JEE Conic Sections Questions

The centre of the circle $$x^{2}+y^{2}-2x-4y-11=0$$ is obtained by completing the squares.
$$x^{2}-2x+1+y^{2}-4y+4=11+1+4$$
$$\Rightarrow (x-1)^{2}+(y-2)^{2}=16$$

Hence the centre is $$C(1,2)$$ and the radius is $$r=4$$.

The ellipse is $$3x^{2}+py^{2}=4$$ and it passes through the point $$C(1,2)$$, so

$$3(1)^{2}+p(2)^{2}=4$$
$$\Rightarrow 3+4p=4$$
$$\Rightarrow 4p=1 \;\Longrightarrow\; p=\frac14$$

Therefore the ellipse is $$3x^{2}+\frac14\,y^{2}=4$$.

Divide by $$4$$ to get the standard form:

$$\frac{3}{4}x^{2}+\frac{1}{16}y^{2}=1$$
$$\Rightarrow \frac{x^{2}}{\frac{4}{3}}+\frac{y^{2}}{16}=1$$

Comparing with $$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$$ (major axis along $$y$$-axis), we have

$$a^{2}=16,\;a=4,\qquad b^{2}=\frac{4}{3}$$

The focal length satisfies $$c^{2}=a^{2}-b^{2}$$, hence

$$c^{2}=16-\frac{4}{3}=\frac{48-4}{3}=\frac{44}{3},\qquad c=\sqrt{\frac{44}{3}}$$

The foci are $$F_{1}(0,c)$$ and $$F_{2}(0,-c)$$.

Let the focal distances of the point $$C(1,2)$$ be $$f_{1}=CF_{1}$$ and $$f_{2}=CF_{2}$$.

Since $$C$$ lies on the ellipse, the sum of its focal distances equals the major axis length:

$$f_{1}+f_{2}=2a=8 \quad -(1)$$

Now compute the squares of the distances.

$$f_{1}^{2}=(1-0)^{2}+(2-c)^{2}=1+(2-c)^{2} =1+4-4c+c^{2}=5-4c+c^{2}$$
$$f_{2}^{2}=(1-0)^{2}+(2+c)^{2}=1+(2+c)^{2} =1+4+4c+c^{2}=5+4c+c^{2}$$

Adding:

$$f_{1}^{2}+f_{2}^{2}=(5-4c+c^{2})+(5+4c+c^{2}) =10+2c^{2} \quad -(2)$$

From identities,

$$(f_{1}+f_{2})^{2}=f_{1}^{2}+f_{2}^{2}+2f_{1}f_{2} \quad -(3)$$

Substitute $$-(1)$$ and $$-(2)$$ into $$-(3)$$:

$$8^{2}=10+2c^{2}+2f_{1}f_{2}$$
$$64=10+2\left(\frac{44}{3}\right)+2f_{1}f_{2}$$
$$64=10+\frac{88}{3}+2f_{1}f_{2}$$
$$64=\frac{30}{3}+\frac{88}{3}+2f_{1}f_{2} =\frac{118}{3}+2f_{1}f_{2}$$

$$\Rightarrow 2f_{1}f_{2}=64-\frac{118}{3} =\frac{192-118}{3}=\frac{74}{3}$$

$$\therefore\; f_{1}f_{2}=\frac{37}{3}$$

Finally,

$$6f_{1}f_{2}-r=6\left(\frac{37}{3}\right)-4 =2\cdot37-4 =74-4 =70$$

Hence $$6f_{1}f_{2}-r=70$$, which matches Option C.

We have two ellipses with the same eccentricity $$e = \frac{1}{\sqrt{3}}$$, product of latus rectum lengths $$= \frac{32}{\sqrt{3}}$$, and distance between foci of $$E_1$$ is 4.

