracket gives

$$A(t)=5\Bigl(-t^{2}+t+6\Bigr).$$

To maximise $$A(t)$$ we consider it as a quadratic function of $$t$$. Writing

$$A(t)=5\bigl(-t^{2}+t+6\bigr)$$

the coefficient of $$t^{2}$$ is negative, so the maximum occurs at the vertex, where

$$t=-\dfrac{b}{2a}=-\dfrac{1}{2(-1)}=\dfrac12.$$

This critical value $$t=\dfrac12$$ indeed lies in the interval $$[-2,3]$$. We now compute the corresponding maximum area:

$$\begin{aligned} A_{\max}&=5\,(3-\tfrac12)\,(\tfrac12+2)\\[4pt] &=5\left(\dfrac{5}{2}\right)\left(\dfrac{5}{2}\right)\\[4pt] &=5\cdot\dfrac{25}{4}=\dfrac{125}{4}\;\text{square units}. \end{aligned}$$

Hence, the correct answer is Option C.

First recall the standard facts about the given curves.

The parabola is $$y^{2}=4x$$. Comparing this with the general form $$y^{2}=4ax$$, we see that the parameter is $$a=1$$.

For a parabola $$y^{2}=4ax$$, the equation of a tangent having slope $$m$$ is given by the slope-form formula

$$y=mx+\dfrac{a}{m}.$$

Stating the value of $$a$$ and substituting it, we get

$$y = mx + \dfrac{1}{m} \quad\text{(1)}$$

This straight line touches the parabola by construction. We now require the same line to be a tangent to the hyperbola $$xy = 2$$ as well.

To impose this condition, we substitute the expression for $$y$$ from equation (1) into the hyperbola’s equation:

$$x\left(mx+\dfrac{1}{m}\right)=2.$$

Multiplying out, we obtain a quadratic in $$x$$:

$$m x^{2} + \dfrac{1}{m}\,x - 2 = 0.$$

For the line to be tangent to the hyperbola, this quadratic must possess exactly one real root. The condition for a quadratic $$Ax^{2}+Bx+C=0$$ to have equal (coincident) roots is that its discriminant vanishes, that is

$$B^{2}-4AC = 0.$$

Here $$A = m,\; B = \dfrac{1}{m},\; C = -2.$$ Substituting these values, we write the discriminant:

$$\left(\dfrac{1}{m}\right)^{2} - 4\,(m)\,(-2) = 0.$$

Simplifying term by term,

$$\dfrac{1}{m^{2}} + 8m = 0.$$

To clear the fraction, multiply the entire equation by $$m^{2}$$:

$$1 + 8m^{3} = 0.$$

Thus

$$8m^{3} = -1 \quad\Longrightarrow\quad m^{3} = -\dfrac{1}{8}.$$

Taking the real cube root on both sides, we find

$$m = -\dfrac{1}{2}.$$

Now we substitute this value of $$m$$ back into the tangent equation (1):

$$y = \left(-\dfrac{1}{2}\right)x + \dfrac{1}{\, -\dfrac{1}{2}\,}.$$

Calculating the constant term,

$$\dfrac{1}{\, -\dfrac{1}{2}\,} = -2,$$

so the equation becomes

$$y = -\dfrac{1}{2}x - 2.$$

To express this in the usual linear form, multiply through by $$2$$:

$$2y = -x - 4.$$

Bringing all terms to the left side gives

$$x + 2y + 4 = 0.$$

This matches Option C.

Hence, the correct answer is Option C.

We are asked to find that point $$P(x,y)$$ on the parabola $$x^{2}=4y$$ which is nearest to the centre of the circle $$x^{2}+y^{2}+6x+8=0$$, and then to write the equation of the tangent to the parabola at that particular point.

First we rewrite the equation of the circle in its standard (centre-radius) form. We have

$$x^{2}+y^{2}+6x+8=0.$$

To complete the square in $$x$$ we add and subtract $$9$$ (because $$(x+3)^{2}=x^{2}+6x+9$$). So

$$x^{2}+6x+9+y^{2}= -8+9,$$

which becomes

$$(x+3)^{2}+y^{2}=1.$$

Hence the centre of the circle is $$C(-3,0)$$ and its radius is $$r=1$$ (though we shall actually need only the centre).

Now we choose a general point $$P$$ on the parabola. For the parabola $$x^{2}=4y$$ it is convenient to take

$$x=t,\qquad y=\dfrac{t^{2}}{4},$$

where $$t$$ is a real parameter. Thus

$$P(t,\;t^{2}/4).$$

The squared distance between $$P$$ and the fixed point $$C(-3,0)$$ is

$$D^{2}=(x+3)^{2}+(y-0)^{2}.$$

Substituting $$x=t$$ and $$y=\dfrac{t^{2}}{4}$$ gives

$$D^{2}=(t+3)^{2}+\left(\dfrac{t^{2}}{4}\right)^{2}.$$

Let us expand and tidy this expression because minimising $$D^{2}$$ is the same as minimising $$D$$ itself.

First expand $$(t+3)^{2}$$:

$$(t+3)^{2}=t^{2}+6t+9.$$

Next, square the second term:

$$\left(\dfrac{t^{2}}{4}\right)^{2}=\dfrac{t^{4}}{16}.$$

Therefore

$$D^{2}=t^{2}+6t+9+\dfrac{t^{4}}{16}.$$

Define

$$f(t)=\dfrac{t^{4}}{16}+t^{2}+6t+9.$$

We differentiate $$f(t)$$ with respect to $$t$$ and set the derivative to zero to locate the critical point. Using the power rule $$\dfrac{d}{dt}(t^{n})=nt^{n-1}$$, we obtain

$$\frac{d}{dt}\!\left(\dfrac{t^{4}}{16}\right)=\dfrac{4t^{3}}{16}=\dfrac{t^{3}}{4},$$ $$\frac{d}{dt}(t^{2})=2t,$$ $$\frac{d}{dt}(6t)=6,$$ and of course the derivative of the constant $$9$$ is $$0$$.

Hence

$$f'(t)=\dfrac{t^{3}}{4}+2t+6.$$

Setting $$f'(t)=0$$ gives

$$\dfrac{t^{3}}{4}+2t+6=0.$$

Multiply through by $$4$$ (to clear the denominator) and we have

$$t^{3}+8t+24=0.$$

To solve this cubic we look for simple integral roots. Try $$t=-2$$:

$$(-2)^{3}+8(-2)+24=-8-16+24=0.$$

Because $$t=-2$$ satisfies the equation, $$(t+2)$$ is a factor. Carrying out polynomial division (or synthetic division) gives

$$t^{3}+8t+24=(t+2)\bigl(t^{2}-2t+12\bigr).$$

The quadratic factor $$t^{2}-2t+12$$ has discriminant $$\Delta=(-2)^{2}-4(1)(12)=4-48=-44<0,$$ so it yields no real roots. Therefore the only real critical point is

$$t=-2.$$

This critical point indeed minimises $$f(t)$$ (and hence $$D^{2}$$), so the required point $$P$$ is

$$x=-2,\qquad y=\dfrac{(-2)^{2}}{4}=\dfrac{4}{4}=1.$$

Thus

$$P(-2,1).$$

Next we need the tangent to the parabola $$x^{2}=4y$$ at this point. First we compute the slope using calculus. Differentiating the parabola implicitly with respect to $$x$$ gives

$$\frac{d}{dx}(x^{2})=\frac{d}{dx}(4y).$$

Because $$\frac{d}{dx}(x^{2})=2x$$ and $$\frac{d}{dx}(4y)=4\frac{dy}{dx},$$ we obtain

$$2x=4\frac{dy}{dx}.$$

Hence

$$\frac{dy}{dx}=\frac{x}{2}.$$

At the point $$(-2,1)$$ the slope is

$$m=\frac{-2}{2}=-1.$$

Using the point-slope form of a straight line, $$y-y_{1}=m(x-x_{1}),$$ with $$(x_{1},y_{1})=(-2,1)$$ and $$m=-1,$$ we write

$$y-1=-1\,(x+2).$$

Expand the right‐hand side:

$$y-1=-x-2.$$

Now bring all terms to one side:

$$x+y+1=0.$$

Thus the tangent to the parabola at the nearest point is

$$x+y+1=0.$$

Comparing with the options, this corresponds to Option A.

Hence, the correct answer is Option A.

The parabola is $$y^{2}=16x$$. Comparing with the standard form $$y^{2}=4ax$$ we have $$4a=16\; \Rightarrow\; a=4$$. Its axis is the $$x$$-axis, i.e. the line $$y=0$$.

The given point is $$P(16,16)$$. First we confirm that it lies on the parabola:

$$y^{2}=16^{2}=256,\qquad 16x=16\cdot16=256;\qquad\text{hence }P\text{ is on the curve.}$$

We now require the equations of the tangent and the normal at $$P$$.

For $$y^{2}=16x$$ we differentiate implicitly:

$$2y\,\dfrac{dy}{dx}=16\quad\Longrightarrow\quad\dfrac{dy}{dx}=\dfrac{8}{y}.$$

At $$P(16,16)$$ the slope of the tangent is

$$m_{t}=\dfrac{8}{16}=\dfrac12.$$

The tangent therefore has the equation

$$y-16=\dfrac12\,(x-16).$$

To find its intersection with the axis $$y=0$$ we put $$y=0$$:

$$0-16=\dfrac12\,(x-16)\;\Longrightarrow\;-16=\dfrac{x-16}{2}$$

$$\Longrightarrow\;-32=x-16\;\Longrightarrow\;x=-16.$$

Thus the tangent meets the axis at $$A(-16,0).$$

The slope of the normal, being the negative reciprocal of the tangent’s slope, is

$$m_{n}=-\dfrac{1}{m_{t}}=-\dfrac{1}{\tfrac12}=-2.$$

The normal passes through $$P(16,16)$$, so

$$y-16=-2\,(x-16).$$

Setting $$y=0$$ to locate its intersection with the axis:

$$0-16=-2\,(x-16)\;\Longrightarrow\;-16=-2x+32$$

$$\Longrightarrow\;-48=-2x\;\Longrightarrow\;x=24.$$

Hence the normal meets the axis at $$B(24,0).$$

We now have the three points

$$A(-16,0),\; B(24,0),\; P(16,16).$$

Let $$C(h,k)$$ be the centre of the circle through $$A,B,P$$. Because $$A$$ and $$B$$ lie on the $$x$$-axis and are symmetric with respect to the centre, the perpendicular bisector of $$AB$$ is vertical.

The midpoint of $$AB$$ is

$$M_{1}\left(\dfrac{-16+24}{2},\dfrac{0+0}{2}\right)=(4,0).$$

Since $$AB$$ is horizontal, its perpendicular bisector is the vertical line $$x=4$$, so $$h=4$$. Thus $$C=(4,k).$$

Next we take the perpendicular bisector of $$BP$$ to determine $$k$$. The midpoint of $$BP$$ is

$$M_{2}\left(\dfrac{24+16}{2},\dfrac{0+16}{2}\right)=(20,8).$$

The slope of $$BP$$ is

$$m_{BP}=\dfrac{0-16}{24-16}=-\dfrac{16}{8}=-2,$$

so the slope of its perpendicular bisector is $$\dfrac12$$.

The equation of this bisector is therefore

$$y-8=\dfrac12\,(x-20).$$

Substituting $$x=4$$ into this equation (because $$C$$ lies on both bisectors) gives

$$y-8=\dfrac12\,(4-20)=\dfrac12(-16)=-8 \;\Longrightarrow\;y=0.$$

Hence $$k=0$$ and the centre is

$$C(4,0).$$

We now compute the angle $$\theta=\angle CPB$$, i.e. the angle at $$P$$ formed by the segments $$PC$$ and $$PB$$. For this we form their direction vectors:

$$\overrightarrow{PC}=C-P=(4-16,\,0-16)=(-12,\,-16),$$

$$\overrightarrow{PB}=B-P=(24-16,\,0-16)=(8,\,-16).$$

Using the dot-product formula for the angle between two vectors,

$$\cos\theta=\dfrac{\overrightarrow{PC}\cdot\overrightarrow{PB}} {|\overrightarrow{PC}|\;|\overrightarrow{PB}|}.$$

The dot product is

$$\overrightarrow{PC}\cdot\overrightarrow{PB}=(-12)(8)+(-16)(-16) =-96+256=160.$$

The magnitudes are

$$|\overrightarrow{PC}|=\sqrt{(-12)^{2}+(-16)^{2}} =\sqrt{144+256}=20,$$

$$|\overrightarrow{PB}|=\sqrt{8^{2}+(-16)^{2}} =\sqrt{64+256}=8\sqrt5.$$

Therefore

$$\cos\theta=\dfrac{160}{20\cdot8\sqrt5} =\dfrac{160}{160\sqrt5} =\dfrac1{\sqrt5}.$$

Next, using $$\sin^{2}\theta=1-\cos^{2}\theta$$ we get

$$\sin\theta=\sqrt{1-\dfrac1{5}} =\sqrt{\dfrac{4}{5}} =\dfrac{2}{\sqrt5}.$$

Finally,

$$\tan\theta=\dfrac{\sin\theta}{\cos\theta} =\dfrac{\dfrac{2}{\sqrt5}}{\dfrac1{\sqrt5}} =2.$$

Thus an admissible value of $$\tan\theta$$ is $$2$$.

Hence, the correct answer is Option C.

The two parabolas have a common vertex at the origin, with axes along the x-axis and y-axis respectively, and both have latus rectum of length 3.

For a parabola $$y^2 = 4ax$$, the latus rectum is $$4a$$. Setting $$4a = 3$$ gives $$a = \frac{3}{4}$$, so the first parabola is $$y^2 = 3x$$.

For a parabola $$x^2 = 4by$$, the latus rectum is $$4b$$. Setting $$4b = 3$$ gives $$b = \frac{3}{4}$$, so the second parabola is $$x^2 = 3y$$.

The equation of a tangent to $$y^2 = 3x$$ in slope form is $$y = mx + \frac{3}{4m}$$ (using the standard tangent $$y = mx + \frac{a}{m}$$ with $$a = \frac{3}{4}$$).

For this line to also be tangent to $$x^2 = 3y$$, substitute $$y = mx + \frac{3}{4m}$$ into $$x^2 = 3y$$: $$x^2 = 3mx + \frac{9}{4m}$$, i.e., $$x^2 - 3mx - \frac{9}{4m} = 0$$.

For tangency, the discriminant must be zero: $$\Delta = 9m^2 + 4 \cdot \frac{9}{4m} = 0$$, which gives $$9m^2 + \frac{9}{m} = 0$$.

Multiplying through by $$m$$: $$9m^3 + 9 = 0$$, so $$m^3 = -1$$, giving $$m = -1$$.

Substituting $$m = -1$$ back into the tangent equation: $$y = -x + \frac{3}{4(-1)} = -x - \frac{3}{4}$$.

Multiplying both sides by 4: $$4y = -4x - 3$$, which gives $$4x + 4y + 3 = 0$$, or equivalently $$4(x + y) + 3 = 0$$.

The correct answer is Option D: $$4(x + y) + 3 = 0$$.

