JEE Parabola Questions

The given parabola is $$y^{2}=16x$$.
Write it in the standard form $$y^{2}=4ax$$ to identify $$a$$.

Comparing, $$4a = 16 \implies a = 4$$.
Hence the focus is $$S(4,0)$$.

For a parabola $$y^{2}=4ax$$, any point can be written parametrically as $$\bigl(at^{2},\,2at\bigr)$$.
With $$a = 4$$, a point corresponding to parameter $$t$$ is $$P(t)\; \equiv\; \bigl(4t^{2},\,8t\bigr)$$.

The point $$P(1,-4)$$ lies on the parabola, so find its parameter:

Set $$4t^{2}=1 \implies t^{2}=\tfrac14 \implies t=\pm\tfrac12$$.
Set $$8t=-4 \implies t=-\tfrac12$$.
Therefore the parameter for point $$P$$ is $$t_{1} = -\tfrac12$$.

Property of a focal chord: if the end-points have parameters $$t_{1}$$ and $$t_{2}$$, then $$t_{1}t_{2} = -1$$.

Hence $$t_{2} = -\dfrac{1}{t_{1}} = -\dfrac{1}{-\tfrac12} = 2$$.

Coordinates of the other end $$Q$$ of the focal chord (using $$t_{2}=2$$):

$$Q \; \equiv\; \bigl(4(2)^{2},\,8(2)\bigr) \;=\; (16,\,16).$$

Let the focus $$S(4,0)$$ divide the chord $$PQ$$ internally in the ratio $$m:n$$ (with $$\gcd(m,n)=1$$).
For internal division of $$P(x_{1},y_{1})$$ and $$Q(x_{2},y_{2})$$ in the ratio $$m:n$$, the coordinates are

$$\biggl(\dfrac{mx_{2}+n x_{1}}{m+n},\; \dfrac{my_{2}+n y_{1}}{m+n}\biggr).$$

Substitute $$P(1,-4)$$, $$Q(16,16)$$ and $$S(4,0)$$:

For the $$x$$-coordinate:

$$4 = \dfrac{m\cdot16 + n\cdot1}{m+n}.$$

Multiply: $$4(m+n) = 16m + n \implies 4m + 4n = 16m + n \implies 12m = 3n \implies n = 4m.$$

For the $$y$$-coordinate (check consistency):

$$0 = \dfrac{m\cdot16 + n(-4)}{m+n} \implies 16m = 4n \implies n = 4m,$$
which matches the previous result.

Thus $$m:n = m:4m = 1:4$$ after cancelling the common factor.
So $$m = 1, \; n = 4$$.

Required value:

$$m^{2}+n^{2} = 1^{2} + 4^{2} = 1 + 16 = 17.$$

Therefore, $$m^{2}+n^{2}=17,$$ corresponding to Option A.

The shortest distance from (a, 0) with a > 0 to the parabola $$y^2 = 4x$$ is specified as 4. A generic point on the parabola can be written as $$(t^2, 2t)$$, so the squared distance to $$(a,0)$$ is $$D^2 = (t^2 - a)^2 + 4t^2.$$ Differentiating and setting $$\frac{d(D^2)}{dt} = 0$$ gives $$t\bigl(4t^2 - 4a + 8\bigr) = 0.$$ Excluding the trivial solution $$t = 0$$ and requiring $$a > 2$$ yields $$t^2 = a - 2$$. Substituting back gives $$D^2 = 4 + 4(a - 2) = 4a - 4,$$ and imposing $$D = 4$$ leads to $$4a - 4 = 16$$, hence $$a = 5$$.

Having found $$a = 5$$, the required circle must pass through $$(5,0)$$ and the focus $$(1,0)$$ of the parabola, with its center on the x-axis at $$(h,0)$$. Equating distances from the center to these points, $$(5 - h)^2 = (1 - h)^2$$ leads to $$24 = 8h$$ and thus $$h = 3$$. The radius squared is then $$(5 - 3)^2 = 4$$, so the circle is $$(x - 3)^2 + y^2 = 4,$$ which expands to $$x^2 + y^2 - 6x + 5 = 0.$$

The correct answer is Option 2: $$x^2 + y^2 - 6x + 5 = 0$$.

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.

Let the given parabola be:

$$y^2 = 8x$$

Comparing with the standard form $$y^2 = 4ax$$, we get $$4a = 8 \implies a = 2$$. Any parametric point on this parabola can be represented as $$P(2t^2, 4t)$$.

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Let the given circle be:

$$x^2 + y^2 + 12y + 35 = 0$$

By completing the square, we can rewrite the equation of the circle as:

$$x^2 + (y + 6)^2 - 36 + 35 = 0 \implies x^2 + (y + 6)^2 = 1$$

Thus, the center of the circle is $$C(0, -6)$$ and its radius is $$R = 1$$.

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The shortest distance between a parabola and a circle occurs along the common normal line, which must pass through the center of the circle $$C(0, -6)$$.

The equation of the normal to the parabola $$y^2 = 4ax$$ at the parametric point $$t$$ is:

$$y = -tx + 2at + at^3$$

Substituting $$a = 2$$ into the normal equation:

$$y = -tx + 4t + 2t^3$$

Since this normal passes through the center of the circle $$(0, -6)$$, substituting these coordinates gives:

$$-6 = -t(0) + 4t + 2t^3$$

$$2t^3 + 4t + 6 = 0 \implies t^3 + 2t + 3 = 0$$

Testing for integer roots, we find that $$t = -1$$ satisfies the equation because $$(-1)^3 + 2(-1) + 3 = 0$$. Since the derivative of the polynomial expression $$3t^2 + 2$$ is always positive, $$t = -1$$ is the only real root.

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

$$P = (2(-1)^2, 4(-1)) = (2, -4)$$

Now, we calculate the distance between the point $$P(2, -4)$$ on the parabola and the center of the circle $$C(0, -6)$$:

$$PC = \sqrt{(2 - 0)^2 + (-4 - (-6))^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$$

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The shortest distance between the two curves is obtained by subtracting the radius of the circle from this distance:

$$\text{Shortest Distance} = PC - R = 2\sqrt{2} - 1$$

Therefore, the shortest distance between the curves is equal to $$2\sqrt{2} - 1$$.

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 given parabola is $$y^{2}=4x$$.
For $$y^{2}=4ax$$ we have $$a=1$$, so its focus is $$S(1,0)$$.

Let the focal chord $$PQ$$ pass through the focus $$S$$ and make an angle $$60^{\circ}$$ with the positive $$x$$-axis.
Hence the slope of $$PQ$$ is $$\tan 60^{\circ}= \sqrt{3}$$.

Equation of the chord through $$S(1,0)$$ with this slope:
$$y-0=\sqrt{3}(x-1)\; \Longrightarrow\; y=\sqrt{3}(x-1) \; -(1)$$

The points $$P(x_1,y_1)$$ and $$Q(x_2,y_2)$$ lie on the parabola, so substitute $$y$$ from $$(1)$$ into $$y^{2}=4x$$:

$$\bigl[\sqrt{3}(x-1)\bigr]^{2}=4x$$
$$3(x-1)^{2}=4x$$
$$3x^{2}-6x+3-4x=0$$
$$3x^{2}-10x+3=0 \; -(2)$$

Solving $$(2)$$:
Discriminant $$D=(-10)^{2}-4\cdot3\cdot3=100-36=64$$.
$$x=\dfrac{10\pm8}{2\cdot3}=\dfrac{10\pm8}{6}$$ gives
$$x_1=3,\; x_2=\dfrac{1}{3}$$.

Since point $$P$$ lies in the first quadrant, we take $$x_1=3$$.
Using $$(1)$$, $$y_1=\sqrt{3}(3-1)=2\sqrt{3}$$.
Thus $$P(3,\,2\sqrt{3})$$.