For the ellipse $$E_1$$ given by $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ with $$a \gt b$$ and eccentricity $$e = \frac{1}{\sqrt{3}}$$, the distance between the foci equals $$2ae = 4$$, so $$ae = 2$$, yielding $$a = 2\sqrt{3}$$. Then

$$b^2 = a^2(1 - e^2) = 12\left(1 - \frac{1}{3}\right) = 12 \times \frac{2}{3} = 8$$

and hence $$b = 2\sqrt{2}$$. The latus rectum of $$E_1$$ is

$$\ell_1 = \frac{2b^2}{a} = \frac{16}{2\sqrt{3}} = \frac{8}{\sqrt{3}}.$$

For the ellipse $$E_2$$ defined by $$\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$$ with $$A \lt B$$ (so the major axis is along the y-axis) and the same eccentricity $$e = \frac{1}{\sqrt{3}}$$, we have

$$A^2 = B^2(1 - e^2) = \frac{2B^2}{3}.$$

Its latus rectum is

$$\ell_2 = \frac{2A^2}{B} = \frac{2 \cdot \frac{2B^2}{3}}{B} = \frac{4B}{3}.$$

Since the product of the latus recta is

$$\ell_1 \cdot \ell_2 = \frac{8}{\sqrt{3}} \cdot \frac{4B}{3} = \frac{32B}{3\sqrt{3}} = \frac{32}{\sqrt{3}},$$

it follows that $$B = 3$$ and hence

$$A^2 = \frac{2 \times 9}{3} = 6.$$

The equations of the ellipses become

$$E_1: \frac{x^2}{12} + \frac{y^2}{8} = 1,\qquad E_2: \frac{x^2}{6} + \frac{y^2}{9} = 1.$$

From the first,

$$x^2 = 12\left(1 - \frac{y^2}{8}\right) = 12 - \frac{3y^2}{2},$$

which substituted into the second gives

$$\frac{12 - \frac{3y^2}{2}}{6} + \frac{y^2}{9} = 1$$

$$2 - \frac{y^2}{4} + \frac{y^2}{9} = 1$$

$$y^2\left(\frac{1}{9} - \frac{1}{4}\right) = -1$$

$$y^2 \times \left(\frac{-5}{36}\right) = -1 \implies y^2 = \frac{36}{5},$$

$$x^2 = 12 - \frac{3}{2} \times \frac{36}{5} = 12 - \frac{54}{5} = \frac{6}{5}.$$

Thus the four intersection points are

$$\left(\pm\sqrt{\frac{6}{5}},\; \pm\sqrt{\frac{36}{5}}\right),$$

forming a rectangle by symmetry. Its side lengths are $$2\sqrt{\frac{6}{5}}$$ and $$2\sqrt{\frac{36}{5}}$$, so the area is

$$2\sqrt{\frac{6}{5}} \times 2\sqrt{\frac{36}{5}} = 4\sqrt{\frac{216}{25}} = 4 \times \frac{6\sqrt{6}}{5} = \frac{24\sqrt{6}}{5}.$$

The correct answer is Option 4: $$\frac{24\sqrt{6}}{5}$$.

The hyperbola is $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$, with foci $$F_{1}(-c,0)$$ and $$F_{2}(c,0)$$, where $$c^{2}=a^{2}+b^{2}$$.

For any point on this hyperbola, the difference of its focal distances equals the length of the transverse axis: $$|PF_{2}-PF_{1}|=2a$$.

Let
$$d_{1}=PF_{1},\qquad d_{2}=PF_{2}$$
for the point $$P(4,3)$$. The data give the following:

1. Sum of focal distances
$$d_{1}+d_{2}=8\sqrt{\frac{5}{3}}$$ $$-(1)$$

2. Using coordinates,
$$d_{1}^{2}=(4-c)^{2}+3^{2}=25-8c+c^{2}$$
$$d_{2}^{2}=(4+c)^{2}+3^{2}=25+8c+c^{2}$$

Hence
$$d_{2}^{2}-d_{1}^{2}=16c$$ $$-(2)$$

3. From the property of a hyperbola,
$$d_{2}-d_{1}=2a$$ $$-(3)$$

Using $$(1)$$ and $$(3)$$,
$$d_{2}=\frac{(d_{1}+d_{2})+(d_{2}-d_{1})}{2}=\frac{S}{2}+a,\quad d_{1}=\frac{S}{2}-a,$$ where $$S=d_{1}+d_{2}=8\sqrt{\frac{5}{3}}$$.