We are given the parabola $$y^{2}=8x$$. A standard result is that any parabola of the form $$y^{2}=4ax$$ has its focus at $$(a,0)$$. Here, $$4a=8$$, so $$a=2$$ and therefore the focus is

$$F(2,0).$$

For the same parabola, a convenient parametric representation of points on the curve is

$$\bigl(at^{2},\,2at\bigr),$$

where $$t$$ is a real parameter. Substituting $$a=2$$ gives every point on the curve as

$$\bigl(2t^{2},\,4t\bigr).$$

A standard formula for the tangent to $$y^{2}=4ax$$ at the parametric point $$(at^{2},2at)$$ is

$$ty = x + at^{2}.$$

Putting $$a=2$$, the tangent at the point corresponding to the parameter $$t$$ is

$$ty = x + 2t^{2}. \quad -(1)$$

We are told that this tangent must pass through the external point $$(-8,0)$$. Hence we substitute $$x=-8$$ and $$y=0$$ into (1):

$$t\cdot 0 = -8 + 2t^{2}.$$

This simplifies step by step as

$$0 = -8 + 2t^{2},$$

$$2t^{2}=8,$$

$$t^{2}=4,$$

$$t=\pm 2.$$

Thus the two tangents from $$(-8,0)$$ touch the parabola at the parameters

$$t_{1}=2,\qquad t_{2}=-2.$$

Let $$P$$ correspond to $$t=2$$ and $$Q$$ to $$t=-2$$. Using the parametric form $$(2t^{2},4t)$$ we obtain

For $$t=2$$:

$$P\;=\;\bigl(2(2)^{2},\,4(2)\bigr)=\bigl(2\cdot 4,\,8\bigr)=(8,8).$$

For $$t=-2$$:

$$Q\;=\;\bigl(2(-2)^{2},\,4(-2)\bigr)=\bigl(2\cdot 4,\,-8\bigr)=(8,-8).$$

We now have the three vertices of the triangle:

$$P(8,8),\quad F(2,0),\quad Q(8,-8).$$

To find the area of triangle $$PFQ$$ we use the determinant (shoelace) formula:

For points $$(x_{1},y_{1}), (x_{2},y_{2}), (x_{3},y_{3}),$$ the area is

$$\text{Area}=\dfrac12\left|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right|.$$

Assigning $$(x_{1},y_{1})=(8,8),\;(x_{2},y_{2})=(2,0),\;(x_{3},y_{3})=(8,-8),$$ we substitute:

$$\begin{aligned} \text{Area}&=\dfrac12\Bigl|\,8\bigl(0-(-8)\bigr)+2\bigl((-8)-8\bigr)+8\bigl(8-0\bigr)\Bigr|\\[4pt] &=\dfrac12\Bigl|\,8(8)+2(-16)+8(8)\Bigr|\\[4pt] &=\dfrac12\Bigl|\,64-32+64\Bigr|\\[4pt] &=\dfrac12\bigl|\,96\bigr|\\[4pt] &=\dfrac{96}{2}=48. \end{aligned}$$

Hence, the correct answer is Option A.

We have two curves, the parabola $$y^{2}=6x$$ and the conic $$9x^{2}+by^{2}=16$$. To find the condition for them to cut at right angles we shall first obtain the slopes of their tangents (i.e. $$\dfrac{dy}{dx}$$) at any common point and then use the fact that for two curves to intersect orthogonally, the product of their slopes at the point of intersection must be $$-1$$.

Step 1 : Slope of the tangent to the parabola.
Starting with $$y^{2}=6x$$, we differentiate implicitly with respect to $$x$$. Using the rule $$\dfrac{d}{dx}(y^{2})=2y\dfrac{dy}{dx}$$ and $$\dfrac{d}{dx}(6x)=6$$, we get $$2y\dfrac{dy}{dx}=6.$$ Solving for $$\dfrac{dy}{dx}$$, $$\dfrac{dy}{dx}=\dfrac{6}{2y}=\dfrac{3}{y}.$$ So the slope of the tangent to the first curve is $$m_{1}=\dfrac{3}{y}.$$

Step 2 : Slope of the tangent to the second curve.
For $$9x^{2}+by^{2}=16$$ we again differentiate implicitly with respect to $$x$$. Using $$\dfrac{d}{dx}(9x^{2})=18x$$ and $$\dfrac{d}{dx}(by^{2})=2by\dfrac{dy}{dx}$$, we have $$18x+2by\dfrac{dy}{dx}=0.$$ Isolating $$\dfrac{dy}{dx}$$ gives $$2by\dfrac{dy}{dx}=-18x$$ $$\dfrac{dy}{dx}=\dfrac{-18x}{2by}=\dfrac{-9x}{by}.$$ Thus the slope of the tangent to the second curve is $$m_{2}=\dfrac{-9x}{by}.$$

Step 3 : Condition for orthogonality.
For two curves to meet at right angles we must have $$m_{1}\,m_{2}=-1.$$ Substituting the expressions of $$m_{1}$$ and $$m_{2}$$ obtained above, $$\left(\dfrac{3}{y}\right)\left(\dfrac{-9x}{by}\right)=-1.$$ Multiplying the numerators and denominators, $$\dfrac{-27x}{by^{2}}=-1.$$ Cancelling the negative sign on both sides, $$\dfrac{27x}{by^{2}}=1.$$ Cross-multiplying, $$27x=by^{2}.$$

Step 4 : Eliminating $$y^{2}$$ using the first curve.
From the parabola we already have $$y^{2}=6x.$$ Substituting this value of $$y^{2}$$ into the relation $$27x=by^{2}$$, we obtain $$27x=b(6x).$$ Dividing both sides by $$x$$ (the intersection point cannot have $$x=0$$ because that would force $$y=0$$, which does not satisfy the ellipse equation), $$27=6b.$$ Finally, solving for $$b$$ gives $$b=\dfrac{27}{6}=\dfrac{9}{2}.$$

Hence, the correct answer is Option A.

We begin with the standard parabola equation $$x^2 = 4y$$. For a parabola of the form $$x^2 = 4ay$$ the slope form of a tangent is

$$y = mx + \frac{a}{m},$$

where $$m$$ is the slope of the tangent and $$a$$ is the parameter of the parabola. In the given parabola $$4a = 4$$, so $$a = 1$$. Substituting $$a = 1$$, every tangent to the parabola can be written as

$$y = mx + \frac{1}{m}\qquad\text{(1)}$$

Next we want the same line to be tangent to the circle $$x^2 + y^2 = 4$$. A line in the form $$y = mx + c$$ can be rewritten as $$mx - y + c = 0$$, and its perpendicular distance from the origin $$(0,0)$$ is given by the formula

$$\text{Distance} = \frac{|c|}{\sqrt{m^2 + 1}}.$$

For tangency to the circle, this distance must equal the radius of the circle. The circle has radius $$2$$. Hence we must satisfy

$$\frac{|c|}{\sqrt{m^2 + 1}} = 2.$$

From equation (1) we identify $$c = \dfrac{1}{m}$$, and so

$$\frac{\left|\dfrac{1}{m}\right|}{\sqrt{m^2 + 1}} = 2.$$

Removing the absolute value (because we will square both sides) and squaring, we have

$$\frac{1}{m^2}\cdot\frac{1}{m^2 + 1} = 4.$$

Multiplying both sides by $$m^2(m^2 + 1)$$ gives

$$1 = 4m^2(m^2 + 1).$$

Expanding the right-hand side,

$$1 = 4m^4 + 4m^2.$$

Bringing all terms to one side,

$$4m^4 + 4m^2 - 1 = 0.$$

Letting $$t = m^2$$ turns this into a quadratic in $$t$$:

$$4t^2 + 4t - 1 = 0.$$

Using the quadratic formula

$$t = \frac{-4 \pm \sqrt{(4)^2 - 4\cdot4\cdot(-1)}}{2\cdot4} = \frac{-4 \pm \sqrt{16 + 16}}{8} = \frac{-4 \pm 4\sqrt{2}}{8} = \frac{-1 \pm \sqrt{2}}{2}.$$

Since $$t = m^2 \ge 0$$, we keep the positive root:

$$m^2 = \frac{\sqrt{2} - 1}{2}.$$

Therefore there are exactly two slopes,

$$m_1 = +\sqrt{\frac{\sqrt{2} - 1}{2}}, \qquad m_2 = -\sqrt{\frac{\sqrt{2} - 1}{2}}.$$

These two slopes give the two common tangents to both the parabola and the circle:

$$\begin{aligned} \text{(i)}\; & y = m_1x + \frac{1}{m_1},\\[4pt] \text{(ii)}\; & y = m_2x + \frac{1}{m_2}. \end{aligned}$$

We now find their point of intersection $$P(x_P,\,y_P)$$. Because the two tangents meet at this point, their $$y$$-values are equal:

$$m_1x_P + \frac{1}{m_1} = m_2x_P + \frac{1}{m_2}.$$

Substituting $$m_2 = -m_1$$ and $$\dfrac{1}{m_2} = -\dfrac{1}{m_1}$$ gives

$$m_1x_P + \frac{1}{m_1} = -m_1x_P - \frac{1}{m_1}.$$

Collecting like terms,

$$m_1x_P + m_1x_P = -\frac{1}{m_1} - \frac{1}{m_1},$$

$$2m_1x_P = -\frac{2}{m_1}.$$

Dividing by $$2m_1$$ yields the $$x$$-coordinate of $$P$$:

$$x_P = -\frac{1}{m_1^2}.$$

But $$m_1^2 = \dfrac{\sqrt{2} - 1}{2}$$, so

$$x_P = -\,\frac{1}{\dfrac{\sqrt{2} - 1}{2}} = -\frac{2}{\sqrt{2} - 1}.$$

To rationalise the denominator we multiply numerator and denominator by $$\sqrt{2} + 1$$:

$$x_P = -\frac{2(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = -\frac{2(\sqrt{2} + 1)}{2 - 1} = -2(\sqrt{2} + 1).$$

To find $$y_P$$ we substitute $$x_P$$ in, say, the first tangent:

$$y_P = m_1x_P + \frac{1}{m_1} = m_1\!\Bigl(-\frac{1}{m_1^2}\Bigr) + \frac{1}{m_1} = -\frac{1}{m_1} + \frac{1}{m_1} = 0.$$

Thus the point of intersection is

$$P\bigl(-2(\sqrt{2} + 1),\,0\bigr).$$

The distance of $$P$$ from the origin is simply the absolute value of its $$x$$-coordinate (because $$y_P = 0$$):

$$OP = \left| -2(\sqrt{2} + 1) \right| = 2(\sqrt{2} + 1).$$

Hence, the correct answer is Option D.

We have the parabola $$y^{2}=8x$$. A general point on this parabola can be written as $$P(x,y)$$ with the relation $$x=\dfrac{y^{2}}{8}$$.

The circle given in the question is $$x^{2}+(y+6)^{2}=1$$. Its centre is clearly obtained by comparing with $$(x-h)^{2}+(y-k)^{2}=r^{2}$$; we get $$C(0,-6)$$.

We require the point $$P$$ on the parabola that is at the minimum distance from the fixed point $$C$$. Let us denote that (squared) distance by $$D^{2}$$. Using the distance‐formula,

$$D^{2}=(x-0)^{2}+(y+6)^{2}=x^{2}+(y+6)^{2}.$$

Because $$x=\dfrac{y^{2}}{8}$$, we substitute to obtain a single‐variable function:

$$D^{2}(y)=\left(\dfrac{y^{2}}{8}\right)^{2}+(y+6)^{2}=\dfrac{y^{4}}{64}+(y+6)^{2}.$$

To find the minimum we differentiate with respect to $$y$$ and equate to zero. The rule used is “minimum or maximum occurs where the derivative vanishes.”

First derivative:

$$\frac{d}{dy}\!\left[D^{2}(y)\right] =\frac{d}{dy}\left(\dfrac{y^{4}}{64}\right)+\frac{d}{dy}(y+6)^{2} =\dfrac{4y^{3}}{64}+2(y+6) =\dfrac{y^{3}}{16}+2y+12.$$

Setting the derivative to zero gives the cubic equation

$$\dfrac{y^{3}}{16}+2y+12=0 \;\;\Longrightarrow\;\; y^{3}+32y+192=0.$$ Multiplying through by 16 clears the denominator.

We now solve $$y^{3}+32y+192=0$$. A quick integral root test shows $$y=-4$$ is a root because

$$(-4)^{3}+32(-4)+192=-64-128+192=0.$$

Dividing the cubic by $$(y+4)$$ gives the quadratic remainder $$y^{2}-4y+48$$ whose discriminant is negative $$(\Delta=-176<0)$$, so $$y=-4$$ is the only real solution. Hence at the required point we have

$$y_{P}=-4.$$

Using $$x=\dfrac{y^{2}}{8}$$ we get

$$x_{P}=\dfrac{(-4)^{2}}{8}=\dfrac{16}{8}=2.$$

Thus the point of minimum distance is $$P(2,-4).$$

The distance $$PC$$ itself (which will become the radius of the required circle) is found using the distance formula again:

$$PC=\sqrt{(2-0)^{2}+(-4+6)^{2}} =\sqrt{2^{2}+2^{2}} =\sqrt{4+4} =\sqrt{8} =2\sqrt{2}.$$

Now we construct the circle whose centre is $$P(2,-4)$$ and which passes through $$C(0,-6)$$; thus its radius is $$2\sqrt{2}$$. The standard form $$(x-h)^{2}+(y-k)^{2}=r^{2}$$ therefore reads

$$\bigl(x-2\bigr)^{2}+\bigl(y+4\bigr)^{2}=(2\sqrt{2})^{2}=8.$$

We expand to match the options:

$$\begin{aligned} (x-2)^{2}+(y+4)^{2}&=8\\[4pt] \bigl(x^{2}-4x+4\bigr)+\bigl(y^{2}+8y+16\bigr)&=8\\[4pt] x^{2}+y^{2}-4x+8y+20&=8\\[4pt] x^{2}+y^{2}-4x+8y+12&=0. \end{aligned}$$

Comparing with the given alternatives we see that this is exactly Option C.

Hence, the correct answer is Option C.

We have the standard parabola $$y^{2}=4x$$.

For this curve, a convenient parametric form is obtained by putting $$y=2t$$. Substituting in $$y^{2}=4x$$ gives $$4t^{2}=4x$$, so $$x=t^{2}$$. Hence a general point on the parabola can be written as

$$P(t)\;:\;\bigl(t^{2},\,2t\bigr).$$

Let $$P$$ correspond to the parameter $$t$$ and $$Q$$ correspond to the parameter $$t_{1}$$, so that

$$P\equiv\bigl(t^{2},\,2t\bigr),\qquad Q\equiv\bigl(t_{1}^{2},\,2t_{1}\bigr).$$

First we need the equation of the normal to the parabola at the point $$P$$. To do that, we recall the formula for the slope of the tangent to $$y^{2}=4x$$ at a point:

The curve is given implicitly by $$y^{2}=4x$$. Differentiating, we obtain

$$2y\frac{dy}{dx}=4 \;\;\Longrightarrow\;\; \frac{dy}{dx}=\frac{2}{y}.$$

At the point $$P(t)$$ we have $$y=2t$$, so the slope of the tangent is

$$m_{\text{tangent}}=\frac{2}{\,2t\,}=\frac{1}{t}.$$

The slope of the normal is the negative reciprocal of the tangent’s slope. Thus

$$m_{\text{normal}}=-t.$$

The point-slope form of a straight line with slope $$m_{\text{normal}}$$ passing through $$P(t^{2},2t)$$ is

$$y-2t=-t\bigl(x-t^{2}\bigr).$$

This is the required normal at $$P$$. By hypothesis, the point $$Q\bigl(t_{1}^{2},2t_{1}\bigr)$$ lies on this normal. Therefore we substitute $$x=t_{1}^{2}$$ and $$y=2t_{1}$$ into the normal’s equation:

$$2t_{1}-2t=-t\bigl(t_{1}^{2}-t^{2}\bigr).$$

We expand the right-hand side:

$$2t_{1}-2t=-t\left(t_{1}^{2}-t^{2}\right) =-t\bigl(t_{1}-t\bigr)\bigl(t_{1}+t\bigr).$$

Because $$P$$ and $$Q$$ are distinct, $$t_{1}\neq t$$, so we can safely divide both sides by the factor $$\bigl(t_{1}-t\bigr)$$. First we factor out $$2$$ on the left:

$$2\,(t_{1}-t)=-t\bigl(t_{1}-t\bigr)\bigl(t_{1}+t\bigr).$$

Now cancelling the common factor $$t_{1}-t$$ from both sides gives

$$2=-t\bigl(t_{1}+t\bigr).$$

We rearrange this relation to express $$t_{1}$$ in terms of $$t$$:

$$t_{1}+t=-\frac{2}{t}\quad\Longrightarrow\quad t_{1}=-\,\frac{2}{t}-t.$$

Our aim is to find the minimum possible value of $$t_{1}^{2}$$. Substituting the above expression, we obtain

$$t_{1}^{2}=\left(-\frac{2}{t}-t\right)^{2} =\left(t+\frac{2}{t}\right)^{2}.$$

To minimise $$t_{1}^{2}$$ we can instead minimise the non-negative quantity

$$g(t)=\left(t+\frac{2}{t}\right)^{2}.$$

Let us set $$h(t)=t+\dfrac{2}{t}$$ and first minimise $$h(t)^{2}$$ by minimising $$|h(t)|$$. The derivative of $$h(t)$$ is

$$h'(t)=1-\frac{2}{t^{2}}.$$

Setting $$h'(t)=0$$ gives

$$1-\frac{2}{t^{2}}=0\quad\Longrightarrow\quad t^{2}=2.$$

Thus the critical points are $$t=\sqrt{2}$$ and $$t=-\sqrt{2}$$ (both are admissible because the parameter $$t$$ may be positive or negative, but never zero).