Consider the circle having $$PS$$ as a diameter.
If two points $$A(x_1,y_1)$$ and $$B(x_2,y_2)$$ are the ends of a diameter, the centre is the midpoint $$M\bigl(\tfrac{x_1+x_2}{2},\tfrac{y_1+y_2}{2}\bigr)$$ and the radius is $$\tfrac{1}{2}AB$$.

For $$P(3,2\sqrt{3})$$ and $$S(1,0)$$:
Centre $$M\left(\dfrac{3+1}{2},\dfrac{2\sqrt{3}+0}{2}\right)=(2,\sqrt{3})$$.
Radius $$R=\dfrac{1}{2}\sqrt{(3-1)^{2}+(2\sqrt{3}-0)^{2}}=\dfrac{1}{2}\sqrt{4+12}=2$$.

The circle is to touch the $$y$$-axis.
For a circle with centre $$(h,k)$$ and radius $$R$$, tangency to the line $$x=0$$ (the $$y$$-axis) requires $$|h|=R$$ and the point of contact is $$(0,k)$$.

Here $$h=2$$ and $$R=2$$, so the condition is satisfied.
Hence the touching point is $$(0,\alpha)=(0,k)=(0,\sqrt{3})$$, giving $$\alpha=\sqrt{3}$$.

Finally,
$$5\alpha^{2}=5(\sqrt{3})^{2}=5\cdot3=15$$.

Therefore, the value of $$5\alpha^{2}$$ is $$15$$, which corresponds to Option A.

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.

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.

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 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 $$\tfrac{343\sqrt{3}}{8}$$.

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.

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: $$20$$.

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.

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 $$(0,0)$$:

$$p = \frac{|-3\sin\theta|}{\sqrt{\sin^2\theta + \cos^2\theta}} = 3\sin\theta$$

$$p^2 = 9\sin^2\theta = 9 \left(\frac{4}{5}\right) = \frac{36}{5}$$

 $$10p^2 = 10 \left(\frac{36}{5}\right) = \mathbf{72}$$.

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.

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 $$9x^{2}+12x+18y-14=0$$.
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 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: $$(-3, 43)$$.

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 $$4x = 3$$.

We need to find the locus of midpoints of line segments $$AB$$ where a tangent to $$y^2 = 24x$$ meets $$xy = 2$$.

For the parabola $$y^2 = 24x$$, we have $$4a = 24$$ so that $$a = 6$$, and its tangent in slope form is given by $$y = mx + \frac{6}{m}$$.

The curve $$xy = 2$$ can be written as $$y = \frac{2}{x}$$. Substituting this into the tangent equation yields

$$\frac{2}{x} = mx + \frac{6}{m},$$

which simplifies to

$$2 = mx^2 + \frac{6x}{m}$$

and then to the quadratic

$$m^2x^2 + 6x - 2m = 0.$$

Let the points of intersection be $$A = (x_1,y_1)$$ and $$B = (x_2,y_2)$$. By Vieta’s formulas, the roots of the quadratic satisfy

$$x_1 + x_2 = \frac{-6}{m^2},\qquad x_1 x_2 = \frac{-2}{m}.$$

If $$(h,k)$$ is the midpoint of $$AB$$, then

$$h = \frac{x_1 + x_2}{2} = \frac{-3}{m^2}.$$

Since $$y_i = mx_i + \frac{6}{m}$$ for each intersection point, the midpoint’s $$y$$-coordinate is

$$k = \frac{y_1 + y_2}{2} = m\cdot h + \frac{6}{m} = m\cdot\frac{-3}{m^2} + \frac{6}{m} = \frac{-3}{m} + \frac{6}{m} = \frac{3}{m}.$$

From $$k = \frac{3}{m}$$ we have $$m = \frac{3}{k}$$. Substituting into the expression for $$h$$ gives

$$h = \frac{-3}{m^2} = \frac{-3}{\bigl(\tfrac{3}{k}\bigr)^2} = \frac{-3k^2}{9} = \frac{-k^2}{3}.$$

Hence the coordinates of the midpoint satisfy

$$k^2 = -3h,$$

or in the usual $$(x,y)$$-form,

$$y^2 = -3x.$$

This is a left-opening parabola with equation $$y^2 = -3x$$, for which $$4a = 3$$ so that $$a = \tfrac{3}{4}$$ and the vertex is at the origin. Its directrix is

$$x = \frac{3}{4},$$

or equivalently $$4x = 3$$, and the length of the latus rectum is $$4a = 3$$.

The answer is Option 1: directrix $$4x = 3$$.

We need to find the common tangent $$y = mx + c$$ with $$m > 0$$ to the curves $$x = 2y^2$$ and $$x = 1 + y^2$$.

For the curve $$x = 2y^2$$, at the point $$(2t^2, t)$$ the derivative is $$\frac{dy}{dx} = \frac{1}{4t}$$. Hence the tangent line can be written as $$y = \frac{x}{4t} + \frac{t}{2}$$, which shows that $$m = \frac{1}{4t}$$ and $$c = \frac{t}{2}$$.

For the curve $$x = 1 + y^2$$, at the point $$(1 + s^2, s)$$ the derivative is $$\frac{dy}{dx} = \frac{1}{2s}$$. The tangent line is then $$y = \frac{x}{2s} - \frac{1}{2s} + \frac{s}{2}$$, giving $$m = \frac{1}{2s}$$ and $$c = -\frac{1}{2s} + \frac{s}{2}$$.

In order for these tangents to coincide, their slopes must be equal, so $$\frac{1}{4t} = \frac{1}{2s}$$ which implies $$s = 2t$$. Their intercepts must also match, so

$$\frac{t}{2} = -\frac{1}{2s} + \frac{s}{2} = -\frac{1}{4t} + t$$

Rearranging gives $$-\frac{t}{2} = -\frac{1}{4t}$$, and hence $$2t^2 = 1$$, so $$t = \frac{1}{\sqrt{2}}$$.

Substituting back, we find $$m = \frac{1}{4t} = \frac{\sqrt{2}}{4}$$ and $$c = \frac{t}{2} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$$.

Thus the common tangent is $$y = \frac{\sqrt{2}}{4}x + \frac{\sqrt{2}}{4}$$, which can also be written as $$\sqrt{2}\,x - 4y + \sqrt{2} = 0$$.

The distance from the point $$(6, -2\sqrt{2})$$ to this line is

$$d = \frac{\bigl|\sqrt{2}\,(6) - 4(-2\sqrt{2}) + \sqrt{2}\bigr|}{\sqrt{2 + 16}} = \frac{|6\sqrt{2} + 8\sqrt{2} + \sqrt{2}|}{\sqrt{18}} = \frac{15\sqrt{2}}{3\sqrt{2}} = 5\,. $$

Therefore, the correct answer is Option B: $$5$$.

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): $$\boxed{d, e, f \text{ are in G.P.}}$$

We need to find $$(QR)^2$$ where $$Q$$ and $$R$$ are the points where a common tangent from a point $$A$$ on the $$x$$-axis touches the circle $$x^2 + y^2 = 8$$ and the parabola $$y^2 = 16x$$ respectively.

A tangent to the parabola $$y^2 = 16x$$ (where $$a = 4$$) has the form $$y = mx + \dfrac{4}{m}$$ and for it to also be tangent to the circle $$x^2 + y^2 = 8$$ (radius $$= 2\sqrt{2}$$), the distance from the origin to the line must equal $$2\sqrt{2}$$:

$$\dfrac{|4/m|}{\sqrt{1 + m^2}} = 2\sqrt{2}$$

Squaring gives $$\dfrac{16}{m^2(1 + m^2)} = 8$$, which leads to $$16 = 8m^2(1 + m^2) \Rightarrow m^4 + m^2 - 2 = 0$$ and hence $$(m^2 + 2)(m^2 - 1) = 0 \Rightarrow m^2 = 1 \Rightarrow m = \pm 1$$.