Evaluating $$h(t)$$ at these values:

For $$t=\sqrt{2}$$,

$$h(\sqrt{2})=\sqrt{2}+\frac{2}{\sqrt{2}} =\sqrt{2}+\sqrt{2} =2\sqrt{2}.$$

For $$t=-\sqrt{2}$$,

$$h(-\sqrt{2})=-\sqrt{2}+\frac{2}{-\sqrt{2}} =-\sqrt{2}-\sqrt{2} =-2\sqrt{2}.$$

In magnitude both values are equal, so $$|h(t)|_{\min}=2\sqrt{2}$$. Squaring this gives the minimum of $$h(t)^{2}$$ and therefore of $$g(t)$$:

$$t_{1,\min}^{2}= \bigl(2\sqrt{2}\bigr)^{2}=8.$$

Hence, the minimum possible value of $$t_{1}^{2}$$ is $$8$$.

Hence, the correct answer is Option A.

Let us denote $$O(0,0)$$ as the vertex and take an arbitrary point on the given parabola $$x^{2}=8y$$ as $$Q(x,y)$$, so that $$Q$$ automatically satisfies the relation $$x^{2}=8y$$.

We are told that the point $$P$$ divides the line segment $$OQ$$ internally in the ratio $$1:3$$, that is

$$OP:PQ = 1:3.$$

First, recall the coordinate-division formula. For two points $$A(x_1,y_1)$$ and $$B(x_2,y_2)$$, a point $$R$$ that divides $$AB$$ internally in the ratio $$m:n$$ (measured from $$A$$ to $$B$$) has coordinates

$$\left(\dfrac{nx_1+mx_2}{m+n},\;\dfrac{ny_1+my_2}{m+n}\right).$$

Here we take $$A=O(0,0)$$ and $$B=Q(x,y)$$, with the ratio $$m:n = 1:3$$, where $$m$$ is attached to $$B(Q)$$ and $$n$$ to $$A(O)$$. Substituting $$x_1=0,\;y_1=0,\;x_2=x,\;y_2=y,\;m=1,\;n=3$$, we get

$$P\Bigl(X,Y\Bigr)=\left(\dfrac{3\cdot 0+1\cdot x}{1+3},\;\dfrac{3\cdot 0+1\cdot y}{1+3}\right)=\left(\dfrac{x}{4},\;\dfrac{y}{4}\right).$$

Thus the coordinates of $$P$$ can be expressed in terms of those of $$Q$$ as

$$X=\dfrac{x}{4},\qquad Y=\dfrac{y}{4}.$$

Rewriting these gives the reverse substitutions

$$x = 4X,\qquad y = 4Y.$$

Because $$Q$$ lies on the parabola $$x^{2}=8y$$, we have

$$x^{2}=8y.$$

Now substitute $$x=4X$$ and $$y=4Y$$ into this relation:

$$\bigl(4X\bigr)^{2}=8\bigl(4Y\bigr).$$

Carrying out the algebra step by step,

$$16X^{2}=32Y.$$

Divide every term by $$16$$:

$$X^{2}=2Y.$$

The symbols $$X$$ and $$Y$$ represent the running coordinates of the point $$P$$, so its locus is given by

$$x^{2}=2y.$$

Hence, the correct answer is Option A.

We are given the parabola $$ y^2 = -4x $$. Since $$ PQ $$ is a double ordinate, it is perpendicular to the axis of the parabola. Given that the parabola is symmetric about the x-axis, $$ PQ $$ is horizontal. Let $$ P $$ be a point in the second quadrant, so its x-coordinate is negative and y-coordinate is positive. Let the coordinates of $$ P $$ be $$ (x_1, y_1) $$. Since $$ P $$ lies on the parabola, it satisfies the equation:

$$ y_1^2 = -4x_1 $$

Because $$ PQ $$ is a double ordinate, $$ Q $$ will have the same x-coordinate as $$ P $$ but the negative y-coordinate due to symmetry. Thus, the coordinates of $$ Q $$ are $$ (x_1, -y_1) $$.

Now, $$ R $$ divides $$ PQ $$ in the ratio 2:1. This means that the ratio $$ PR:RQ = 2:1 $$. Using the section formula, the coordinates of $$ R $$ are calculated as follows:

Let $$ R $$ have coordinates $$ (x_R, y_R) $$. The section formula for dividing the line segment joining $$ P(x_1, y_1) $$ and $$ Q(x_1, -y_1) $$ in the ratio $$ m:n = 2:1 $$ is:

$$ x_R = \frac{m \cdot x_Q + n \cdot x_P}{m + n} $$

$$ y_R = \frac{m \cdot y_Q + n \cdot y_P}{m + n} $$

Substituting $$ m = 2 $$, $$ n = 1 $$, $$ x_P = x_1 $$, $$ y_P = y_1 $$, $$ x_Q = x_1 $$, and $$ y_Q = -y_1 $$:

$$ x_R = \frac{2 \cdot x_1 + 1 \cdot x_1}{2 + 1} = \frac{2x_1 + x_1}{3} = \frac{3x_1}{3} = x_1 $$

$$ y_R = \frac{2 \cdot (-y_1) + 1 \cdot y_1}{3} = \frac{-2y_1 + y_1}{3} = \frac{-y_1}{3} = -\frac{y_1}{3} $$

So, $$ R $$ has coordinates $$ \left( x_1, -\frac{y_1}{3} \right) $$.

To find the locus of $$ R $$, we eliminate the parameters $$ x_1 $$ and $$ y_1 $$. Let $$ R $$ be denoted by $$ (h, k) $$, so:

$$ h = x_1 $$

$$ k = -\frac{y_1}{3} \quad \Rightarrow \quad y_1 = -3k $$

Since $$ P(x_1, y_1) $$ lies on the parabola $$ y^2 = -4x $$, substitute $$ x_1 = h $$ and $$ y_1 = -3k $$:

$$ (-3k)^2 = -4h $$

$$ 9k^2 = -4h $$

Replacing $$ h $$ and $$ k $$ with $$ x $$ and $$ y $$ respectively, the locus is:

$$ 9y^2 = -4x $$

Comparing with the given options:

A. $$ 3y^2 = -2x $$

B. $$ 9y^2 = 4x $$

C. $$ 9y^2 = -4x $$

D. $$ 3y^2 = 2x $$

The locus matches option C. Hence, the correct answer is Option C.

The given parabola is $$ y^2 = 6x $$. Comparing this with the standard form $$ y^2 = 4ax $$, we find $$ 4a = 6 $$, so $$ a = \frac{3}{2} $$. The focus of the parabola is at $$ \left( \frac{3}{2}, 0 \right) $$, and the vertex is at $$ (0, 0) $$. A chord is drawn through the focus $$ \left( \frac{3}{2}, 0 \right) $$ with slope $$ m $$. The equation of this chord is: $$ y - 0 = m \left( x - \frac{3}{2} \right) $$ which simplifies to: $$ y = m \left( x - \frac{3}{2} \right) $$ Rewriting this in the standard form $$ ax + by + c = 0 $$: $$ m x - y - \frac{3m}{2} = 0 $$ Here, $$ a = m $$, $$ b = -1 $$, and $$ c = -\frac{3m}{2} $$. The perpendicular distance from the vertex $$ (0, 0) $$ to this line is given by the formula: $$ \text{distance} = \frac{|a \cdot 0 + b \cdot 0 + c|}{\sqrt{a^2 + b^2}} = \frac{| -\frac{3m}{2} |}{\sqrt{m^2 + (-1)^2}} = \frac{\frac{3|m|}{2}}{\sqrt{m^2 + 1}} = \frac{3|m|}{2\sqrt{m^2 + 1}} $$ This distance is given as $$ \frac{\sqrt{5}}{2} $$. So: $$ \frac{3|m|}{2\sqrt{m^2 + 1}} = \frac{\sqrt{5}}{2} $$ Multiplying both sides by 2: $$ \frac{3|m|}{\sqrt{m^2 + 1}} = \sqrt{5} $$ Since the options are positive, assume $$ m > 0 $$ (the solution will account for both signs later). Thus, $$ |m| = m $$: $$ \frac{3m}{\sqrt{m^2 + 1}} = \sqrt{5} $$ Squaring both sides to eliminate the square root: $$ \left( \frac{3m}{\sqrt{m^2 + 1}} \right)^2 = (\sqrt{5})^2 $$ $$ \frac{9m^2}{m^2 + 1} = 5 $$ Cross-multiplying: $$ 9m^2 = 5(m^2 + 1) $$ $$ 9m^2 = 5m^2 + 5 $$ Bringing all terms to one side: $$ 9m^2 - 5m^2 - 5 = 0 $$ $$ 4m^2 - 5 = 0 $$ $$ 4m^2 = 5 $$ $$ m^2 = \frac{5}{4} $$ $$ m = \pm \frac{\sqrt{5}}{2} $$ The slopes are $$ \frac{\sqrt{5}}{2} $$ and $$ -\frac{\sqrt{5}}{2} $$. Now, verify these satisfy the original distance condition. For $$ m = \frac{\sqrt{5}}{2} $$: $$ \text{distance} = \frac{3 \left| \frac{\sqrt{5}}{2} \right|}{2 \sqrt{ \left( \frac{\sqrt{5}}{2} \right)^2 + 1 }} = \frac{3 \cdot \frac{\sqrt{5}}{2}}{2 \sqrt{ \frac{5}{4} + 1 }} = \frac{ \frac{3\sqrt{5}}{2} }{2 \sqrt{ \frac{5}{4} + \frac{4}{4} }} = \frac{ \frac{3\sqrt{5}}{2} }{2 \sqrt{ \frac{9}{4} }} = \frac{ \frac{3\sqrt{5}}{2} }{2 \cdot \frac{3}{2} } = \frac{ \frac{3\sqrt{5}}{2} }{3} = \frac{3\sqrt{5}}{2} \cdot \frac{1}{3} = \frac{\sqrt{5}}{2} $$ For $$ m = -\frac{\sqrt{5}}{2} $$: $$ \text{distance} = \frac{3 \left| -\frac{\sqrt{5}}{2} \right|}{2 \sqrt{ \left( -\frac{\sqrt{5}}{2} \right)^2 + 1 }} = \frac{3 \cdot \frac{\sqrt{5}}{2}}{2 \sqrt{ \frac{5}{4} + 1 }} = \text{same as above} = \frac{\sqrt{5}}{2} $$ Both slopes satisfy the distance condition. Now, check if the chord intersects the parabola. For $$ m = \frac{\sqrt{5}}{2} $$, substitute into the parabola $$ y^2 = 6x $$: $$ y = \frac{\sqrt{5}}{2} \left( x - \frac{3}{2} \right) $$ $$ \left[ \frac{\sqrt{5}}{2} \left( x - \frac{3}{2} \right) \right]^2 = 6x $$ $$ \frac{5}{4} \left( x - \frac{3}{2} \right)^2 = 6x $$ Multiply by 4: $$ 5 \left( x - \frac{3}{2} \right)^2 = 24x $$ $$ 5 \left( x^2 - 3x + \frac{9}{4} \right) = 24x $$ $$ 5x^2 - 15x + \frac{45}{4} = 24x $$ $$ 5x^2 - 15x - 24x + \frac{45}{4} = 0 $$ $$ 5x^2 - 39x + \frac{45}{4} = 0 $$ Multiply by 4: $$ 20x^2 - 156x + 45 = 0 $$ Discriminant: $$ d = (-156)^2 - 4 \cdot 20 \cdot 45 = 24336 - 3600 = 20736 $$ $$ \sqrt{20736} = 144 \quad (144^2 = 20736) $$ Real roots exist. Similarly, for $$ m = -\frac{\sqrt{5}}{2} $$, the chord intersects the parabola. Among the options, $$ \frac{\sqrt{5}}{2} $$ is option A. The other options do not satisfy the distance condition: - Option B: $$ m = \frac{2}{\sqrt{3}} $$ gives distance $$ \approx 1.134 \neq \frac{\sqrt{5}}{2} \approx 1.118 $$ - Option C: $$ m = \frac{\sqrt{3}}{2} $$ gives distance $$ \approx 0.982 \neq \frac{\sqrt{5}}{2} $$ - Option D: $$ m = \frac{2}{\sqrt{5}} $$ gives distance $$ = 1 \neq \frac{\sqrt{5}}{2} $$ Hence, the slope can be $$ \frac{\sqrt{5}}{2} $$, which is option A. So, the answer is Option A.

We have two curves. The first is the circle $$x^2 + y^2 = 9$$ whose centre is at the origin and whose radius is $$r = 3$$ because, for a circle $$x^2 + y^2 = R^2$$, the radius is $$R$$. The second curve is the parabola $$y^2 = 8x$$; this is of the standard form $$y^2 = 4ax$$ where $$a$$ is the focal length.

First we find the length $$L_2$$ of the latus rectum of the parabola. For any parabola of the form $$y^2 = 4ax$$, the length of the latus rectum is given by the formula $$4a$$. Comparing $$y^2 = 8x$$ with $$y^2 = 4ax$$ we see

$$4a = 8 \; \Longrightarrow \; a = 2.$$

Therefore the length of the latus rectum is

$$L_2 = 4a = 4 \times 2 = 8.$$

Now we calculate $$L_1$$, the length of the common chord of the circle and the parabola. Points on both curves must satisfy both equations simultaneously. From the parabola we have $$y^2 = 8x$$, so we substitute $$y^2$$ from this relation into the circle’s equation.