Taking $$m = 1$$, the tangent is $$y = x + 4$$. Substituting into $$x^2 + y^2 = 8$$ gives $$x^2 + (x+4)^2 = 8 \Rightarrow 2x^2 + 8x + 8 = 0 \Rightarrow (x+2)^2 = 0$$, so $$x = -2, y = 2$$ and thus $$Q = (-2, 2)$$.

Substituting $$y = x + 4$$ into $$y^2 = 16x$$ yields $$(x+4)^2 = 16x \Rightarrow x^2 - 8x + 16 = 0 \Rightarrow (x-4)^2 = 0$$, so $$x = 4, y = 8$$ and hence $$R = (4, 8)$$.

Finally, $$(QR)^2 = (4-(-2))^2 + (8-2)^2 = 36 + 36 = 72$$, so the answer is $$(QR)^2 = 72$$, which corresponds to Option D.

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

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

·  Find Vertices: Midpoints are $$A(0,1), B(1,1), C(1,0)$$. The vertices of the large triangle are $$V_1(0,0), V_2(0,2), V_3(2,0)$$.

• This is a right-angled isosceles triangle with legs of length 2.
• With $$a=2$$: $$D = (\frac{2}{2+\sqrt{2}}, \frac{2}{2+\sqrt{2}}) = (2-\sqrt{2}, 2-\sqrt{2})$$.
• $$(2-\sqrt{2})^2 = 4a(2-\sqrt{2}) \implies 4a = 2-\sqrt{2} \implies a = \frac{2-\sqrt{2}}{4}$$.
• $$\alpha = \frac{1}{2}$$, $$\beta = -\frac{1}{4}$$.
• $$\frac{1/2}{(-1/4)^2} = \frac{1/2}{1/16} = \mathbf{8}$$.

·  Find Incentre $$D$$: For a right-angled triangle $$(0,0), (a,0), (0,a)$$, the incentre is $$(\frac{a}{2+\sqrt{2}}, \frac{a}{2+\sqrt{2}})$$.

·  Find Parabola $$a$$: $$y^2 = 4ax$$ passes through $$D$$:

·  Identify Focus: Focus is $$(a, 0) = (\frac{1}{2} - \frac{1}{4}\sqrt{2}, 0)$$.

·  Calculate $$\frac{\alpha}{\beta^2}$$:

• $$\frac{1/2}{(-1/4)^2} = \frac{1/2}{1/16} = \mathbf{8}$$.

To solve for $$PQ^2$$, we use the properties of the parabola and the centroid:

• Focus $$R$$: For $$y^2 = 20x$$, $$4a = 20 \implies a = 5$$. Thus, $$R = (5, 0)$$.
• Centroid $$G(10, 10)$$: Let $$P(x_1, y_1)$$ and $$Q(x_2, y_2)$$.
  • $$x_1 + x_2 + 5 = 3(10) \implies \mathbf{x_1 + x_2 = 25}$$
  • $$y_1 + y_2 + 0 = 3(10) \implies \mathbf{y_1 + y_2 = 30}$$

Substitute the points into $$y = mx + c$$:

$$(y_1 + y_2) = m(x_1 + x_2) + 2c \implies 30 = 25m + 2c$$

Given $$c - m = 6 \implies c = m + 6$$.

$$30 = 25m + 2(m + 6) \implies 18 = 27m \implies \mathbf{m = \frac{2}{3}, c = \frac{20}{3}}$$

The distance $$PQ$$ on a line with slope $$m$$ is $$PQ^2 = (x_1 - x_2)^2(1 + m^2)$$.

Find the intersection by substituting the line into $$y^2 = 20x$$:

$$\left[\frac{2}{3}(x + 10)\right]^2 = 20x \implies \frac{4}{9}(x^2 + 20x + 100) = 20x$$

$$x^2 - 25x + 100 = 0$$

Using roots $$(x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1x_2$$:

$$(x_1 - x_2)^2 = 25^2 - 4(100) = \mathbf{225}$$

Final Result:

$$PQ^2 = 225 \left(1 + \left(\frac{2}{3}\right)^2\right) = 225 \left(\frac{13}{9}\right) = 25 \times 13 = \mathbf{325}$$

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)

We need to find the parabola $$y = f(x)$$ with focus $$(-1/2, 0)$$ and directrix $$y = -1/2$$, then determine the set $$S$$.

The vertex of a parabola is the midpoint between the focus and the nearest point on the directrix. The focus is at $$(-1/2, 0)$$ and the directrix is $$y = -1/2$$. The point on the directrix closest to the focus (dropping a perpendicular) is $$(-1/2, -1/2)$$. The vertex is:

$$V = \left(-\frac{1}{2}, -\frac{1}{4}\right)$$

Since the focus is above the directrix, the parabola opens upward. The distance from vertex to focus is $$a = \bigl|0 - (-\tfrac{1}{4})\bigr| = \tfrac{1}{4}$$. The standard form (with vertex at $$(h,k)$$) is:

$$ (x-h)^2 = 4a(y-k) \implies \left(x+\frac{1}{2}\right)^2 = 4 \times \frac{1}{4}\!\left(y+\frac{1}{4}\right) = y + \frac{1}{4}$$

So $$y = \left(x + \frac{1}{2}\right)^2 - \frac{1}{4} = x^2 + x + \frac{1}{4} - \frac{1}{4} = x^2 + x$$. Therefore $$f(x) = x^2 + x = x(x+1)$$.

$$S = \{x \in \mathbb{R} : \tan^{-1}(\sqrt{f(x)}) + \sin^{-1}(\sqrt{f(x)+1}) = \tfrac{\pi}{2}\}$$

For $$\tan^{-1}(\sqrt{f(x)})$$ to be defined, we need $$f(x) \ge 0$$, i.e., $$x(x+1) \ge 0$$. For $$\sin^{-1}(\sqrt{f(x)+1})$$ to be defined, we need $$0 \le f(x) + 1 \le 1$$, i.e., $$-1 \le f(x) \le 0$$.

Combining these conditions gives $$f(x) = 0$$.

When $$f(x) = 0$$ we have $$\tan^{-1}(0) + \sin^{-1}(1) = 0 + \tfrac{\pi}{2} = \tfrac{\pi}{2}$$, which satisfies the equation.

Setting $$x(x+1)=0$$ yields $$x = 0$$ or $$x = -1$$, so $$S = \{0, -1\}$$, which contains exactly two elements.

The correct answer is Option 1: contains exactly two elements.

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

We need to find the length of a focal chord of the parabola $$y^2 = 8x + 4y + 4$$ whose x-intercept is 3.

Rewriting the given equation as $$y^2 - 4y = 8x + 4$$ and completing the square gives $$(y - 2)^2 = 8(x + 1)$$, which is of the standard form $$Y^2 = 8X$$ with the substitutions $$Y = y - 2$$ and $$X = x + 1$$.

Since $$4a = 8$$, it follows that $$a = 2$$, the vertex of the parabola is at $$(-1,2)$$, and the focus is at $$(1,2)$$.

A focal chord must pass through the focus at $$(1,2)$$ and also through the point of x-intercept $$(3,0)$$. The slope of this line is $$m = \frac{2 - 0}{1 - 3} = -1$$, so its equation is $$y - 2 = -1 \cdot (x - 1)$$ or equivalently $$y = 3 - x$$.

Substituting $$y = 3 - x$$ into the standard form $$(y - 2)^2 = 8(x + 1)$$ gives $$(3 - x - 2)^2 = 8(x + 1)$$, which simplifies to $$(1 - x)^2 = 8x + 8$$.