Circle equation: $$x^2 + y^2 = 9.$$

Substituting $$y^2 = 8x$$ gives

$$x^2 + 8x = 9.$$

Bringing all terms to the left side we obtain a quadratic in $$x$$,

$$x^2 + 8x - 9 = 0.$$

To solve this quadratic we use the quadratic formula. For $$ax^2 + bx + c = 0$$, the roots are

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$

Here $$a = 1, \; b = 8, \; c = -9$$, so

$$x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1} = \frac{-8 \pm \sqrt{64 + 36}}{2} = \frac{-8 \pm \sqrt{100}}{2} = \frac{-8 \pm 10}{2}.$$

This gives two algebraic values

$$x_1 = \frac{-8 + 10}{2} = \frac{2}{2} = 1, \quad x_2 = \frac{-8 - 10}{2} = \frac{-18}{2} = -9.$$

However, for the parabola $$y^2 = 8x$$ we must have $$x \ge 0$$ because $$y^2$$ is non-negative. Thus $$x = -9$$ is inadmissible (it would force $$y^2 = -72$$, which is impossible for real $$y$$). We therefore keep only

$$x = 1.$$

For this $$x$$-coordinate, the corresponding $$y$$-coordinates come from $$y^2 = 8x$$:

$$y^2 = 8(1) = 8 \; \Longrightarrow \; y = \pm \sqrt{8} = \pm 2\sqrt{2}.$$

Hence the two real points of intersection are

$$(1,\, 2\sqrt{2}) \quad \text{and} \quad (1,\,-2\sqrt{2}).$$

The common chord is simply the straight line segment joining these two points. Because their $$x$$-coordinates are identical (both equal to 1), the chord is vertical. The distance between two points $$(x, y_1)$$ and $$(x, y_2)$$ on a vertical line is $$|y_1 - y_2|.$$ Therefore the length of the chord is

$$L_1 = |\,2\sqrt{2} - (-2\sqrt{2})| = |\,4\sqrt{2}\,| = 4\sqrt{2}.$$

We now compare $$L_1$$ and $$L_2$$:

$$L_1 = 4\sqrt{2}, \qquad L_2 = 8.$$

Dividing gives

$$\frac{L_1}{L_2} = \frac{4\sqrt{2}}{8} = \frac{\sqrt{2}}{2} \approx 0.707,$$

which is clearly less than 1. That is, $$L_1 < L_2.$$

Hence, the correct answer is Option C.

We are given a point $$(-2, -1)$$ and the parabola $$y^2 = 4x$$. We need to find $$|\tan \alpha|$$, where $$\alpha$$ is the angle between the two tangents drawn from this point to the parabola.

First, recall that for a parabola $$y^2 = 4ax$$, the equation of a tangent with slope $$m$$ is $$y = mx + \frac{a}{m}$$. Comparing $$y^2 = 4x$$ with $$y^2 = 4ax$$, we get $$4a = 4$$, so $$a = 1$$. Therefore, the equation of a tangent to the parabola $$y^2 = 4x$$ with slope $$m$$ is:

$$y = mx + \frac{1}{m}$$

Since the tangent passes through the point $$(-2, -1)$$, we substitute $$x = -2$$ and $$y = -1$$ into the tangent equation:

$$-1 = m \cdot (-2) + \frac{1}{m}$$

Simplify the equation:

$$-1 = -2m + \frac{1}{m}$$

To eliminate the denominator, multiply both sides by $$m$$:

$$-1 \cdot m = \left(-2m + \frac{1}{m}\right) \cdot m$$

This gives:

$$-m = -2m^2 + 1$$

Bring all terms to one side to form a quadratic equation:

$$2m^2 - m - 1 = 0$$

Solve for $$m$$ using the quadratic formula. Here, $$a = 2$$, $$b = -1$$, and $$c = -1$$. The discriminant $$D$$ is:

$$D = b^2 - 4ac = (-1)^2 - 4 \cdot 2 \cdot (-1) = 1 + 8 = 9$$

So, the solutions are:

$$m = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-1) \pm \sqrt{9}}{2 \cdot 2} = \frac{1 \pm 3}{4}$$

Thus, the two slopes are:

$$m_1 = \frac{1 + 3}{4} = \frac{4}{4} = 1$$

$$m_2 = \frac{1 - 3}{4} = \frac{-2}{4} = -\frac{1}{2}$$

The angle $$\alpha$$ between two lines with slopes $$m_1$$ and $$m_2$$ is given by:

$$\tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$

Substitute $$m_1 = 1$$ and $$m_2 = -\frac{1}{2}$$:

$$m_1 - m_2 = 1 - \left(-\frac{1}{2}\right) = 1 + \frac{1}{2} = \frac{3}{2}$$

$$1 + m_1 m_2 = 1 + (1) \cdot \left(-\frac{1}{2}\right) = 1 - \frac{1}{2} = \frac{1}{2}$$

Therefore,

$$\tan \alpha = \left| \frac{\frac{3}{2}}{\frac{1}{2}} \right| = \left| \frac{3}{2} \times \frac{2}{1} \right| = |3| = 3$$

Hence, $$|\tan \alpha| = 3$$.

Comparing with the options, we see that 3 corresponds to option D.

Hence, the correct answer is Option D.

Let us take an arbitrary straight line with slope $$m$$ and $$y$$-intercept $$c$$. In point-slope form this line is written as

$$y = m x + c.$$

We want this single line to touch both parabolas one after the other, so we shall impose the condition of tangency separately for each curve and finally make the two conditions compatible.

First parabola  $$y^{2}=4x$$

Substituting $$y = m x + c$$ in $$y^{2}=4x$$ we obtain

$$(m x + c)^{2}=4x.$$

This expands to

$$m^{2}x^{2}+2mcx+c^{2}-4x=0.$$

The above is a quadratic in $$x$$. For the line to be a tangent, both roots must coincide, i.e. the discriminant must vanish. Recall the quadratic discriminant formula $$\Delta = b^{2}-4ac.$$ Here $$a=m^{2},\; b=2mc-4,\; c=c^{2}.$$ So

$$\Delta=(2mc-4)^{2}-4(m^{2})(c^{2}).$$

Evaluating,

$$$\begin{aligned} \Delta &= (2mc-4)^{2}-4m^{2}c^{2}\\ &= 4m^{2}c^{2}-16mc+16-4m^{2}c^{2}\\ &= -16mc+16. \end{aligned}$$$

Setting $$\Delta=0$$ gives

$$-16mc+16=0 \;\Longrightarrow\; mc=1 \;\Longrightarrow\; c=\frac1m.$$

Second parabola  $$x^{2}=-32y$$

Again substitute the same line $$y = m x + c$$ into $$x^{2}=-32y$$:

$$x^{2} = -32(m x + c).$$

This rearranges to a quadratic in $$x$$ as

$$x^{2}+32m x+32c=0.$$

Here $$a=1,\; b=32m,\; c=32c.$$ Applying the discriminant condition $$\Delta = b^{2}-4ac = 0$$ we have

$$(32m)^{2}-4(1)(32c)=0.$$

Simplifying step by step,

$$$\begin{aligned} 1024m^{2}-128c &=0\\ \Rightarrow\; 32m^{2}-4c&=0\\ \Rightarrow\; c&=8m^{2}. \end{aligned}$$$

Common tangency condition

The same line must satisfy both relations for $$c$$. Therefore we equate

$$\frac1m = 8m^{2}.$$

Multiplying both sides by $$m$$ gives

$$1 = 8m^{3}.$$

Hence

$$m^{3}=\frac18 \;\Longrightarrow\; m = \sqrt[3]{\frac18}= \frac12.$$

Thus the slope of the line which touches both parabolas is

$$m = \frac12.$$

Hence, the correct answer is Option C.

The given parabola is $$ y^2 = 4x $$. Comparing this with the standard form $$ y^2 = 4ax $$, we find that $$ 4a = 4 $$, so $$ a = 1 $$. The focus of the parabola is at $$ (a, 0) = (1, 0) $$. The latus rectum is the line segment perpendicular to the axis passing through the focus, so it is the line $$ x = a $$, which is $$ x = 1 $$.

To find the endpoints of the latus rectum, substitute $$ x = 1 $$ into the parabola equation: $$ y^2 = 4 \times 1 = 4 $$. Solving for $$ y $$, we get $$ y = \pm 2 $$. Thus, the endpoints are $$ (1, 2) $$ and $$ (1, -2) $$.

Next, we need the equations of the normals to the parabola at these points. Using the parametric form, a point on the parabola $$ y^2 = 4ax $$ can be written as $$ (at^2, 2at) $$. Since $$ a = 1 $$, the points are $$ (t^2, 2t) $$.

For the point $$ (1, 2) $$, we have $$ 2t = 2 $$, so $$ t = 1 $$. For the point $$ (1, -2) $$, we have $$ 2t = -2 $$, so $$ t = -1 $$.

The equation of the normal to the parabola $$ y^2 = 4ax $$ at parameter $$ t $$ is given by $$ y = tx - 2at - at^3 $$. Substituting $$ a = 1 $$, this simplifies to $$ y = tx - 2t - t^3 $$.

For $$ t = 1 $$: $$ y = (1)x - 2(1) - (1)^3 = x - 2 - 1 = x - 3 $$. So, the normal at $$ (1, 2) $$ is $$ y = x - 3 $$.

For $$ t = -1 $$: $$ y = (-1)x - 2(-1) - (-1)^3 = -x + 2 - (-1) = -x + 2 + 1 = -x + 3 $$. So, the normal at $$ (1, -2) $$ is $$ y = -x + 3 $$.

To find the point of intersection of these two normals, solve the system of equations:

Equation 1: $$ y = x - 3 $$

Equation 2: $$ y = -x + 3 $$

Set them equal: $$ x - 3 = -x + 3 $$. Add $$ x $$ to both sides: $$ x + x - 3 = 3 $$, which simplifies to $$ 2x - 3 = 3 $$. Add 3 to both sides: $$ 2x = 6 $$. Divide by 2: $$ x = 3 $$.

Substitute $$ x = 3 $$ into Equation 1: $$ y = 3 - 3 = 0 $$. Thus, the point of intersection is $$ (3, 0) $$.

Verifying with Equation 2: when $$ x = 3 $$, $$ y = -3 + 3 = 0 $$, which matches.

Comparing with the options:

A. $$ (0, 2) $$

B. $$ (3, 0) $$

C. $$ (0, 3) $$

D. $$ (2, 0) $$

Hence, the correct answer is Option B.

First, consider Statement-1: The slope of the tangent at any point P on a parabola with its axis along the x-axis and vertex at the origin is inversely proportional to the ordinate (y-coordinate) of point P.

The standard equation of such a parabola is $$ y^2 = 4ax $$, where $$ a $$ is a constant. To find the slope of the tangent at any point $$ (x, y) $$ on this parabola, differentiate the equation implicitly with respect to $$ x $$.

Differentiate both sides of $$ y^2 = 4ax $$:

$$ \frac{d}{dx}(y^2) = \frac{d}{dx}(4ax) $$

$$ 2y \frac{dy}{dx} = 4a $$

Solve for $$ \frac{dy}{dx} $$:

$$ \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} $$

So, the slope $$ m = \frac{2a}{y} $$. Since $$ a $$ is a constant, $$ m $$ is inversely proportional to $$ y $$, the ordinate. Therefore, Statement-1 is true.

Now, consider Statement-2: The system of parabolas $$ y^2 = 4ax $$ satisfies a differential equation of degree 1 and order 1.

The equation $$ y^2 = 4ax $$ represents a family of parabolas with parameter $$ a $$. To form a differential equation, eliminate the arbitrary constant $$ a $$. Differentiate both sides of $$ y^2 = 4ax $$ with respect to $$ x $$:

$$ \frac{d}{dx}(y^2) = \frac{d}{dx}(4ax) $$

$$ 2y \frac{dy}{dx} = 4a $$

Solve for $$ a $$:

$$ a = \frac{2y \frac{dy}{dx}}{4} = \frac{y \frac{dy}{dx}}{2} $$

Substitute this expression for $$ a $$ back into the original equation $$ y^2 = 4ax $$:

$$ y^2 = 4 \left( \frac{y \frac{dy}{dx}}{2} \right) x $$

Simplify:

$$ y^2 = 2 y x \frac{dy}{dx} $$

Assuming $$ y \neq 0 $$ (valid for points not on the x-axis), divide both sides by $$ y $$:

$$ y = 2x \frac{dy}{dx} $$

Rearrange to standard form:

$$ 2x \frac{dy}{dx} - y = 0 $$

The highest derivative is $$ \frac{dy}{dx} $$, which is of order 1. The exponent of this derivative is 1, so the degree is 1. Thus, the differential equation is of order 1 and degree 1. Therefore, Statement-2 is true.

Now, check if Statement-2 correctly explains Statement-1. The differential equation $$ 2x \frac{dy}{dx} - y = 0 $$ gives $$ \frac{dy}{dx} = \frac{y}{2x} $$. Statement-1 requires the slope to be inversely proportional to $$ y $$, but this expression depends on both $$ x $$ and $$ y $$. To relate it to Statement-1, substitute $$ x = \frac{y^2}{4a} $$ from the parabola equation:

$$ \frac{dy}{dx} = \frac{y}{2 \cdot \frac{y^2}{4a}} = \frac{y \cdot 4a}{2 y^2} = \frac{4a}{2y} = \frac{2a}{y} $$

This shows the slope is inversely proportional to $$ y $$, but the substitution relies on the specific parabola equation, not solely on the differential equation from Statement-2. The differential equation describes the family and relates slope to both $$ x $$ and $$ y $$, but without the original equation, it does not directly show inverse proportionality to $$ y $$ alone. Thus, Statement-2 is true but does not directly explain Statement-1.

Hence, the correct answer is Option B.

To find the equation of the line passing through the points of intersection of the parabola $$x^2 = 8y$$ and the ellipse $$\frac{x^2}{3} + y^2 = 1$$, we need to solve these equations simultaneously. The points where both curves intersect will satisfy both equations, and the line joining these points can be found.

First, write the equations clearly:

Parabola: $$x^2 - 8y = 0$$ ...(1)

Ellipse: $$\frac{x^2}{3} + y^2 - 1 = 0$$ ...(2)

From equation (1), express $$x^2$$ in terms of $$y$$:

$$x^2 = 8y$$

Substitute this into equation (2) to eliminate $$x^2$$:

$$\frac{8y}{3} + y^2 = 1$$

Multiply both sides by 3 to clear the denominator:

$$8y + 3y^2 = 3$$

Bring all terms to one side to form a quadratic equation:

$$3y^2 + 8y - 3 = 0$$

Solve this quadratic equation for $$y$$. The discriminant $$D$$ is given by $$D = b^2 - 4ac$$, where $$a = 3$$, $$b = 8$$, and $$c = -3$$:

$$D = 8^2 - 4 \cdot 3 \cdot (-3) = 64 + 36 = 100$$

Now, find the roots:

$$y = \frac{-b \pm \sqrt{D}}{2a} = \frac{-8 \pm \sqrt{100}}{6} = \frac{-8 \pm 10}{6}$$

So, the two solutions are:

$$y = \frac{-8 + 10}{6} = \frac{2}{6} = \frac{1}{3}$$

$$y = \frac{-8 - 10}{6} = \frac{-18}{6} = -3$$

For each $$y$$, find the corresponding $$x$$ using $$x^2 = 8y$$ from equation (1).

First, for $$y = \frac{1}{3}$$:

$$x^2 = 8 \cdot \frac{1}{3} = \frac{8}{3}$$

$$x = \pm \sqrt{\frac{8}{3}} = \pm \frac{2\sqrt{2}}{\sqrt{3}} = \pm \frac{2\sqrt{6}}{3}$$

Second, for $$y = -3$$:

$$x^2 = 8 \cdot (-3) = -24$$

Since $$x^2$$ cannot be negative for real $$x$$, there are no real solutions for $$y = -3$$.

Thus, the only real points of intersection are $$\left( \frac{2\sqrt{6}}{3}, \frac{1}{3} \right)$$ and $$\left( -\frac{2\sqrt{6}}{3}, \frac{1}{3} \right)$$. Both points have the same $$y$$-coordinate, $$y = \frac{1}{3}$$.

The line joining these points is horizontal, given by $$y = \frac{1}{3}$$. Rewriting this in standard form:

$$y - \frac{1}{3} = 0$$

Multiply both sides by 3 to clear the fraction:

$$3y - 1 = 0$$

Now, verify that these points satisfy both original equations.

For the point $$\left( \frac{2\sqrt{6}}{3}, \frac{1}{3} \right)$$:

Parabola: $$x^2 = \left( \frac{2\sqrt{6}}{3} \right)^2 = \frac{4 \cdot 6}{9} = \frac{24}{9} = \frac{8}{3}$$ and $$8y = 8 \cdot \frac{1}{3} = \frac{8}{3}$$, so equal.