Expanding and rearranging yields $$x^2 - 2x + 1 = 8x + 8$$ and hence $$x^2 - 10x - 7 = 0$$, whose solutions are $$x = \frac{10 \pm \sqrt{100 + 28}}{2} = 5 \pm 4\sqrt{2}$$.

The corresponding y-values follow from $$y = 3 - x$$, giving $$y_1 = 3 - (5 + 4\sqrt{2}) = -2 - 4\sqrt{2}$$ and $$y_2 = 3 - (5 - 4\sqrt{2}) = -2 + 4\sqrt{2}$$.

The horizontal and vertical differences between these two points are $$\Delta x = (5 + 4\sqrt{2}) - (5 - 4\sqrt{2}) = 8\sqrt{2}$$ and $$\Delta y = (-2 - 4\sqrt{2}) - (-2 + 4\sqrt{2}) = -8\sqrt{2}$$.

Therefore, the length of the chord is $$\sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(8\sqrt{2})^2 + (-8\sqrt{2})^2} = \sqrt{128 + 128} = \sqrt{256} = 16$$.

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.

The given parabola is $$y^{2}=4x$$, which is of the form $$y^{2}=4ax$$ with $$a=1$$.
Its focus is therefore $$S(1,0)$$.

1. Tangents from the external point $$P(-2,1)$$

For $$y^{2}=4ax$$, the tangent at the point $$(t^{2},2t)$$ on the parabola is
$$ty=x+t^{2} \; -(1)$$

Since this tangent passes through $$P(-2,1)$$, substitute $$x=-2,\;y=1$$ in $$(1)$$:
$$t(1)= -2+t^{2}\;\Longrightarrow\;t^{2}-t-2=0$$
$$\Rightarrow\; t=\dfrac{1\pm\sqrt{1+8}}{2}=2,\,-1$$

Hence the two points of contact are
$$P_{1}(4,4)\quad (t=2),\qquad P_{2}(1,-2)\quad (t=-1)$$

2. Lines $$SP_{1}$$ and $$SP_{2}$$

$$S(1,0),\;P_{1}(4,4) \;\Rightarrow\; SP_{1}: \dfrac{x-1}{3}=\dfrac{y}{4}$$
Direction vector $$\mathbf{d}_{1}=(3,4)$$ with $$|\mathbf{d}_{1}|^{2}=3^{2}+4^{2}=25$$.

$$S(1,0),\;P_{2}(1,-2) \;\Rightarrow\; SP_{2}: x=1$$ (a vertical line).

3. Foot of the perpendicular from $$P$$ to $$SP_{1}$$  ($$Q_{1}$$)

The vector $$\overrightarrow{SP}=(-3,\,1)$$.
Scalar projection of $$\overrightarrow{SP}$$ on $$\mathbf{d}_{1}$$ is
$$\dfrac{\overrightarrow{SP}\cdot\mathbf{d}_{1}}{|\mathbf{d}_{1}|^{2}} =\dfrac{(-3)(3)+1\cdot4}{25}=\dfrac{-5}{25}=-\dfrac15$$

Therefore
$$Q_{1}=S+\left(-\dfrac15\right)\mathbf{d}_{1} =(1,0)+\left(-\dfrac15\right)(3,4) =\left(1-\dfrac35,\;0-\dfrac45\right) =\left(\dfrac25,\;-\dfrac45\right)$$

Lengths involving $$Q_{1}$$

$$SQ_{1}= \left|\,-\dfrac15\right|\,|\mathbf{d}_{1}|=\dfrac15\cdot5=1$$

Vector $$\overrightarrow{PQ_{1}}=(-2-\tfrac25,\;1+\tfrac45)=(-\tfrac{12}{5},\,\tfrac{9}{5})$$
$$PQ_{1}= \sqrt{\left(\tfrac{12}{5}\right)^{2}+\left(\tfrac{9}{5}\right)^{2}} =\sqrt{\dfrac{144+81}{25}}=\sqrt{\dfrac{225}{25}}=3$$

4. Foot of the perpendicular from $$P$$ to $$SP_{2}$$  ($$Q_{2}$$)

The line $$SP_{2}$$ is $$x=1$$. The perpendicular to a vertical line is horizontal, so $$y$$ remains $$1$$.

Hence $$Q_{2}=(1,1)$$.

Length $$SQ_{2}$$

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

5. Distance $$Q_{1}Q_{2}$$

$$Q_{1}\left(\dfrac25,-\dfrac45\right),\;Q_{2}(1,1)$$
$$\overrightarrow{Q_{1}Q_{2}}=\left(1-\dfrac25,\;1+\dfrac45\right) =\left(\dfrac35,\;\dfrac95\right)$$
$$Q_{1}Q_{2}= \sqrt{\left(\dfrac35\right)^{2}+\left(\dfrac95\right)^{2}} =\sqrt{\dfrac{9+81}{25}} =\dfrac{\sqrt{90}}{5} =\dfrac{3\sqrt{10}}{5}$$

6. Verification of the options

Option A: $$SQ_{1}=2$$  →  False  ($$SQ_{1}=1$$)
Option B: $$Q_{1}Q_{2}=\dfrac{3\sqrt{10}}5$$  →  True
Option C: $$PQ_{1}=3$$  →  True
Option D: $$SQ_{2}=1$$  →  True

Therefore, the correct statements are:
Option B, Option C, and Option D.

We need to find $$8|m_1 m_2|$$ where $$y = m_1x + c_1$$ and $$y = m_2x + c_2$$ are the common tangents to the circle $$x^2 + y^2 = 2$$ and the parabola $$y^2 = x$$.

The parabola can be expressed as $$y^2 = 4 \cdot \tfrac{1}{4} \cdot x$$ with $$a = \tfrac{1}{4}$$, so its tangent in slope form is $$y = mx + \frac{1}{4m}$$.

For this line to be tangent to the circle $$x^2 + y^2 = 2$$, its perpendicular distance from the origin must equal $$\sqrt{2}$$. Hence,
$$\frac{\bigl|1/(4m)\bigr|}{\sqrt{1 + m^2}} = \sqrt{2}$$
and squaring both sides gives
$$\frac{1}{16m^2} = 2(1 + m^2)$$
so
$$1 = 32m^2 + 32m^4$$.

Solving the equation $$32m^4 + 32m^2 - 1 = 0$$ by setting $$u = m^2$$ leads to
$$32u^2 + 32u - 1 = 0$$
which yields
$$u = \frac{-32 \pm \sqrt{1024 + 128}}{64} = \frac{-32 \pm \sqrt{1152}}{64} = \frac{-32 \pm 24\sqrt{2}}{64}$$. Since $$u = m^2 > 0$$, we select the positive root to obtain
$$u = \frac{-32 + 24\sqrt{2}}{64} = \frac{-4 + 3\sqrt{2}}{8}$$.

Because both tangent lines correspond to $$m^2 = \frac{3\sqrt{2} - 4}{8}$$, it follows that $$m_1 = -m_2$$ and thus
$$|m_1 m_2| = m^2 = \frac{3\sqrt{2} - 4}{8}$$.

Finally, multiplying by 8 gives
$$8|m_1 m_2| = 8 \times \frac{3\sqrt{2} - 4}{8} = 3\sqrt{2} - 4$$.

Option A: $$3\sqrt{2} - 4$$

The vertex of the parabola is $$V = (5, 4)$$ and the directrix is $$3x + y - 29 = 0$$.

The axis of the parabola is perpendicular to the directrix with direction $$(3, 1)$$.

Distance from $$V$$ to the directrix: $$\dfrac{|3(5) + 4 - 29|}{\sqrt{9+1}} = \dfrac{|-10|}{\sqrt{10}} = \sqrt{10}$$.

Since $$3(5)+4-29 = -10 < 0$$, the vertex is on the side where $$3x+y < 29$$. The focus lies on the same side, at distance $$\sqrt{10}$$ from the vertex along the direction $$-(3,1)/\sqrt{10}$$.

Focus: $$S = (5,4) - \sqrt{10} \cdot \dfrac{(3,1)}{\sqrt{10}} = (5-3, 4-1) = (2, 3)$$.

By the definition of a parabola, for any point $$P(x,y)$$: distance to focus = distance to directrix.

$$(x-2)^2 + (y-3)^2 = \dfrac{(3x+y-29)^2}{10}$$

Multiplying both sides by 10:

$$10(x^2 - 4x + 4 + y^2 - 6y + 9) = (3x+y-29)^2$$

$$10x^2 - 40x + 40 + 10y^2 - 60y + 90 = 9x^2 + 6xy - 174x + y^2 - 58y + 841$$

Rearranging: $$x^2 + 9y^2 - 6xy + 134x - 2y - 711 = 0$$

So $$a = 9$$, $$b = -6$$, $$c = 134$$, $$d = -2$$, $$k = -711$$.

$$a + b + c + d + k = 9 - 6 + 134 - 2 - 711 = -576$$.

The answer is Option D: $$-576$$.

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 need to find the ordinate of point $$Q$$ where the tangent at $$P$$ on the parabola $$y^2 = 6x$$ meets the directrix.

Since $$y^2 = 6x$$ implies $$4a = 6$$ and hence $$a = \frac{3}{2}$$, the directrix is $$x = -\frac{3}{2}$$ and a general point on the parabola can be written as $$P = \left(\frac{3t^2}{2},3t\right)$$ using the parametric form $$(at^2,2at)$$.

Now the equation of the normal at parameter $$t$$ is given by $$y + tx = 2at + at^3$$ which simplifies to $$y + tx = 3t + \frac{3t^3}{2}$$. Since this normal passes through $$(5,-8)$$, substituting these coordinates gives $$-8 + 5t = 3t + \frac{3t^3}{2}$$. This leads to $$-8 + 2t = \frac{3t^3}{2}$$, and multiplying by 2 yields $$-16 + 4t = 3t^3$$ or $$3t^3 - 4t + 16 = 0$$. Testing $$t = -2$$ confirms the equation, since $$3(-8) - 4(-2) + 16 = -24 + 8 + 16 = 0$$. Therefore $$t = -2$$ and $$P = (6,-6)$$.

Next, the equation of the tangent at $$P$$ follows from the general tangent $$ty = x + at^2$$ which for $$t = -2$$ becomes $$-2y = x + \frac{3\cdot4}{2} = x + 6$$, or equivalently $$x + 2y + 6 = 0$$.

From the directrix $$x = -\frac{3}{2}$$, substituting into the tangent equation gives $$-\frac{3}{2} + 2y + 6 = 0$$, which simplifies to $$2y = -\frac{9}{2}$$ and hence $$y = -\frac{9}{4}$$.

The ordinate of point $$Q$$ is $$-\frac{9}{4}$$.

The correct answer is Option A.

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 need to find the equation of the parabola with vertex $$(5, 4)$$ and directrix $$3x + y - 29 = 0$$.

Find the axis of the parabola

The axis is perpendicular to the directrix. The directrix has slope $$-3$$, so the axis has slope $$\frac{1}{3}$$.

Find the focus

The vertex is the midpoint of the focus and the foot of the perpendicular from the focus to the directrix.

Distance from vertex $$(5, 4)$$ to the directrix $$3x + y - 29 = 0$$:

$$d = \frac{|3(5) + 4 - 29|}{\sqrt{9 + 1}} = \frac{|15 + 4 - 29|}{\sqrt{10}} = \frac{|-10|}{\sqrt{10}} = \frac{10}{\sqrt{10}} = \sqrt{10}$$

So $$a = \sqrt{10}$$ (the distance from vertex to directrix).

The focus is at distance $$a = \sqrt{10}$$ from the vertex on the opposite side of the directrix. The direction from vertex toward the directrix is along $$(3, 1)/\sqrt{10}$$. Since $$3(5) + 1(4) - 29 = -10 < 0$$, the vertex is on the side away from the directrix (toward the origin), so the focus is at:

$$(5, 4) - \sqrt{10} \cdot \frac{(3, 1)}{\sqrt{10}} = (5 - 3, 4 - 1) = (2, 3)$$

Use the definition of a parabola

For any point $$(x, y)$$ on the parabola, the distance to the focus equals the distance to the directrix:

$$(x - 2)^2 + (y - 3)^2 = \frac{(3x + y - 29)^2}{10}$$

Expand

$$10[(x-2)^2 + (y-3)^2] = (3x + y - 29)^2$$

$$10[x^2 - 4x + 4 + y^2 - 6y + 9] = 9x^2 + y^2 + 841 + 6xy - 174x - 58y$$

$$10x^2 - 40x + 40 + 10y^2 - 60y + 90 = 9x^2 + y^2 + 841 + 6xy - 174x - 58y$$

$$10x^2 + 10y^2 - 40x - 60y + 130 = 9x^2 + y^2 + 6xy - 174x - 58y + 841$$

Rearrange to standard form

$$x^2 + 9y^2 - 6xy + 134x - 2y - 711 = 0$$

Compare with the given form

The equation is $$x^2 + ay^2 + bxy + cx + dy + k = 0$$

$$a = 9, \quad b = -6, \quad c = 134, \quad d = -2, \quad k = -711$$

Calculate $$a + b + c + d + k$$

$$a + b + c + d + k = 9 + (-6) + 134 + (-2) + (-711) = 9 - 6 + 134 - 2 - 711 = -576$$

Therefore, $$a + b + c + d + k = -576$$.

The correct answer is Option D: $$-576$$.

The tangent at the vertex is $$x + y = a$$, which has slope $$-1$$. Since the axis of the parabola is perpendicular to this tangent, its slope is $$1$$.

Because the axis passes through the focus $$(a, a)$$ with slope 1, its equation is

$$y - a = 1 \cdot (x - a) \implies y = x$$

The vertex lies on both the axis $$y = x$$ and the tangent at the vertex $$x + y = a$$, so

$$x + x = a \implies x = \frac{a}{2}$$

Hence the vertex is $$\left(\dfrac{a}{2}, \dfrac{a}{2}\right)$$.

The distance between the vertex and the focus is

$$VF = \sqrt{\left(a - \frac{a}{2}\right)^2 + \left(a - \frac{a}{2}\right)^2} = \sqrt{\frac{a^2}{4} + \frac{a^2}{4}} = \frac{|a|}{\sqrt{2}}$$

The length of the latus rectum is $$4 \times VF$$, so

$$4 \times \frac{|a|}{\sqrt{2}} = 16$$

$$\frac{|a|}{\sqrt{2}} = 4$$

$$|a| = 4\sqrt{2}$$

Therefore, $$\boxed{|a| = 4\sqrt{2}}$$. The answer is Option C.

First rewrite the parabola as $$y^2=2\left(x-\frac32\right)$$ or $$y^2=4a(x-h)$$, where $$a=\frac12$$ and $$h=\frac32$$.

A point on the parabola corresponding to parameter $$t$$ is $$P\left(\frac32+\frac{t^2}{2},t\right)$$, and the tangent at parameter $$t$$ is $$ty=x-\frac32+\frac{t^2}{2}$$.

Since the tangents at $$P(t_1)$$ and $$Q(t_2)$$ intersect at $$R(0,1)$$, the point $$R$$ must satisfy both tangent equations. Thus $$t=-\frac32+\frac{t^2}{2}$$.

Multiplying by $$2$$, we get $$t^2-2t-3=0$$, which gives $$(t-3)(t+1)=0$$. Hence $$t_1=3$$ and $$t_2=-1$$.

Therefore, $$P=(6,3)$$ and $$Q=(2,-1)$$.