Ellipse: $$\frac{x^2}{3} + y^2 = \frac{\frac{8}{3}}{3} + \left( \frac{1}{3} \right)^2 = \frac{8}{9} + \frac{1}{9} = \frac{9}{9} = 1$$, so satisfied.

The same holds for $$\left( -\frac{2\sqrt{6}}{3}, \frac{1}{3} \right)$$ due to symmetry.

Comparing with the options:

A. $$y - 3 = 0$$

B. $$y + 3 = 0$$

C. $$3y + 1 = 0$$

D. $$3y - 1 = 0$$

The equation $$3y - 1 = 0$$ matches our result.

Hence, the correct answer is Option D.

Given Equations:

Circle: $$2x^2 + 2y^2 = 5 \implies x^2 + y^2 = \frac{5}{2}$$

Center $$(0, 0)$$ and radius $$r = \sqrt{\frac{5}{2}}$$.

Parabola: $$y^2 = 4\sqrt{5}x$$

Comparing with $$y^2 = 4ax$$, we get $$a = \sqrt{5}$$.

Condition for Common Tangent: The equation of any tangent to the parabola $$y^2 = 4ax$$ in slope form is:

$$y = mx + \frac{a}{m} \implies y = mx + \frac{\sqrt{5}}{m}$$

$$mx - y + \frac{\sqrt{5}}{m} = 0 \quad \text{--- (i)}$$

For this line to be tangent to the circle $$x^2 + y^2 = r^2$$, the perpendicular distance from the center $$(0, 0)$$ to the line must equal the radius $$r$$.

$$\left| \frac{m(0) - 0 + \frac{\sqrt{5}}{m}}{\sqrt{m^2 + (-1)^2}} \right| = \sqrt{\frac{5}{2}}$$

$$\frac{\sqrt{5}}{|m|\sqrt{m^2 + 1}} = \sqrt{\frac{5}{2}}$$

$$\frac{5}{m^2(m^2 + 1)} = \frac{5}{2}$$

$$m^2(m^2 + 1) = 2 \implies m^4 + m^2 - 2 = 0$$

$$(m^2 + 2)(m^2 - 1) = 0$$

Since $$m^2$$ cannot be negative, $$m^2 = 1 \implies m = \pm 1$$.

Thus, Statement - I is true.

Statement - II:

The $$m$$ values of the common tangents satisfy the equation given in Statement - II. Thus, Statement - II is true.

However, the actual condition for the common tangent is $$m^4 + m^2 - 2 = 0$$. Statement - II provides a different equation that happens to share the same roots ($$m^2=1$$), but it does not represent the logical derivation or the complete set of conditions for Statement - I. Therefore, it is not the correct explanation.

We have the parabola $$y^{2}+2x=0$$ which can be rewritten as $$x=-\dfrac{y^{2}}{2}$$. The given straight line is $$x-2y=2$$, that is $$x=2y+2$$.

To find their points of intersection we substitute the value of $$x$$ from the line into the parabola:

$$-\dfrac{y^{2}}{2}=2y+2$$

Multiplying both sides by $$2$$ to clear the denominator,

$$-y^{2}=4y+4$$

Now bring every term to the left side:

$$y^{2}+4y+4=0$$

Notice that the left-hand side is a perfect square:

$$(y+2)^{2}=0$$

So,

$$y=-2$$

Substituting this value of $$y$$ back in $$x=2y+2$$,

$$x=2(-2)+2=-4+2=-2$$

Hence, the only common point of the line and the parabola is $$(-2,-2)$$, and because the quadratic equation in $$y$$ gives a repeated root, the line merely touches (is tangent to) the parabola there. Therefore Statement-1 is true.

Now we examine Statement-2. First recall the slope-form of the tangent to the standard parabola $$y^{2}=4ax$$. It is stated in textbooks as

$$y=mx+\dfrac{a}{m},\qquad m\neq 0$$

where $$m$$ is the slope and $$\left(\dfrac{a}{m^{2}},\,\dfrac{2a}{m}\right)$$ is the corresponding point of contact.

For the parabola in the present question, $$y^{2}=-2x$$. We equate this to the standard form $$y^{2}=4ax$$ to find $$a$$:

$$4a=-2\;\;\Longrightarrow\;\;a=-\dfrac12$$

Using the above formula for the tangent with this value of $$a$$, we get

$$y=mx+\dfrac{a}{m}=mx+\dfrac{-\dfrac12}{m}=mx-\dfrac1{2m},\qquad m\neq 0$$

which is exactly the line quoted in Statement-2.

Further, the point of contact should be

$$\left(\dfrac{a}{m^{2}},\,\dfrac{2a}{m}\right)=\left(\dfrac{-\dfrac12}{m^{2}},\,\dfrac{2\!\left(-\dfrac12\right)}{m}\right)=\left(-\dfrac1{2m^{2}},\,-\dfrac1{m}\right)$$

This is exactly the point written in Statement-2. Hence Statement-2 is also true.

Finally, observe that by putting $$m=\dfrac12$$ in the general tangent $$y=mx-\dfrac1{2m}$$ we obtain

$$y=\dfrac12x-\dfrac1{2\left(\tfrac12\right)}=\dfrac12x-1$$

which, after multiplying by $$2$$, gives $$x-2y=2$$ - the very line of Statement-1. The corresponding point of contact becomes

$$\left(-\dfrac1{2\left(\tfrac12\right)^{2}},\,-\dfrac1{\tfrac12}\right)=\left(-2,\,-2\right)$$

Thus Statement-2 supplies the general fact from which Statement-1 follows as a direct special case. Therefore Statement-2 is indeed a correct explanation for Statement-1.

Hence, the correct answer is Option B.

We are given two conics: $$x^2 = 6y$$ (a parabola) and $$2x^2 - 4y^2 = 9$$ (a hyperbola). We need to find a common tangent from the given options. A tangent line touches a conic at exactly one point, so we will substitute each line into both conics and set the discriminant to zero for tangency.

First, rewrite the options in slope-intercept form $$y = mx + c$$ for easier substitution:

• Option A: $$x - y = \frac{3}{2}$$ → $$y = x - \frac{3}{2}$$ (slope $$m = 1$$, intercept $$c = -\frac{3}{2}$$)
• Option B: $$x + y = 1$$ → $$y = -x + 1$$ (slope $$m = -1$$, intercept $$c = 1$$)
• Option C: $$x + y = \frac{9}{2}$$ → $$y = -x + \frac{9}{2}$$ (slope $$m = -1$$, intercept $$c = \frac{9}{2}$$)
• Option D: $$x - y = 1$$ → $$y = x - 1$$ (slope $$m = 1$$, intercept $$c = -1$$)

Start with the parabola $$x^2 = 6y$$. Substitute $$y = mx + c$$:

$$x^2 = 6(mx + c)$$

$$x^2 = 6mx + 6c$$

$$x^2 - 6mx - 6c = 0$$

For tangency, the discriminant must be zero. Discriminant $$D = (-6m)^2 - 4(1)(-6c) = 36m^2 + 24c$$. Set $$D = 0$$:

$$36m^2 + 24c = 0$$

$$36m^2 = -24c$$

$$c = -\frac{36m^2}{24} = -\frac{3m^2}{2}$$

So, for the parabola, the condition is $$c = -\frac{3}{2}m^2$$.

Check each option against this condition:

• Option A: $$m = 1$$, $$c = -\frac{3}{2}$$. Compute $$-\frac{3}{2}(1)^2 = -\frac{3}{2}$$. Matches.
• Option B: $$m = -1$$, $$c = 1$$. Compute $$-\frac{3}{2}(-1)^2 = -\frac{3}{2}$$. But $$1 \neq -\frac{3}{2}$$, does not match.
• Option C: $$m = -1$$, $$c = \frac{9}{2}$$. Compute $$-\frac{3}{2}(-1)^2 = -\frac{3}{2}$$. But $$\frac{9}{2} \neq -\frac{3}{2}$$, does not match.
• Option D: $$m = 1$$, $$c = -1$$. Compute $$-\frac{3}{2}(1)^2 = -\frac{3}{2}$$. But $$-1 \neq -\frac{3}{2}$$, does not match.

Only Option A satisfies the tangency condition for the parabola.

Now, check if Option A is also tangent to the hyperbola $$2x^2 - 4y^2 = 9$$. Substitute $$y = x - \frac{3}{2}$$:

$$2x^2 - 4\left(x - \frac{3}{2}\right)^2 = 9$$

First, expand $$\left(x - \frac{3}{2}\right)^2 = x^2 - 3x + \frac{9}{4}$$:

$$2x^2 - 4\left(x^2 - 3x + \frac{9}{4}\right) = 9$$

$$2x^2 - 4x^2 + 12x - 9 = 9$$ (since $$4 \times \frac{9}{4} = 9$$)

$$-2x^2 + 12x - 9 = 9$$

Bring all terms to one side:

$$-2x^2 + 12x - 9 - 9 = 0$$

$$-2x^2 + 12x - 18 = 0$$

Multiply by $$-1$$ to simplify:

$$2x^2 - 12x + 18 = 0$$

Divide by 2:

$$x^2 - 6x + 9 = 0$$

Discriminant $$D = (-6)^2 - 4(1)(9) = 36 - 36 = 0$$. Since discriminant is zero, there is exactly one solution, so the line is tangent to the hyperbola.

Thus, Option A is tangent to both conics. For completeness, we can verify that the other options are not tangent to the hyperbola (though they already fail the parabola condition):

• Option B: Substitute $$y = -x + 1$$ into hyperbola: $$2x^2 - 4(-x + 1)^2 = 9$$. Expand $$(-x + 1)^2 = x^2 - 2x + 1$$, so $$2x^2 - 4(x^2 - 2x + 1) = 9$$ → $$2x^2 - 4x^2 + 8x - 4 = 9$$ → $$-2x^2 + 8x - 13 = 0$$. Discriminant $$8^2 - 4(-2)(-13) = 64 - 104 = -40 < 0$$, no real solution.
• Option C: Substitute $$y = -x + \frac{9}{2}$$ into hyperbola: $$2x^2 - 4\left(-x + \frac{9}{2}\right)^2 = 9$$. Expand $$\left(-x + \frac{9}{2}\right)^2 = x^2 - 9x + \frac{81}{4}$$, so $$2x^2 - 4\left(x^2 - 9x + \frac{81}{4}\right) = 9$$ → $$2x^2 - 4x^2 + 36x - 81 = 9$$ → $$-2x^2 + 36x - 90 = 0$$. Multiply by $$-1$$: $$2x^2 - 36x + 90 = 0$$ → divide by 2: $$x^2 - 18x + 45 = 0$$. Discriminant $$(-18)^2 - 4(1)(45) = 324 - 180 = 144 > 0$$, two solutions.
• Option D: Substitute $$y = x - 1$$ into hyperbola: $$2x^2 - 4(x - 1)^2 = 9$$. Expand $$(x - 1)^2 = x^2 - 2x + 1$$, so $$2x^2 - 4(x^2 - 2x + 1) = 9$$ → $$2x^2 - 4x^2 + 8x - 4 = 9$$ → $$-2x^2 + 8x - 13 = 0$$. Discriminant same as Option B: $$64 - 104 = -40 < 0$$, no real solution.

Hence, only Option A is a common tangent to both conics. So, the answer is Option A.

The given parabola is $$x^2 = 8y$$.
Comparing with the standard form $$x^2 = 4ay$$, we get $$4a = 8 \Rightarrow a = 2$$.

Hence:
Vertex $$V(0,0)$$,
Focus $$S(0,a) = (0,2)$$,
Axis is the $$y$$-axis.

The latus rectum is the line through the focus perpendicular to the axis, i.e. $$y = a = 2$$.
Its extremities are obtained by putting $$y = 2$$ in $$x^2 = 8y$$:

$$x^2 = 8(2) \Rightarrow x^2 = 16 \Rightarrow x = \pm4.$$
Therefore the extremities are $$L_1(4,2)$$ and $$L_2(-4,2)$$.

The required triangle has vertices $$V(0,0),\,L_1(4,2),\,L_2(-4,2)$$.
Using the coordinate‐geometry area formula for a triangle with vertices $$(x_1,y_1),(x_2,y_2),(x_3,y_3)$$:

$$\text{Area} = \frac12 \left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.$$
Substituting $$V(0,0),\,L_1(4,2),\,L_2(-4,2)$$:

$$\text{Area} = \frac12 \left|0(2-2)+4(2-0)+(-4)(0-2)\right|$$
$$= \frac12 \left|0 + 8 + 8\right|$$
$$= \frac12 (16)$$
$$= 8.$$

Hence, the area of the triangle is $$8$$.

Option B which is: $$8$$

The parabola is $$x^2 = 8y$$, i.e. $$y = \frac{x^2}{8}$$.

For any differentiable curve, the slope of the tangent at a point is $$\frac{dy}{dx}$$. Differentiate:

$$2x = 8\frac{dy}{dx} \;\; \Longrightarrow \;\; \frac{dy}{dx} = \frac{x}{4}$$

At the specified abscissa $$x = 4$$, the ordinate is

$$y = \frac{4^2}{8} = \frac{16}{8} = 2$$

Thus the point of tangency (and normality) is $$(4,\,2)$$. The tangent slope there is

$$m_{\text{tangent}} = \left.\frac{dy}{dx}\right|_{x=4} = \frac{4}{4} = 1$$

The normal is perpendicular to the tangent, so

$$m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -1$$

Using point-slope form through $$(4,2)$$:

$$y - 2 = -1\,(x - 4)$$

Simplify:

$$y - 2 = -x + 4 \;\; \Longrightarrow \;\; x + y = 6$$

Therefore, the required normal is Option D which is: $$x + y = 6$$.

The ellipse is $$\frac{x^2}{16}+\frac{y^2}{3}=1$$ and the given point on it is $$P\left(2,\frac32\right)$$.

1. Slope of the tangent at $$P$$
Differentiate the ellipse implicitly:
$$\frac{x^2}{16}+\frac{y^2}{3}=1 \;\Rightarrow\; \frac{x}{8}+\frac{2y}{3}\,\frac{dy}{dx}=0$$
$$\therefore\;\frac{dy}{dx}= -\,\frac{3x}{16y}$$

At $$P(2,\tfrac32)$$,
$$m_{\text{tangent}} = -\,\frac{3(2)}{16\left(\tfrac32\right)} = -\,\frac{6}{24} = -\frac14$$

2. Equation of the normal at $$P$$
Slope of the normal: $$m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}=4$$
Using point-slope form:
$$y-\frac32 = 4(x-2)$$
$$\Rightarrow\; y = 4x - \frac{13}{2}$$
This is the line which must be a tangent to the required parabola.

3. Slope form of a tangent to a standard parabola
• For $$y^2 = 4ax$$ (opening to the right): the tangent with slope $$m$$ is $$y = mx + \frac{a}{m}$$.
• For $$y^2 = -4ax$$ (opening to the left): the tangent with slope $$m$$ is $$y = mx - \frac{a}{m}$$.

Our line has slope $$m = 4$$ and intercept $$c = -\frac{13}{2} = -6.5$$. Comparing:

• If the parabola were $$y^2 = 4ax$$: intercept should be $$\frac{a}{m} = \frac{a}{4}$$, which is positive. Our intercept is negative, so this case is impossible.

• If the parabola is $$y^2 = -4ax$$: intercept should be $$-\frac{a}{m} = -\frac{a}{4}$$.
Set $$-\frac{a}{4} = -6.5 \Longrightarrow a = 26$$.

Hence $$4a = 104$$ and the parabola is $$y^2 = -104x$$.

4. Checking the options
Option A $$y^2 = -104x$$ matches exactly.
The other three options do not give the required intercept for $$m = 4$$.

Therefore, the parabola is represented by:

Option A which is: $$y^2 = -104x$$

Let the parabola be $$y^{2}=4x$$, i.e. $$a = 1$$.