Now $$m_{PQ}=\frac{3-(-1)}{6-2}=1$$. Hence the altitude through $$R$$ has slope $$-1$$ and its equation is $$y=-x+1$$.

Also, $$m_{RQ}=\frac{1-(-1)}{0-2}=-1$$. Hence the altitude through $$P$$ has slope $$1$$ and its equation is $$y-3=x-6$$, i.e., $$y=x-3$$.

Intersecting the two altitudes, we get $$x-3=-x+1$$, which gives $$2x=4$$ and hence $$x=2$$. Substituting in $$y=-x+1$$, we obtain $$y=-1$$.

Therefore, the orthocentre of $$\triangle PQR$$ is $$(2,-1)$$.

Final Answer : $$(2,-1)$$

Given: Vertex of the parabola is $$(2, -1)$$ and the equation of the directrix is $$4x - 3y = 21$$.

Find the distance from the vertex to the directrix:

The perpendicular distance from a point $$(x_0, y_0)$$ to the line $$ax + by + c = 0$$ is:

$$d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$$

Rewriting the directrix as $$4x - 3y - 21 = 0$$:

$$d = \frac{|4(2) - 3(-1) - 21|}{\sqrt{16 + 9}} = \frac{|8 + 3 - 21|}{5} = \frac{|-10|}{5} = 2$$

Relate the distance to the latus rectum:

For a parabola, the distance from the vertex to the directrix equals $$a$$ (the focal distance). Therefore $$a = 2$$.

Find the length of the latus rectum:

$$\text{Length of latus rectum} = 4a = 4 \times 2 = 8$$

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

We are given the parabola $$P: y^2 = 4ax$$, $$a > 0$$, with focus $$S = (a, 0)$$, and we seek the tangents to $$P$$ that make an angle of $$\frac{\pi}{4}$$ with the line $$y = 3x + 5$$, touching $$P$$ at points $$A$$ and $$B$$. The line $$y = 3x + 5$$ has slope 3, so if a tangent has slope $$m$$ and makes an angle of $$\frac{\pi}{4}$$ with this line, then $$\tan\frac{\pi}{4} = \left|\frac{m - 3}{1 + 3m}\right| = 1$$. Case 1: $$\frac{m - 3}{1 + 3m} = 1 \Rightarrow m - 3 = 1 + 3m \Rightarrow -2m = 4 \Rightarrow m = -2$$. Case 2: $$\frac{m - 3}{1 + 3m} = -1 \Rightarrow m - 3 = -1 - 3m \Rightarrow 4m = 2 \Rightarrow m = \frac{1}{2}$$.

For the parabola $$y^2 = 4ax$$, a tangent with slope $$m$$ touches at the point $$\left(\frac{a}{m^2}, \frac{2a}{m}\right)$$, yielding for $$m = -2$$ the point $$A = \left(\frac{a}{4}, -a\right)$$ and for $$m = \frac{1}{2}$$ the point $$B = (4a, 4a)$$.

Three points are collinear if the area of the triangle they form is zero: Area $$= \frac{1}{2}|x_A(y_B - y_S) + x_B(y_S - y_A) + x_S(y_A - y_B)| = \frac{1}{2}\left|\frac{a}{4}(4a - 0) + 4a(0 - (-a)) + a(-a - 4a)\right| = \frac{1}{2}\left|a^2 + 4a^2 - 5a^2\right| = \frac{1}{2}|0| = 0$$, so the area is always zero regardless of the value of $$a$$.

Hence, $$A$$, $$B$$, and $$S$$ are collinear for any $$a > 0$$. Therefore, the answer is Option D: any $$a > 0$$.

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 need to find the point $$P$$ on the parabola $$y = x^2 + 4$$ that is closest to the straight line $$y = 4x - 1$$.

The line $$y = 4x - 1$$ can be rewritten as $$4x - y - 1 = 0$$. The distance from any point $$(x_0, y_0)$$ on the parabola to this line is $$d = \frac{|4x_0 - y_0 - 1|}{\sqrt{16 + 1}} = \frac{|4x_0 - y_0 - 1|}{\sqrt{17}}$$.

Since $$y_0 = x_0^2 + 4$$, the distance becomes $$d = \frac{|4x_0 - x_0^2 - 4 - 1|}{\sqrt{17}} = \frac{|{-x_0^2 + 4x_0 - 5}|}{\sqrt{17}} = \frac{x_0^2 - 4x_0 + 5}{\sqrt{17}}$$, where we dropped the absolute value because the discriminant of $$-x_0^2 + 4x_0 - 5$$ is $$16 - 20 = -4 < 0$$, so $$x_0^2 - 4x_0 + 5 > 0$$ for all real $$x_0$$.

To minimize $$d$$, we minimize $$g(x_0) = x_0^2 - 4x_0 + 5$$. Differentiating, $$g'(x_0) = 2x_0 - 4 = 0$$, giving $$x_0 = 2$$. Since $$g''(x_0) = 2 > 0$$, this is indeed a minimum.

Substituting $$x_0 = 2$$ into the parabola equation, $$y_0 = (2)^2 + 4 = 4 + 4 = 8$$.

We can verify: the minimum distance is $$\frac{4 - 8 + 5}{\sqrt{17}} = \frac{1}{\sqrt{17}}$$. Also, the slope of the tangent to the parabola at $$x = 2$$ is $$\frac{dy}{dx}\bigg|_{x=2} = 2(2) = 4$$, which equals the slope of the given line, confirming that the tangent at the closest point is parallel to the line.

Therefore, the coordinates of $$P$$ are $$(2, 8)$$.

We consider the parabola $$y^{2}=16\,(x-3)$$.

First we rewrite it in the standard form $$y^{2}=4a\,(x-3)$$ so that we can recognise the parameter $$a$$. Comparing with $$y^{2}=4aX$$ we obtain $$4a=16 \Rightarrow a=4$$. Hence the vertex is at $$(3,0)$$ and the axis is the X-axis (rightwards).

For a parabola of the type $$y^{2}=4aX$$ the parametric coordinates of a general point are

$$X=a\,t^{2},\qquad Y=2a\,t,$$

where $$t$$ is the parameter. After shifting back to our actual variable $$x$$ (because $$X=x-3$$) we get

$$x-3=a\,t^{2},\qquad y=2a\,t.$$

Substituting $$a=4$$ gives the parametric point

$$x=4t^{2}+3,\qquad y=8t.$$

The next step is to write the tangent to the parabola at this parametric point. The standard tangent formula to $$y^{2}=4aX$$ is

$$t\,Y=X+a\,t^{2}.$$

Replacing $$X$$ by $$x-3$$ and $$Y$$ by $$y$$, and again taking $$a=4$$, we have

$$t\,y=(x-3)+4t^{2}.$$

This linear equation represents the tangent at the parameter value $$t$$. We rearrange it to slope-intercept form so that the slope is explicit:

$$$ t\,y = x-3+4t^{2} \;\Longrightarrow\; y=\frac{1}{t}\,x-\frac{3-4t^{2}}{t}. $$$

Thus the slope of this tangent is

$$m=\frac{1}{t}.$$

Let the external point from which we draw the tangents be $$P(h,k)$$. Because $$P$$ lies on each tangent, we substitute $$x=h,\;y=k$$ into the tangent equation:

$$$ t\,k = (h-3)+4t^{2}. $$$

Bringing all terms to one side gives a quadratic in $$t$$,

$$4t^{2}-k\,t+(h-3)=0.$$

Its two roots $$t_{1},t_{2}$$ correspond to the two tangents from $$P$$. For a quadratic $$at^{2}+bt+c=0$$ we know from Vieta’s formulae that

$$t_{1}+t_{2}=-\frac{b}{a},\qquad t_{1}t_{2}=\frac{c}{a}.$$

Here $$a=4,\;b=-k,\;c=h-3$$, so we obtain

$$t_{1}+t_{2}=\frac{k}{4},\qquad t_{1}t_{2}=\frac{h-3}{4}.$$

We are told that the two tangents are at right angles. If two lines with slopes $$m_{1},m_{2}$$ are perpendicular, the condition is $$m_{1}m_{2}=-1$$. We already noted that the slope of a tangent corresponding to parameter $$t$$ is $$m=\dfrac1t$$. Therefore

$$$ m_{1}m_{2}=\frac{1}{t_{1}}\cdot\frac{1}{t_{2}}=\frac{1}{t_{1}t_{2}}=-1. $$$

Hence

$$$ t_{1}t_{2}=-1. $$$

But we have also found that $$t_{1}t_{2}=\dfrac{h-3}{4}$$. Equating the two expressions for the product,

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

The y-coordinate $$k$$ does not appear in this final condition, so the locus of $$P(h,k)$$ is the vertical straight line

$$x=-1,$$

which can be rewritten as $$x+1=0$$.