Using the slope form, the equation of a tangent to $$y^{2}=4ax$$ with slope $$m$$ is
$$y = mx + \frac{a}{m} \;.$$

For our parabola $$a = 1$$, so every tangent can be written as
$$y = mx + \frac{1}{m}\;.\tag{-1}$$

Suppose a point $$P(h,k)$$ lies outside the curve and two tangents can be drawn from it. Because each tangent passes through $$P$$, substitute $$x = h,\; y = k$$ in $$-(1)$$:

$$k = mh + \frac{1}{m}\;.$$

Re-arrange into a quadratic in $$m$$:

$$mh - k + \frac{1}{m} = 0 \;\;\Longrightarrow\;\; h m^{2} - k m + 1 = 0\;.\tag{-2}$$

The roots $$m_{1}, m_{2}$$ of $$-(2)$$ are the slopes of the two tangents. For two lines to be perpendicular, the product of their slopes must equal $$-1$$:

$$m_{1}m_{2} = -1\;.\tag{-3}$$

From a quadratic $$Am^{2}+Bm+C=0$$, the product of its roots is $$\dfrac{C}{A}$$. For $$-(2)$$ we have $$A = h,\; C = 1$$, hence

$$m_{1}m_{2} = \frac{1}{h}\;.\tag{-4}$$

Equating $$-(3)$$ and $$-(4)$$:

$$\frac{1}{h} = -1 \;\;\Longrightarrow\;\; h = -1\;.$$

Thus every point $$P(h,k)$$ from which the two tangents are at right angles must satisfy $$h = -1$$. Therefore the locus of $$P$$ is

$$x = -1\;.$$

Option B which is: $$x = -1$$

A parabola is the locus of a point $$P(x,y)$$ that remains equidistant from a fixed point (focus) and a fixed straight line (directrix).

Here the focus is the origin $$F(0,0)$$ and the directrix is the vertical line $$x = 2$$.

Distance of $$P(x,y)$$ from the focus:
$$PF = \sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2 + y^2}$$

Perpendicular distance of $$P(x,y)$$ from the directrix $$x = 2$$:
$$PD = |x - 2|$$

Equating the two distances, because every point on a parabola satisfies $$PF = PD$$:
$$\sqrt{x^2 + y^2} = |x - 2|$$

Square both sides (squaring eliminates the absolute value as well):
$$x^2 + y^2 = (x - 2)^2$$

Expand the right-hand side:
$$x^2 + y^2 = x^2 - 4x + 4$$

Simplify (the $$x^2$$ terms cancel):
$$y^2 = -4x + 4 = 4(1 - x)$$

Rewrite to highlight the standard horizontal-axis form $$ (y - k)^2 = 4p(x - h) $$:
$$y^2 = -4(x - 1)$$

Comparing with $$ (y - 0)^2 = 4p(x - 1) $$ gives $$h = 1,\; k = 0$$.
Thus the vertex $$V(h,k)$$ of the parabola is $$V(1,0)$$.

Therefore, the vertex is at (1, 0).

Option B which is: (1, 0)

The standard form of the given parabola is $$y^{2}=4ax$$.

Comparing $$y^{2}=8x$$ with $$y^{2}=4ax$$ gives $$a=2$$.

For $$y^{2}=4ax$$ the tangent in slope form is

$$y = mx + \frac{a}{m} \qquad -(1)$$

1. Tangent already given
    The equation $$y = x + 2$$ has slope $$m_1 = 1$$.
    With $$m = 1$$, (1) gives $$c = \dfrac{a}{m}= \dfrac{2}{1}=2$$, so $$y = x + 2$$ is indeed a tangent.

2. Slope of the required second tangent
    The second tangent must be perpendicular to the first.
    If $$m_1 = 1$$, then $$m_2 = -\dfrac{1}{m_1} = -1$$.

3. Equation of the perpendicular tangent
    Put $$m = -1$$ in (1):

$$y = (-1)x + \frac{2}{-1} \; \Rightarrow \; y = -x - 2 \qquad -(2)$$

4. Point of intersection of the two tangents
    Solve $$y = x + 2$$ and $$y = -x - 2$$ simultaneously:

$$x + 2 = -x - 2 \;\Longrightarrow\; 2x = -4 \;\Longrightarrow\; x = -2$$

With $$x = -2$$, $$y = x + 2 = 0$$.

Thus the point common to both tangents (and lying on the given tangent) is $$(-2,0)$$.

Hence, the required point is $$(-2,0)$$.

Option D which is: $$(-2, 0)$$

The equations of the two given parabolas are:

$$ y^2 = 4ax $$

$$ x^2 = 4ay $$

To find their points of intersection, express $$ y $$ from the second equation:

$$ y = \frac{x^2}{4a} $$

Substitute this expression for $$ y $$ into the first parabola equation:

$$ \left(\frac{x^2}{4a}\right)^2 = 4ax $$

$$ \frac{x^4}{16a^2} = 4ax $$

Multiply both sides by the denominator to clear the fraction:

$$ x^4 = 64a^3x $$

Move all terms to one side to solve the polynomial equation:

$$ x^4 - 64a^3x = 0 $$

$$ x(x^3 - 64a^3) = 0 $$

This equation gives two possible solutions for $$ x $$:

$$ x = 0 $$

$$ x^3 = 64a^3 $$

$$ x = 4a $$

Now substitute these $$ x $$ values back to find the corresponding $$ y $$ coordinates.

Case 1: When $$ x = 0 $$, the value of $$ y $$ is:

$$ y = \frac{0^2}{4a} = 0 $$

So, the first intersection point is the origin:

$$ (0, 0) $$

Case 2: When $$ x = 4a $$, the value of $$ y $$ is:

$$ y = \frac{(4a)^2}{4a} = 4a $$

So, the second intersection point is:

$$ (4a, 4a) $$

The problem states that the straight line passes through the points of intersection:

$$ 2bx + 3cy + 4d = 0 $$

Since the line passes through the first point $$ (0, 0) $$, substitute these coordinates into the line equation:

$$ 2b(0) + 3c(0) + 4d = 0 $$

$$ 4d = 0 $$

$$ d = 0 $$

Since the line also passes through the second point $$ (4a, 4a) $$, substitute these coordinates and the value of $$ d $$ into the line equation:

$$ 2b(4a) + 3c(4a) + 4(0) = 0 $$

$$ 8ab + 12ac = 0 $$

Factor out the common multiplier from the terms:

$$ 4a(2b + 3c) = 0 $$

Since the problem explicitly states that $$ a \neq 0 $$, we can safely divide both sides by $$ 4a $$:

$$ 2b + 3c = 0 $$

Final Answer:

from options , $$d^2 + (2b + 3c)^2 = 0$$ is correct

To find the equation of the parabola, we can use the fundamental geometric properties of parabolas concerning tangents and the directrix.

The focus of the parabola is given as $$S(2, 1)$$, and the two given tangents are:

$$L_1: x - y = 0$$

$$L_2: x + y = 0$$

Notice that the two tangents are perpendicular to each other because the product of their slopes is $$(1)(-1) = -1$$.

A standard property of parabolas states that the locus of the intersection of orthogonal (perpendicular) tangents is the directrix of the parabola.

Solving the equations of the two tangents simultaneously:

$$x - y = 0 \implies x = y$$

Substituting into the second equation $$x + y = 0$$ gives $$2x = 0$$, so $$x = 0$$ and $$y = 0$$.

Thus, the intersection point of the perpendicular tangents is $$(0, 0)$$, which must lie on the directrix.

Another key property of a parabola is that the reflection of the focus with respect to any tangent line always lies on the directrix.

Let us find the reflection of the focus $$S(2, 1)$$ with respect to the first tangent $$x - y = 0$$:

$$\frac{x_1 - 2}{1} = \frac{y_1 - 1}{-1} = \frac{-2(2 - 1)}{1^2 + (-1)^2} = \frac{-2(1)}{2} = -1$$

From this, we get a point on the directrix:

$$x_1 - 2 = -1 \implies x_1 = 1$$

$$y_1 - 1 = 1 \implies y_1 = 2$$

So, $$P_1(1, 2)$$ lies on the directrix.

Now, let us find the reflection of the focus $$S(2, 1)$$ with respect to the second tangent $$x + y = 0$$:

$$\frac{x_2 - 2}{1} = \frac{y_2 - 1}{1} = \frac{-2(2 + 1)}{1^2 + 1^2} = \frac{-2(3)}{2} = -3$$

From this, we get a second point on the directrix:

$$x_2 - 2 = -3 \implies x_2 = -1$$

$$y_2 - 1 = -3 \implies y_2 = -2$$

So, $$P_2(-1, -2)$$ also lies on the directrix.

Since the directrix is a straight line passing through the origin $$(0, 0)$$ and the point $$(1, 2)$$, we can find its slope $$m$$:

$$m = \frac{2 - 0}{1 - 0} = 2$$

Thus, the equation of the directrix passing through $$(0, 0)$$ with a slope of $$2$$ is:

$$y = 2x \implies 2x - y = 0$$

A parabola is defined as the locus of a point $$P(x, y)$$ whose distance from the focus $$S(2, 1)$$ is equal to its perpendicular distance from the directrix $$2x - y = 0$$:

$$\sqrt{(x - 2)^2 + (y - 1)^2} = \frac{\vert{}2x - y\vert{}}{\sqrt{2^2 + (-1)^2}}$$

Squaring both sides:

$$(x - 2)^2 + (y - 1)^2 = \frac{(2x - y)^2}{5}$$

$$5 \left[ (x^2 - 4x + 4) + (y^2 - 2y + 1) \right] = 4x^2 - 4xy + y^2$$

$$5x^2 + 5y^2 - 20x - 10y + 25 = 4x^2 - 4xy + y^2$$

Rearranging and combining like terms gives the equation of the parabola:

$$x^2 + 4xy + 4y^2 - 20x - 10y + 25 = 0$$

The correct option is B.

The general equation of the parabola is $$y = x^2 + px + q$$.

Because the parabola passes through $$(1,-1)$$, substitute $$x = 1,\, y = -1$$:

$$-1 = 1 + p + q \;\;\Longrightarrow\;\; q = -2 - p \;-(1)$$

The vertex of $$y = x^2 + px + q$$ lies at

$$x_v = -\frac{p}{2}, \qquad y_v = -\frac{p^2}{4} + q \;-(2)$$

Using $$(1)$$ in $$(2)$$, express the y-coordinate of the vertex only in terms of $$p$$:

$$y_v = -\frac{p^2}{4} + (-2 - p) = -\frac{p^2}{4} - p - 2 = -\frac{p^2 + 4p + 8}{4}$$

Distance of the vertex from the x-axis is the absolute value of $$y_v$$:

$$D(p) = \left| -\frac{p^2 + 4p + 8}{4} \right| = \frac{p^2 + 4p + 8}{4}$$

Minimising $$D(p)$$ is equivalent to minimising the quadratic numerator

$$f(p) = p^2 + 4p + 8$$

The minimum of a quadratic $$ap^2 + bp + c$$ occurs at $$p = -\frac{b}{2a}$$. Here, $$a = 1,\; b = 4$$, so

$$p_{\text{min}} = -\frac{4}{2} = -2$$

With $$p = -2$$, calculate $$q$$ from $$(1)$$:

$$q = -2 - (-2) = 0$$

Finally, evaluate $$p^2 + q^2$$:

$$p^2 + q^2 = (-2)^2 + 0^2 = 4$$

Therefore, the required value is $$4$$.

Option B which is: $$4$$

The parabola is given by $$y^2 = 8x$$. For a parabola $$y^2 = 4ax$$, the focus is at $$(a, 0)$$. Here, $$4a = 8$$, so $$a = 2$$. Thus, the focus A is at $$(2, 0)$$.

The line $$y = mx + c$$ intersects the parabola at two distinct points B and C. Substituting $$y = mx + c$$ into $$y^2 = 8x$$:

$$(mx + c)^2 = 8x$$

$$m^2 x^2 + 2mcx + c^2 = 8x$$

$$m^2 x^2 + (2mc - 8)x + c^2 = 0 \quad \text{(1)}$$

This quadratic equation in $$x$$ has roots $$x_1$$ and $$x_2$$, the x-coordinates of B and C.

The centroid of triangle ABC is given as $$\left(\frac{7}{3}, \frac{4}{3}\right)$$. Using the centroid formula:

$$\frac{2 + x_1 + x_2}{3} = \frac{7}{3} \quad \Rightarrow \quad 2 + x_1 + x_2 = 7 \quad \Rightarrow \quad x_1 + x_2 = 5 \quad \text{(2)}$$

$$\frac{0 + y_1 + y_2}{3} = \frac{4}{3} \quad \Rightarrow \quad y_1 + y_2 = 4 \quad \text{(3)}$$

Since B and C lie on the line $$y = mx + c$$, $$y_1 = mx_1 + c$$ and $$y_2 = mx_2 + c$$. Using equation (3):

$$mx_1 + c + mx_2 + c = 4 \quad \Rightarrow \quad m(x_1 + x_2) + 2c = 4$$

Substituting $$x_1 + x_2 = 5$$ from equation (2):

$$m(5) + 2c = 4 \quad \Rightarrow \quad 5m + 2c = 4 \quad \text{(4)}$$

From the quadratic equation (1), the sum of roots is:

$$x_1 + x_2 = -\frac{2mc - 8}{m^2}$$

Using equation (2):

$$5 = -\frac{2mc - 8}{m^2}$$

$$5m^2 = -(2mc - 8)$$

$$5m^2 = -2mc + 8$$

$$5m^2 + 2mc - 8 = 0 \quad \text{(5)}$$

Solving equations (4) and (5) simultaneously. From equation (4):

$$c = \frac{4 - 5m}{2} \quad \text{(6)}$$

Substituting into equation (5):

$$5m^2 + 2m\left(\frac{4 - 5m}{2}\right) - 8 = 0$$

$$5m^2 + m(4 - 5m) - 8 = 0$$

$$5m^2 + 4m - 5m^2 - 8 = 0$$

$$4m - 8 = 0$$

$$4m = 8 \quad \Rightarrow \quad m = 2$$

Substituting $$m = 2$$ into equation (6):

$$c = \frac{4 - 5(2)}{2} = \frac{4 - 10}{2} = \frac{-6}{2} = -3$$

Thus, the line is $$y = 2x - 3$$.

Finding intersection points with the parabola $$y^2 = 8x$$:

$$(2x - 3)^2 = 8x$$

$$4x^2 - 12x + 9 = 8x$$

$$4x^2 - 20x + 9 = 0$$

The discriminant is $$(-20)^2 - 4 \cdot 4 \cdot 9 = 400 - 144 = 256$$.

$$x = \frac{20 \pm \sqrt{256}}{8} = \frac{20 \pm 16}{8}$$

$$x_1 = \frac{20 + 16}{8} = \frac{36}{8} = \frac{9}{2}, \quad x_2 = \frac{20 - 16}{8} = \frac{4}{8} = \frac{1}{2}$$

Corresponding y-coordinates:

$$y_1 = 2 \cdot \frac{9}{2} - 3 = 9 - 3 = 6, \quad y_2 = 2 \cdot \frac{1}{2} - 3 = 1 - 3 = -2$$

Thus, points B and C are $$\left(\frac{9}{2}, 6\right)$$ and $$\left(\frac{1}{2}, -2\right)$$.

Finding $$(BC)^2$$, the square of the distance between B and C:

$$BC = \sqrt{\left(\frac{9}{2} - \frac{1}{2}\right)^2 + (6 - (-2))^2} = \sqrt{(4)^2 + (8)^2} = \sqrt{16 + 64} = \sqrt{80}$$

$$(BC)^2 = 80$$

The value of $$(BC)^2$$ is 80, which corresponds to option C.

$$y^2 = 16x \implies 4a = 16 \implies a = 4$$

Any parametric point on this parabola is given by $$(at^2, 2at) = (4t^2, 8t)$$.