Hence, the correct answer is Option D.

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.

We are told that the vertex and the focus both lie on the positive $$x$$-axis. Hence we can assign the vertex the coordinates $$(R,0)$$ and the focus the coordinates $$(S,0)$$ with the given condition $$S > R$$. Because the focus is to the right of the vertex, the parabola opens towards the positive $$x$$-direction.

For any parabola that opens to the right, the standard form of the equation, when its vertex is at $$(h,k)$$, is stated first:

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

In this form, $$a$$ is defined as the distance from the vertex to the focus, and we also know the following fact:

The length of the latus rectum of such a parabola is $$4a$$.

Now, for our particular parabola we have

$$h = R, \quad k = 0$$

and the distance from $$(R,0)$$ to $$(S,0)$$ is simply the difference of their abscissae:

$$ a = S - R. $$

Substituting this value of $$a$$ into the standard statement for the latus-rectum length, we obtain

$$ \text{Length of latus rectum} = 4a = 4(S - R). $$

Hence, the correct answer is Option C.

We begin with the geometric definition of a parabola: every point on the curve is equidistant from its focus and its directrix. The vertex is given as $$\left(\dfrac12,\dfrac34\right)$$ and the directrix is the horizontal line $$y=\dfrac12$$, so the axis of the parabola is vertical.

The distance from the vertex to the directrix is $$\dfrac34-\dfrac12=\dfrac14$$. This same distance, denoted by $$a$$, separates the vertex from the focus but in the opposite (upward) direction. Hence the focus is $$\left(\dfrac12,\dfrac34+\dfrac14\right)=\left(\dfrac12,1\right)$$ and we have $$a=\dfrac14$$.

For a parabola with vertical axis, vertex $$(h,k)$$ and focal length $$a$$, the standard form is $$ (x-h)^2 = 4a\,(y-k). $$ Here $$h=\dfrac12,\;k=\dfrac34,\;4a=4\left(\dfrac14\right)=1$$, so the explicit equation of the parabola is $$ (x-\dfrac12)^2 = y-\dfrac34, \qquad\text{or}\qquad y = (x-\dfrac12)^2 + \dfrac34. $$

We next locate the point $$P$$ where the parabola meets the vertical line $$x=-\dfrac12$$. Substituting $$x=-\dfrac12$$ into the equation of the parabola,

$$ y = \Bigl(-\dfrac12-\dfrac12\Bigr)^2 + \dfrac34 = (-1)^2 + \dfrac34 = 1 + \dfrac34 = \dfrac74. $$

Thus $$ P\left(-\dfrac12,\dfrac74\right). $$

To find the normal at $$P$$, we first need the slope of the tangent. Re-writing the parabola as $$y=(x-\dfrac12)^2+\dfrac34$$, we differentiate:

$$ \frac{dy}{dx}=2(x-\dfrac12). $$

At $$x=-\dfrac12$$, $$ m_{\text{tan}}=2\Bigl(-\dfrac12-\dfrac12\Bigr)=2(-1)=-2. $$ Hence the slope of the normal is the negative reciprocal: $$ m_{\text{norm}}=\dfrac12. $$

The normal through $$P(-\dfrac12,\dfrac74)$$ therefore satisfies $$ y-\dfrac74=\dfrac12\bigl(x+\dfrac12\bigr). $$ Adding $$\dfrac74$$ to both sides gives the straight-line equation $$ y=\dfrac12x+\dfrac14+\dfrac74=\dfrac12x+2. $$

Let the normal meet the parabola again at $$Q(x,y)$$. Setting the expression for $$y$$ on the line equal to the expression for $$y$$ on the parabola, we obtain

$$ (x-\dfrac12)^2+\dfrac34=\dfrac12x+2. $$

Simplifying step by step:

$$ (x-\dfrac12)^2 = \dfrac12x + 2 - \dfrac34 = \dfrac12x + \dfrac54, $$ $$ x^2 - x + \dfrac14 = \dfrac12x + \dfrac54, $$ $$ x^2 - x - \dfrac12x + \dfrac14 - \dfrac54 = 0, $$ $$ x^2 - \dfrac32x - 1 = 0. $$

Multiplying by $$2$$ to clear the fraction,

$$ 2x^2 - 3x - 2 = 0. $$

The quadratic formula, $$ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}, $$ with $$a=2,\;b=-3,\;c=-2$$ gives

$$ x=\dfrac{3\pm\sqrt{(-3)^2-4(2)(-2)}}{2\cdot2} =\dfrac{3\pm\sqrt{9+16}}{4} =\dfrac{3\pm5}{4}. $$

Thus $$x=-\dfrac12$$ or $$x=2$$. The first solution reproduces point $$P$$, so the second root corresponds to $$Q$$ with $$x=2$$. Substituting into $$y=\dfrac12x+2$$, we find

$$ y=\dfrac12\cdot2+2=1+2=3, \qquad Q(2,3). $$

Finally we compute the squared distance between $$P\bigl(-\dfrac12,\dfrac74\bigr)$$ and $$Q(2,3)$$. Using the distance formula $$ (PQ)^2=(x_2-x_1)^2+(y_2-y_1)^2, $$ we have

$$ x_2-x_1 = 2-\Bigl(-\dfrac12\Bigr)=\dfrac52, \qquad y_2-y_1 = 3-\dfrac74 = \dfrac{12}{4}-\dfrac74=\dfrac54. $$

Therefore

$$ (PQ)^2 = \left(\dfrac52\right)^2 + \left(\dfrac54\right)^2 = \dfrac{25}{4}+\dfrac{25}{16} = \dfrac{100}{16}+\dfrac{25}{16} = \dfrac{125}{16}. $$

Hence, the correct answer is Option C.

The parabola is $$y^2 = 2x$$, so comparing with $$y^2 = 4a_0 x$$, we get $$a_0 = \frac{1}{2}$$. A point on the parabola can be written in parametric form as $$P = \left(\frac{t^2}{2}, t\right)$$, since $$y = t$$ gives $$x = t^2/2$$.

The slope of the tangent at $$P$$ is obtained by differentiating $$y^2 = 2x$$: $$2y\frac{dy}{dx} = 2$$, so $$\frac{dy}{dx} = \frac{1}{y} = \frac{1}{t}$$. The slope of the normal at $$P$$ is therefore $$-t$$.

The equation of the normal at $$P\left(\frac{t^2}{2}, t\right)$$ with slope $$-t$$ is $$y - t = -t\left(x - \frac{t^2}{2}\right)$$, which simplifies to $$y = -tx + t + \frac{t^3}{2}$$.

Since this normal passes through the point $$(a, 0)$$, we substitute $$x = a$$ and $$y = 0$$: $$0 = -ta + t + \frac{t^3}{2}$$. Factoring out $$t$$: $$t\left(-a + 1 + \frac{t^2}{2}\right) = 0$$.