The fixed vertex containing the right angle is $$B(4, 8)$$. For $$B$$, $$8t = 8 \implies t = 1$$.

Let the other two variable vertices be: $$A = (4t_1^2, 8t_1)$$, $$C = (4t_2^2, 8t_2)$$

The slope of a line joining two points $$(4t_i^2, 8t_i)$$ and $$(4t_j^2, 8t_j)$$ on the parabola $$y^2 = 16x$$ is given by:

$$m = \frac{8t_i - 8t_j}{4t_i^2 - 4t_j^2} = \frac{2}{t_i + t_j}$$

$$m_{AB} = \frac{2}{t_1 + 1}$$

$$m_{BC} = \frac{2}{t_2 + 1}$$

Since $$\angle ABC = 90^\circ$$, we have $$m_{AB} \cdot m_{BC} = -1$$: $$\left(\frac{2}{t_1 + 1}\right) \cdot \left(\frac{2}{t_2 + 1}\right) = -1$$

$$4 = -(t_1 + 1)(t_2 + 1) \implies t_1 t_2 + t_1 + t_2 + 5 = 0$$

Let the coordinates of the centroid of $$\triangle ABC$$ be $$G(x, y)$$.

$$x = \frac{4t_1^2 + 4t_2^2 + 4}{3} \implies 3x - 4 = 4(t_1^2 + t_2^2)$$

$$y = \frac{8t_1 + 8t_2 + 8}{3} \implies 3y - 8 = 8(t_1 + t_2) \implies t_1 + t_2 = \frac{3y - 8}{8}$$

$$t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1 t_2$$

$$t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2[-(t_1 + t_2 + 5)]$$

$$t_1^2 + t_2^2 = (t_1 + t_2)^2 + 2(t_1 + t_2) + 10$$

$$3x - 4 = 4 \left[ (t_1 + t_2)^2 + 2(t_1 + t_2) + 10 \right]$$

$$3x - 4 = 4 \left[ \left(\frac{3y - 8}{8}\right)^2 + 2\left(\frac{3y - 8}{8}\right) + 10 \right]$$

$$3x - 4 = 4 \left[ \frac{(3y - 8)^2}{64} + \frac{3y - 8}{4} + 10 \right]$$

$$16(3x - 44) = (3y - 8)^2 + 16(3y - 8)$$

$$(3y - 8)^2 + 16(3y - 8) = 48x - 704$$

$$(3y - 8 + 8)^2 - 64 = 48x - 704$$

$$y^2 = \frac{48}{9}\left(x - \frac{40}{3}\right) \implies y^2 = \frac{16}{3}\left(x - \frac{40}{3}\right)$$

$$\text{L.R.} = \frac{16}{3}$$

$$3 \times \text{L.R.} = 3 \times \frac{16}{3} = 16$$

We are asked to find the image of the parabola $$x^2 = 4y$$ in the line $$x - y = 1$$ and express it in the form $$(y + a)^2 = b(x - c)$$, where $$a, b, c \in \mathbb{N}$$, then compute $$a + b + c$$.

For reflecting a point $$(h, k)$$ in the line $$x - y - 1 = 0$$, one uses the formulas $$\frac{h' - h}{1} = \frac{k' - k}{-1} = \frac{-2(h - k - 1)}{1^2 + (-1)^2}$$. From these, simplifying gives $$h' - h = -(h - k - 1) = k - h + 1$$, so $$h' = k + 1$$, and $$k' - k = (h - k - 1)$$, so $$k' = h - 1$$.

Rewriting these relations to express the original coordinates in terms of the reflected coordinates yields $$h = k' + 1$$ and $$k = h' - 1$$.

Since the original parabola satisfies $$h^2 = 4k$$, substituting the expressions for $$h$$ and $$k$$ gives $$(k' + 1)^2 = 4(h' - 1)$$. Renaming $$h'$$ as $$x$$ and $$k'$$ as $$y$$, this becomes $$(y + 1)^2 = 4(x - 1)$$.

Comparing $$(y + 1)^2 = 4(x - 1)$$ with the desired form $$(y + a)^2 = b(x - c)$$, we see that $$a = 1$$, $$b = 4$$, and $$c = 1$$. All of these are natural numbers, so $$a + b + c = 1 + 4 + 1 = 6$$.

The correct answer is Option (2): 6.

$$4x^2 = x^2 + 27 \implies 3x^2 = 27 \implies x = \pm 3$$.

$$A_1 = \int_{-3}^{3} (x^2 + 27 - 4x^2) dx = \int_{-3}^{3} (27 - 3x^2) dx = [27x - x^3]_{-3}^{3} = 54 - (-54) = 108$$

Area between line and parabola ($$A_2$$).

Intersection: $$4x^2 = \alpha x \implies x = 0, x = \alpha/4$$.

$$A_2 = \int_{0}^{\alpha/4} (\alpha x - 4x^2) dx = \left[ \frac{\alpha x^2}{2} - \frac{4x^3}{3} \right]_{0}^{\alpha/4} = \frac{\alpha^3}{32} - \frac{\alpha^3}{48} = \frac{\alpha^3}{96}$$

Set up the ratio.

$$A_1 = 6 \cdot A_2 \implies 108 = 6 \left( \frac{\alpha^3}{96} \right) \implies 108 = \frac{\alpha^3}{16}$$

$$\alpha^3 = 108 \times 16 = 1728 \implies \alpha = \sqrt[3]{1728} = 12$$.

The two parabolas are mirror images of each other across the line $$y = x$$, because inter-changing $$x$$ and $$y$$ in $$y = x^{2}+2$$ gives $$x = y^{2}+2$$. Hence the centre of the smallest circle tangent to both parabolas must lie on the line of symmetry $$y = x$$.
Let the centre be $$C(a,a)$$ and the required radius be $$r$$.

Take an arbitrary point $$P(x,\,x^{2}+2)$$ on the first parabola $$y = x^{2}+2$$.
The squared distance between $$C(a,a)$$ and $$P(x,x^{2}+2)$$ is

$$D^{2}(x,a)= (x-a)^{2}+\left(x^{2}+2-a\right)^{2}\quad -(1)$$

For the circle to be tangent, $$P$$ must be the point on the parabola that is closest to the centre. Thus, for a fixed $$a$$ we first minimise $$D^{2}(x,a)$$ with respect to $$x$$.

Differentiate $$D^{2}$$ w.r.t. $$x$$ and set the result to zero:

$$\frac{\partial D^{2}}{\partial x}=2(x-a)+2\!\left(x^{2}+2-a\right)(2x)=0$$
$$\Longrightarrow\;(x-a)+2x\!\left(x^{2}+2-a\right)=0\quad -(2)$$

Next, we must choose that particular centre $$C(a,a)$$ which makes this minimal distance itself as small as possible. Therefore we also differentiate $$D^{2}(x,a)$$ with respect to $$a$$ and set it to zero:

$$\frac{\partial D^{2}}{\partial a}=-2(x-a)-2\!\left(x^{2}+2-a\right)=0$$
$$\Longrightarrow\;(a-x)+(a-x^{2}-2)=0$$
$$\Longrightarrow\;2a=x+x^{2}+2\quad -(3)$$

Solve equations $$(2)$$ and $$(3)$$ simultaneously.

First write the differences that appear in $$(2)$$ using $$(3)$$:
$$x-a=\frac{-x^{2}+x-2}{2},\qquad x^{2}+2-a=\frac{x^{2}-x+2}{2}$$

Substituting these into $$(2)$$:

$$\frac{-x^{2}+x-2}{2}+2x\!\left(\frac{x^{2}-x+2}{2}\right)=0$$
$$\Longrightarrow\;-x^{2}+x-2+2x\!\left(x^{2}-x+2\right)=0$$
$$\Longrightarrow\;2x^{3}-3x^{2}+5x-2=0$$

Using the Rational Root Theorem, $$x=\tfrac12$$ is a root; dividing out gives
$$(x-\tfrac12)(2x^{2}-2x+4)=0$$
The quadratic factor has negative discriminant, so the only real solution is

$$x=\frac12$$

Insert $$x=\frac12$$ into $$(3)$$ to find the centre:

$$2a=\frac12+\left(\frac12\right)^{2}+2=\frac12+\frac14+2=\frac{11}{4}$$
$$\Longrightarrow\;a=\frac{11}{8}$$

Hence the centre of the required circle is
$$C\!\left(\frac{11}{8},\frac{11}{8}\right)$$

Now compute the radius. Using $$x=\tfrac12$$ and $$a=\tfrac{11}{8}$$ in $$(1)$$:

$$x-a=\frac12-\frac{11}{8}=-\frac78,\qquad x^{2}+2-a=\frac14+2-\frac{11}{8}=+\frac78$$
$$r^{2}=(x-a)^{2}+\left(x^{2}+2-a\right)^{2}=\left(\frac78\right)^{2}+\left(\frac78\right)^{2}=\frac{49}{32}$$
$$\Longrightarrow\;r=\sqrt{\frac{49}{32}}=\frac{7}{4\sqrt2}=\frac{7\sqrt2}{8}$$

The radius of the smallest circle touching both parabolas is therefore
$$\boxed{\dfrac{7\sqrt{2}}{8}}$$

Hence, the correct option is Option D.

We need to find $$\alpha + \beta + \gamma$$ for the parabola with given vertex and directrix.

The vertex is $$V\left(\frac{3}{2}, 3\right)$$ and the directrix is $$x + 2y = 0$$.

Since the axis of the parabola is perpendicular to the directrix and passes through the vertex, and because the directrix $$x + 2y = 0$$ has slope $$-1/2$$, it follows that the axis has slope 2, i.e., direction vector $$(1,2)$$.

Next, the distance from the vertex to the directrix is given by $$d = \frac{|3/2 + 2(3)|}{\sqrt{1^2 + 2^2}} = \frac{|3/2 + 6|}{\sqrt{5}} = \frac{15/2}{\sqrt{5}} = \frac{15}{2\sqrt{5}}$$.

Since the focus is on the axis, on the opposite side of the vertex from the directrix, at the same distance $$d$$, the unit vector along the axis (away from the directrix) is $$\frac{(1,2)}{\sqrt{5}}$$, so

$$F = V + d \cdot \frac{(1,2)}{\sqrt{5}} = \left(\frac{3}{2}, 3\right) + \frac{15}{2\sqrt{5}} \cdot \frac{(1,2)}{\sqrt{5}} = \left(\frac{3}{2}, 3\right) + \frac{15}{10}(1,2) = \left(\frac{3}{2}+\frac{3}{2}, 3+3\right) = (3, 6)$$

Using the definition of a parabola as the locus of points equidistant from the focus and directrix leads to

$$\sqrt{(x-3)^2 + (y-6)^2} = \frac{|x+2y|}{\sqrt{5}}$$

Squaring both sides gives

$$ (x-3)^2 + (y-6)^2 = \frac{(x+2y)^2}{5} $$

and hence

$$ 5[(x-3)^2 + (y-6)^2] = (x+2y)^2. $$

Expanding this equation results in

$$5(x^2 - 6x + 9 + y^2 - 12y + 36) = x^2 + 4xy + 4y^2$$

which simplifies to

$$5x^2 - 30x + 45 + 5y^2 - 60y + 180 = x^2 + 4xy + 4y^2$$

and therefore

$$4x^2 + y^2 - 4xy - 30x - 60y + 225 = 0$$.

Comparing this with the given form $$\alpha x^2 + \beta y^2 - \gamma xy - 30x - 60y + 225 = 0$$ shows that $$\alpha = 4$$, $$\beta = 1$$, and $$\gamma = 4$$, so $$\alpha + \beta + \gamma = 4 + 1 + 4 = 9$$.

The correct answer is Option 2: 9.

The definition of a parabola says that for every point $$P(x,y)$$ on it,

distance $$PF$$ from the focus $$F(-2,1)$$ equals perpendicular distance $$PD$$ to the directrix $$2x + y + 2 = 0$$.

Write both distances using the distance formula.

$$PF = \sqrt{(x+2)^2 + (y-1)^2}$$

The perpendicular distance of $$P(x,y)$$ from the line $$2x + y + 2 = 0$$ is
$$PD = \frac{\lvert 2x + y + 2 \rvert}{\sqrt{2^2 + 1^2}} = \frac{\lvert 2x + y + 2 \rvert}{\sqrt{5}}$$

Equating and squaring (to remove the absolute value):
$$ (x+2)^2 + (y-1)^2 = \frac{(2x + y + 2)^2}{5} $$ $$-(1)$$

The required points have abscissa $$x = -2$$. Substitute $$x = -2$$ in $$(1)$$.

Left side: $$(x+2)^2 = 0$$, so
$$ (y-1)^2 $$

Right side: $$2x + y + 2 = 2(-2) + y + 2 = y - 2$$, hence
$$ \frac{(y-2)^2}{5} $$

Therefore
$$ (y-1)^2 = \frac{(y-2)^2}{5} $$ $$-(2)$$

Cross-multiply:
$$ 5(y-1)^2 = (y-2)^2 $$

Expand both squares:
$$ 5(y^2 - 2y + 1) = y^2 - 4y + 4 $$

Simplify:
$$ 5y^2 - 10y + 5 = y^2 - 4y + 4 $$
$$ 4y^2 - 6y + 1 = 0 $$ $$-(3)$$

Solve the quadratic $$4y^2 - 6y + 1 = 0$$ using the quadratic formula:

$$ y = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4} = \frac{6 \pm \sqrt{36 - 16}}{8} = \frac{6 \pm 2\sqrt{5}}{8} $$

Thus the two ordinates are
$$ y_1 = \frac{3 + \sqrt{5}}{4}, \quad y_2 = \frac{3 - \sqrt{5}}{4} $$

Their sum is
$$ y_1 + y_2 = \frac{3 + \sqrt{5}}{4} + \frac{3 - \sqrt{5}}{4} = \frac{6}{4} = \frac{3}{2} $$

Hence, the sum of the ordinates of the required points is $$\frac{3}{2}$$.

Therefore, Option A is correct.

We have two parabolas with the same focus $$(4,3)$$, where the directrix of the first parabola is the x-axis and the directrix of the second parabola is the y-axis. We need to find $$(AB)^2$$ where $$A$$ and $$B$$ are the intersection points.

For the first parabola with focus $$(4,3)$$ and directrix $$y=0$$, the distance from any point $$(x,y)$$ to the focus equals its distance to the directrix:

$$ \sqrt{(x-4)^2 + (y-3)^2} = |y| $$

Squaring both sides gives

$$ (x-4)^2 + (y-3)^2 = y^2 $$

which expands to

$$ (x-4)^2 + y^2 - 6y + 9 = y^2 $$

and hence

$$ (x-4)^2 = 6y - 9 \quad \cdots (1) $$

For the second parabola with focus $$(4,3)$$ and directrix $$x=0$$, we similarly have

$$ \sqrt{(x-4)^2 + (y-3)^2} = |x| $$

Squaring both sides gives

$$ (x-4)^2 + (y-3)^2 = x^2 $$

which expands to

$$ x^2 - 8x + 16 + (y-3)^2 = x^2 $$

so

$$ (y-3)^2 = 8x - 16 $$

and

$$ (y-3)^2 = 8(x - 2) \quad \cdots (2) $$

At the intersection points the distances to the focus and the two directrices are equal, implying $$|y| = |x|$$. Thus the intersection points lie on $$y = x$$ or $$y = -x$$.

In the case $$y = x$$, substituting into (1) yields

$$ (x-4)^2 = 6x - 9 $$

so

$$ x^2 - 8x + 16 = 6x - 9 $$

and

$$ x^2 - 14x + 25 = 0 $$

giving

$$ x = \frac{14 \pm \sqrt{196 - 100}}{2} = \frac{14 \pm \sqrt{96}}{2} = 7 \pm 2\sqrt{6} $$

Hence the points are $$(7 + 2\sqrt{6},\;7 + 2\sqrt{6})$$ and $$(7 - 2\sqrt{6},\;7 - 2\sqrt{6})$$.