This gives either $$t = 0$$ (which corresponds to the normal along the x-axis through the vertex) or $$\frac{t^2}{2} = a - 1$$, i.e., $$t^2 = 2(a - 1)$$.

For three distinct normals from $$(a, 0)$$ to the parabola, we need the $$t = 0$$ solution plus two additional distinct real values of $$t$$ from $$t^2 = 2(a-1)$$. The equation $$t^2 = 2(a-1)$$ yields two distinct real roots $$t = \pm\sqrt{2(a-1)}$$ if and only if $$2(a-1) > 0$$, which requires $$a > 1$$.

If $$a = 1$$, then $$t^2 = 0$$ gives only $$t = 0$$, and we get just one normal (not three). If $$a < 1$$, there are no real solutions other than $$t = 0$$. Therefore, for three normals to exist, $$a$$ must be greater than 1.

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 is $$y^2 = 6x$$, so $$4a = 6$$ and $$a = \frac{3}{2}$$.

The given line $$2x + y = 1$$ has slope $$-2$$. A line perpendicular to it has slope $$m = \frac{1}{2}$$.

The equation of the tangent to $$y^2 = 6x$$ with slope $$m$$ is $$y = mx + \frac{a}{m} = \frac{x}{2} + \frac{3/2}{1/2} = \frac{x}{2} + 3$$.

Multiplying through by 2: $$2y = x + 6$$, or equivalently $$x - 2y + 6 = 0$$.

Now we check which point does NOT lie on this line. For $$(0, 3)$$: $$0 - 6 + 6 = 0$$ (lies on it). For $$(4, 5)$$: $$4 - 10 + 6 = 0$$ (lies on it). For $$(5, 4)$$: $$5 - 8 + 6 = 3 \neq 0$$ (does NOT lie on it). For $$(-6, 0)$$: $$-6 - 0 + 6 = 0$$ (lies on it).

Therefore, the point that does NOT lie on the tangent is $$(5, 4)$$.

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.

We are given that the straight line $$y = mx + 4$$ touches (is tangent to) both parabolas $$y^2 = 4x$$ and $$x^2 = 2by$$. We shall first impose the condition of tangency with the first parabola to determine the slope $$m$$, and then use this slope to find the unknown parameter $$b$$ for the second parabola.

For the parabola $$y^2 = 4x$$, we can rewrite it as $$x = \dfrac{y^2}{4}$$. Substituting this value of $$x$$ into the line $$y = mx + 4$$ gives

$$y \;=\; m\left(\dfrac{y^2}{4}\right) + 4.$$

Multiplying every term by $$4$$ to clear the denominator, we obtain

$$4y \;=\; my^2 + 16.$$

Rearranging all terms to one side produces a quadratic in $$y$$:

$$my^2 - 4y + 16 \;=\; 0.$$

For a line to be tangent to a parabola, this quadratic must possess exactly one real solution. The quadratic formula tells us that a quadratic $$Ay^2 + By + C = 0$$ has a repeated root when its discriminant $$\Delta = B^2 - 4AC$$ is zero. Here

$$A = m,\quad B = -4,\quad C = 16.$$

So we set

$$\Delta = (-4)^2 - 4(m)(16) = 0.$$

Simplifying, we have

$$16 - 64m = 0 \;\;\Longrightarrow\;\; 64m = 16 \;\;\Longrightarrow\;\; m = \dfrac{16}{64} = \dfrac14.$$

Thus the common tangent must be

$$y = \dfrac14\,x + 4.$$

Now we require this same line to be tangent to the second parabola $$x^2 = 2by$$. Substituting $$y = \dfrac14\,x + 4$$ into that parabola yields

$$x^2 \;=\; 2b\left(\dfrac14\,x + 4\right).$$

Distributing on the right side, we get

$$x^2 \;=\; \frac{2b}{4}\,x + 8b \;=\; \frac{b}{2}\,x + 8b.$$

Collecting all terms to one side gives a quadratic in $$x$$:

$$x^2 - \frac{b}{2}\,x - 8b = 0.$$

Again, tangency requires this quadratic to have exactly one real solution, so its discriminant must vanish. For a general quadratic $$ax^2 + bx + c = 0$$, the discriminant condition is $$b^2 - 4ac = 0$$. Here

$$a = 1,\quad b = -\dfrac{b}{2},\quad c = -8b.$$

Setting the discriminant to zero:

$$\left(-\dfrac{b}{2}\right)^2 - 4(1)(-8b) = 0.$$

Simplifying step by step,

$$\dfrac{b^2}{4} + 32b = 0.$$

Multiplying by $$4$$ to clear the denominator, we get

$$b^2 + 128b = 0.$$

Factoring,

$$b(b + 128) = 0.$$

This gives two possible values: $$b = 0$$ or $$b = -128$$. The case $$b = 0$$ would make the parabola $$x^2 = 2by$$ degenerate (it would collapse to the line $$x = 0$$), so we discard it. Therefore, the only admissible value is

$$b = -128.$$

Hence, the correct answer is Option C.

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 start with the parabola $$y^{2}=8x$$.

First we bring it to the standard form $$y^{2}=4ax$$ so that we can use all known formulae directly.

Comparing $$y^{2}=8x$$ with $$y^{2}=4ax$$, we obtain $$4a = 8 \;\Longrightarrow\; a = 2.$$

For the parabola $$y^{2}=4ax$$, the focus is always at $$(a,0)$$. Hence here the focus is $$(2,0).$$

Any point on this parabola can be expressed parametrically. With the parameter $$t$$, the coordinates are

$$\bigl(at^{2},\,2at\bigr) = \bigl(2t^{2},\,4t\bigr).$$

Let the parameters corresponding to the two ends of the focal chord $$AB$$ be $$t_{1}$$ and $$t_{2}$$. A standard property of a focal chord of a parabola $$y^{2}=4ax$$ is

$$t_{1}\,t_{2} = -1.$$

We are given the coordinates of the point $$A\Bigl(\dfrac12,\,-2\Bigr)$$. We now identify $$t_{1}$$ by equating these coordinates to $$(2t_{1}^{2},\,4t_{1})$$.

From the $$x$$-coordinate:

$$2t_{1}^{2} = \dfrac12 \;\Longrightarrow\; t_{1}^{2} = \dfrac14 \;\Longrightarrow\; t_{1} = \pm \dfrac12.$$

From the $$y$$-coordinate:

$$4t_{1} = -2 \;\Longrightarrow\; t_{1} = -\dfrac12.$$

Both equations agree only for $$t_{1} = -\dfrac12.$$

Using the focal-chord relation $$t_{1} t_{2} = -1$$, we now find $$t_{2}$$:

$$\left(-\dfrac12\right) t_{2} = -1 \;\Longrightarrow\; t_{2} = 2.$$

So the coordinates of the second end $$B$$ are obtained by substituting $$t_{2}=2$$ in $$(2t^{2},4t)$$:

$$x_{B} = 2\,(2)^{2} = 2 \times 4 = 8,$$

$$y_{B} = 4\,(2) = 8.$$

Hence $$B(8,\,8).$$

For a general point $$(at^{2},\,2at)$$ on $$y^{2}=4ax$$, the tangent has the well-known form

$$t\,y = x + a\,t^{2}.$$

We now substitute $$a = 2$$ and the specific parameter $$t = t_{2} = 2$$:

$$2\,y = x + 2\,(2)^{2}.$$

Step by step simplification:

$$(2)^{2} = 4,$$

$$2 \times 4 = 8,$$

so the equation becomes

$$2y = x + 8.$$

Bring every term to the left to obtain the standard linear form:

$$x - 2y + 8 = 0.$$

Thus the required tangent at $$B$$ is

$$x - 2y + 8 = 0.$$

Hence, the correct answer is Option B.

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 $$2$$:

$$2y = -x - 4.$$