In the case $$y = -x$$, substituting into (1) yields

$$ (x-4)^2 = -6x - 9 $$

so

$$ x^2 - 2x + 25 = 0 $$

The discriminant $$4 - 100 = -96 < 0$$ shows there are no real solutions.

Letting $$A = (7 + 2\sqrt{6},\;7 + 2\sqrt{6})$$ and $$B = (7 - 2\sqrt{6},\;7 - 2\sqrt{6})$$, we compute

$$ (AB)^2 = (4\sqrt{6})^2 + (4\sqrt{6})^2 = 96 + 96 = 192 $$

The correct answer is Option 3: 192.

The given parabola is $$4y = 3x^2$$, which can be rewritten as $$y = \frac{3}{4}x^2$$. The vertex of this parabola is at the origin $$(0,0)$$ since it is in the standard form $$x^2 = \frac{4}{3}y$$, implying $$4a = \frac{4}{3}$$, so $$a = \frac{1}{3}$$.

The line is $$3x - 2y + 12 = 0$$, which can be rewritten as $$y = \frac{3}{2}x + 6$$.

To find the points of intersection A and B, substitute $$y = \frac{3}{4}x^2$$ into the line equation:

$$\frac{3}{4}x^2 = \frac{3}{2}x + 6$$

Multiply both sides by 4 to clear the denominator:

$$3x^2 = 6x + 24$$

Rearrange into standard quadratic form:

$$3x^2 - 6x - 24 = 0$$

Divide by 3:

$$x^2 - 2x - 8 = 0$$

Factorize:

$$(x - 4)(x + 2) = 0$$

So, $$x = 4$$ or $$x = -2$$.

Substitute these x-values into $$y = \frac{3}{2}x + 6$$:

For $$x = 4$$: $$y = \frac{3}{2}(4) + 6 = 6 + 6 = 12$$, so point A is $$(4, 12)$$.

For $$x = -2$$: $$y = \frac{3}{2}(-2) + 6 = -3 + 6 = 3$$, so point B is $$(-2, 3)$$.

The vertex is O$$(0,0)$$. The angle subtended by AB at O is the angle between the lines OA and OB.

The slope of OA is $$m_1 = \frac{12 - 0}{4 - 0} = 3$$.

The slope of OB is $$m_2 = \frac{3 - 0}{-2 - 0} = -\frac{3}{2}$$.

The formula for the angle $$\phi$$ between two lines with slopes $$m_1$$ and $$m_2$$ is:

$$\tan\phi = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$

Substitute $$m_1 = 3$$ and $$m_2 = -\frac{3}{2}$$:

$$m_1 - m_2 = 3 - \left(-\frac{3}{2}\right) = 3 + \frac{3}{2} = \frac{9}{2}$$

$$1 + m_1 m_2 = 1 + (3) \left(-\frac{3}{2}\right) = 1 - \frac{9}{2} = -\frac{7}{2}$$

So,

$$\tan\phi = \left| \frac{\frac{9}{2}}{-\frac{7}{2}} \right| = \left| \frac{9}{2} \times -\frac{2}{7} \right| = \left| -\frac{9}{7} \right| = \frac{9}{7}$$

Thus, $$\phi = \tan^{-1}\left(\frac{9}{7}\right)$$.

The dot product of vectors $$\overrightarrow{OA} = (4, 12)$$ and $$\overrightarrow{OB} = (-2, 3)$$ is $$4 \times (-2) + 12 \times 3 = -8 + 36 = 28 > 0$$, confirming the angle is acute, so $$\tan\phi = \frac{9}{7}$$ is appropriate.

Comparing with the options, $$\tan^{-1}\left(\frac{9}{7}\right)$$ corresponds to option B.

Since the parabola is given by $$y^2 = 4ax$$ and the point $$P(4,4\sqrt{3})$$ lies on it, substituting the coordinates into the equation yields: $$(4\sqrt{3})^2 = 4a \cdot 4$$ which simplifies to $$48 = 16a$$ and hence $$a = 3$$. This gives the parabola as $$y^2 = 12x$$.

The focus of this parabola lies at $$(a,0) = (3,0)$$. Since $$PQ$$ is a focal chord, we parametrize the parabola by $$x = a t^2$$ and $$y = 2a t$$; for $$a=3$$ the coordinates become $$(3t^2,6t)$$.

For the point $$P(4,4\sqrt{3})$$ one has $$3t^2 = 4$$ giving $$t^2 = \frac{4}{3}$$ and $$t = \pm \frac{2}{\sqrt{3}} = \pm \frac{2\sqrt{3}}{3}$$. Substituting the $$y$$-coordinate $$6t = 4\sqrt{3}$$ yields $$t = \frac{2\sqrt{3}}{3}$$, and since the $$y$$-coordinate is positive we take $$t_P = \frac{2\sqrt{3}}{3}$$.

From the condition for focal chords $$t_1 t_2 = -1$$, we obtain for the other end $$t_Q$$ that $$\frac{2\sqrt{3}}{3}\,t_Q = -1$$, so $$t_Q = -\frac{3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2}$$ after rationalizing. Substituting into the parametric form gives $$x_Q = 3\bigl(-\tfrac{\sqrt{3}}{2}\bigr)^2 = \tfrac{9}{4}$$ and $$y_Q = 6\bigl(-\tfrac{\sqrt{3}}{2}\bigr) = -3\sqrt{3}$$, hence $$Q\bigl(\tfrac{9}{4},-3\sqrt{3}\bigr)$$.

The directrix of the parabola $$y^2 = 4ax$$ is $$x = -a$$, so here it is $$x = -3$$. Since this line is vertical, the feet of the perpendiculars from $$P$$ and $$Q$$ to the directrix lie on horizontal lines through those points. Thus the foot from $$P$$ is $$M(-3,4\sqrt{3})$$ and that from $$Q$$ is $$N(-3,-3\sqrt{3})$$.

Considering the quadrilateral $$PQMN$$ with vertices $$P(4,4\sqrt{3})$$, $$Q\bigl(\tfrac{9}{4},-3\sqrt{3}\bigr)$$, $$N(-3,-3\sqrt{3})$$, and $$M(-3,4\sqrt{3})$$, we observe that $$PQ$$ connects to $$Q$$, then to $$N$$, then to $$M$$, and back to $$P$$. The segments $$QN$$ and $$MP$$ are horizontal and parallel. The length of $$QN$$ is $$\bigl|\tfrac{9}{4} - (-3)\bigr| = \tfrac{21}{4}$$, while the length of $$MP$$ is $$|4 - (-3)| = 7$$. The vertical separation between these parallel sides is $$\bigl|4\sqrt{3} - (-3\sqrt{3})\bigr| = 7\sqrt{3}$$.

Since the area of a trapezoid is half the sum of the parallel sides times the height, we have $$ \text{Area} = \tfrac12\bigl(\tfrac{21}{4} + 7\bigr)\times 7\sqrt{3} = \tfrac12\times \tfrac{49}{4}\times 7\sqrt{3} = \tfrac{343\sqrt{3}}{8}, $$ which matches option D.

As a verification, applying the shoelace formula to the vertices in order $$P(4,4\sqrt{3})$$, $$Q(\tfrac{9}{4},-3\sqrt{3})$$, $$N(-3,-3\sqrt{3})$$, $$M(-3,4\sqrt{3})$$, and back to $$P$$ yields the same result, $$ \text{Area} = \tfrac12\Bigl|\sum(x_i y_{i+1} - y_i x_{i+1})\Bigr| = \tfrac{343\sqrt{3}}{8}. $$ Thus, the area of quadrilateral $$PQMN$$ is $$\tfrac{343\sqrt{3}}{8}$$.

Final Answer: The area of quadrilateral PQMN is $$\tfrac{343\sqrt{3}}{8}$$.

For parabola $$y^2 = 4ax$$ (here $$4a = 12$$, so $$a = 3$$), the equation of chord with midpoint $$(h, k)$$ is: $$T = S_1$$ where $$T: ky - 2a(x+h) = 0$$ and $$S_1 = k^2 - 4ah$$.

Substituting $$h=4$$ and $$k=1$$ into $$T$$ gives $$T: 1 \cdot y - 6(x+4) = k^2 - 4ah$$ and hence $$y - 6x - 24 = 1 - 48 = -47$$. Since $$S_1 = k^2 - 12h = 1 - 48 = -47$$, equating $$T = S_1$$ yields $$ky - 6(x+h) = k^2 - 12h$$, so $$1 \cdot y - 6(x+4) = 1 - 48$$ which simplifies to $$y - 6x - 24 = -47$$.

Rearranging gives $$y - 6x = -23$$ and thus $$y = 6x - 23$$.

Testing Option A $$(3, -3)$$ leads to $$-3 = 18-23 = -5$$, so it does not lie on the line. Option B $$(2, -9)$$ yields $$-9 = 12-23 = -11$$, and Option C $$(3/2, -16)$$ yields $$-16 = 9-23 = -14$$, both of which fail. However, Option D $$(1/2, -20)$$ gives $$-20 = 3-23 = -20$$.

Therefore, the point $$\left(\frac{1}{2}, -20\right)$$ lies on the line, which matches Option D, so the answer is Option D.

The parabola is $$y^2 = 4x$$ with points $$(t^2, 2t)$$. The circle is $$(x-2)^2 + (y-8)^2 = 4$$ with centre $$C(2, 8)$$.

Squared distance from $$C$$ to a point on the parabola:

$$f(t) = (t^2 - 2)^2 + (2t - 8)^2 = t^4 - 4t^2 + 4 + 4t^2 - 32t + 64 = t^4 - 32t + 68$$

$$f'(t) = 4t^3 - 32 = 0 \implies t = 2$$

$$f(2) = 16 - 64 + 68 = 20$$

So $$d^2 = 20$$.

The correct answer is Option 3: $$20$$.

For $$y^2 = 4ax$$, $$a = 3$$.

Length of focal chord with inclination $$\theta$$ is $$4a \csc^2\theta$$.

$$12 \csc^2\theta = 15 \implies \sin^2\theta = \frac{12}{15} = \frac{4}{5}$$

Equation of focal chord passing through $$(a, 0) = (3, 0)$$:

$$y - 0 = \tan\theta(x - 3) \implies (\sin\theta)x - (\cos\theta)y - 3\sin\theta = 0$$

Distance $$p$$ from $$(0,0)$$:

$$p = \frac{|-3\sin\theta|}{\sqrt{\sin^2\theta + \cos^2\theta}} = 3\sin\theta$$

$$p^2 = 9\sin^2\theta = 9 \left(\frac{4}{5}\right) = \frac{36}{5}$$

 $$10p^2 = 10 \left(\frac{36}{5}\right) = \mathbf{72}$$.

Point $$P(\alpha, \beta)$$ lies on the parabola $$y^2 = 4x$$, so $$\beta^2 = 4\alpha$$.

$$P$$ also lies on the chord of the parabola $$x^2 = 8y$$ whose midpoint is $$\left(1, \frac{5}{4}\right)$$.

Find the equation of the chord: for the parabola $$x^2 = 8y$$, the equation of the chord with midpoint $$(h, k)$$ is given by $$T = S_1$$:

$$xh - 4(y + k) = h^2 - 8k$$

With $$(h, k) = \left(1, \frac{5}{4}\right)$$:

$$x(1) - 4\left(y + \frac{5}{4}\right) = 1 - 8 \cdot \frac{5}{4}$$

$$x - 4y - 5 = 1 - 10 = -9$$

$$x - 4y + 4 = 0 \implies x = 4y - 4$$

Use both conditions to find $$\alpha$$ and $$\beta$$: from the chord equation: $$\alpha = 4\beta - 4$$.

From the parabola: $$\beta^2 = 4\alpha = 4(4\beta - 4) = 16\beta - 16$$.

$$ \beta^2 - 16\beta + 16 = 0 $$

$$ \beta = \frac{16 \pm \sqrt{256 - 64}}{2} = \frac{16 \pm \sqrt{192}}{2} = \frac{16 \pm 8\sqrt{3}}{2} = 8 \pm 4\sqrt{3} $$

Compute $$(\alpha - 28)(\beta - 8)$$: if $$\beta = 8 + 4\sqrt{3}$$: $$\alpha = 4(8 + 4\sqrt{3}) - 4 = 28 + 16\sqrt{3}$$.

$$(\alpha - 28)(\beta - 8) = (16\sqrt{3})(4\sqrt{3}) = 64 \times 3 = 192$$

If $$\beta = 8 - 4\sqrt{3}$$: $$\alpha = 4(8 - 4\sqrt{3}) - 4 = 28 - 16\sqrt{3}$$.

$$(\alpha - 28)(\beta - 8) = (-16\sqrt{3})(-4\sqrt{3}) = 64 \times 3 = 192$$

In both cases, the answer is $$\boxed{192}$$.

$$y^2 = 12x \implies a=3$$. Focus is $$(3,0)$$.

Chord length $$l = 4a \csc^2 \theta = 12(1 + \frac{1}{m^2})$$.

Distance $$d$$ from $$(0,0)$$ to line $$y = m(x-3)$$: $$d = \frac{|-3m|}{\sqrt{m^2+1}}$$.

$$ld^2 = \left[ \frac{12(m^2+1)}{m^2} \right] \cdot \frac{9m^2}{m^2+1} = 108$$.

Answer: 108

The parabola is $$y^2 = 36x$$, so $$4a = 36$$ and $$a = 9$$. The focus is at $$(9, 0)$$.

Finding points P and Q:

Using the parametric form: $$P = (9t^2, 18t)$$ and $$Q = (9/t^2, -18/t)$$ where $$t > 0$$ (since the ordinate of P is positive) and $$t_1 t_2 = -1$$ for a focal chord.

The length of the focal chord is:

$$|PQ| = 9\left(t + \frac{1}{t}\right)^2 = 100$$

Let $$u = t + 1/t$$. Then $$9u^2 = 100$$, so $$u = 10/3$$.

Also $$(t - 1/t)^2 = u^2 - 4 = 100/9 - 4 = 64/9$$, so $$t - 1/t = 8/3$$ (positive for acute angle with positive $$x$$-axis).

Solving: $$t = 3$$, $$1/t = 1/3$$.

So $$P = (81, 54)$$ and $$Q = (1, -6)$$.

Verification: $$|PQ|^2 = (81-1)^2 + (54+6)^2 = 6400 + 3600 = 10000$$, so $$|PQ| = 100$$.

Finding point M:

$$PM : MQ = 3 : 1$$, so M divides PQ in ratio 3:1 from P:

$$M = \frac{1 \cdot P + 3 \cdot Q}{4} = \left(\frac{81 + 3}{4}, \frac{54 - 18}{4}\right) = (21, 9)$$

Line through M perpendicular to PQ:

Slope of PQ: $$\frac{54 - (-6)}{81 - 1} = \frac{60}{80} = \frac{3}{4}$$

Perpendicular slope: $$-\frac{4}{3}$$

Equation: $$y - 9 = -\frac{4}{3}(x - 21)$$, which simplifies to $$y = 37 - \frac{4}{3}x$$.

Checking each option:

Option A: $$(-6, 45)$$: $$y = 37 - \frac{4}{3}(-6) = 37 + 8 = 45$$. Lies on the line.

Option B: $$(6, 29)$$: $$y = 37 - \frac{4}{3}(6) = 37 - 8 = 29$$. Lies on the line.

Option C: $$(3, 33)$$: $$y = 37 - \frac{4}{3}(3) = 37 - 4 = 33$$. Lies on the line.

Option D: $$(-3, 43)$$: $$y = 37 - \frac{4}{3}(-3) = 37 + 4 = 41 \neq 43$$. Does NOT lie on the line.

The point that does NOT lie on the line is Option D: $$(-3, 43)$$.