The product $$(d_{2}+d_{1})(d_{2}-d_{1})$$ equals $$16c$$ by $$(2)$$, so $$(d_{2}+d_{1})(d_{2}-d_{1})=S\,(2a)=16c.$$ Thus $$c=\frac{aS}{8}.$$

Squaring and substituting $$S^{2}=\left(8\sqrt{\frac{5}{3}}\right)^{2}=\frac{320}{3},$$ $$c^{2}=\frac{a^{2}S^{2}}{64}=a^{2}\,\frac{5}{3}.$$ But $$c^{2}=a^{2}+b^{2},$$ therefore $$a^{2}+b^{2}=a^{2}\,\frac{5}{3}\;\;\Longrightarrow\;\;b^{2}=\frac{2}{3}a^{2}.$$ $$-(4)$$

The point $$P(4,3)$$ lies on the hyperbola, so $$\frac{4^{2}}{a^{2}}-\frac{3^{2}}{b^{2}}=1.$$ Using $$(4)$$, $$\frac{16}{a^{2}}-\frac{9}{(2/3)a^{2}}=1 \;\;\Longrightarrow\;\; \frac{16}{a^{2}}-\frac{27}{2a^{2}}=1 \;\;\Longrightarrow\;\; \frac{2.5}{a^{2}}=1.$$ Hence $$a^{2}=\frac{5}{2},\qquad b^{2}=\frac{2}{3}\cdot\frac{5}{2}=\frac{5}{3},\qquad c^{2}=a^{2}+b^{2}=\frac{25}{6}.$$

Length of the latus rectum
For a hyperbola, $$l=\frac{2b^{2}}{a}.$$ Thus $$l=\frac{2(\frac{5}{3})}{\sqrt{\frac{5}{2}}}=\frac{10}{3}\sqrt{\frac{2}{5}},\qquad l^{2}=\frac{40}{9}.$$ Therefore $$9l^{2}=9\cdot\frac{40}{9}=40.$$

Product of the focal distances
Using $$S^{2}=d_{1}^{2}+d_{2}^{2}+2d_{1}d_{2},$$ first note $$d_{1}^{2}+d_{2}^{2}=(25-8c+c^{2})+(25+8c+c^{2})=50+2c^{2}.$$ Hence $$S^{2}=50+2c^{2}+2m \;\;\Longrightarrow\;\; m=\frac{S^{2}-50-2c^{2}}{2}.$$ Substituting $$S^{2}=\frac{320}{3},\;c^{2}=\frac{25}{6},$$ $$m=\frac{\frac{320}{3}-50-\frac{25}{3}}{2} =\frac{\frac{320-150-25}{3}}{2} =\frac{\frac{145}{3}}{2} =\frac{145}{6}.$$ Therefore $$6m=145.$$

Required value
$$9l^{2}+6m=40+145=185.$$

Hence the correct option is Option C (185).

The standard equation of an ellipse with its major axis along the $$x$$-axis is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$, where $$a \gt b \gt 0$$ and its eccentricity is $$e$$.

1. Length of the latus-rectum (the chord through a focus perpendicular to the major axis) is
$$\text{Latus rectum}= \frac{2b^{2}}{a}$$.

2. The eccentricity and the semi-axes satisfy the relation
$$b^{2}=a^{2}(1-e^{2})$$ $$-(1)$$.

3. We are given that the length of the latus-rectum equals $$10$$:
$$\frac{2b^{2}}{a}=10$$ $$-(2)$$.

4. The eccentricity $$e$$ is the minimum value of the quadratic function
$$f(t)=t^{2}+t+\frac{11}{12},\; t\in\mathbb{R}$$.

The minimum of a quadratic $$At^{2}+Bt+C$$ with $$A\gt 0$$ occurs at $$t=-\frac{B}{2A}$$.
Here $$A=1,\; B=1,\; C=\frac{11}{12}$$, so the minimising value of $$t$$ is
$$t=-\frac{1}{2}$$.

Substituting back gives the minimum value (and hence the eccentricity):
$$e=f\!\left(-\frac{1}{2}\right) =\left(-\frac{1}{2}\right)^{2} +\left(-\frac{1}{2}\right) +\frac{11}{12} =\frac{1}{4}-\frac{1}{2}+\frac{11}{12}$$
$$=\frac{1}{4}-\frac{2}{4}+\frac{11}{12} =-\frac{1}{4}+\frac{11}{12} =\frac{-3}{12}+\frac{11}{12} =\frac{8}{12} =\frac{2}{3}$$.

Thus $$e=\frac{2}{3}$$ and $$e^{2}=\frac{4}{9}$$.

5. Using equation $$(1)$$:
$$b^{2}=a^{2}\!\left(1-\frac{4}{9}\right)=a^{2}\!\left(\frac{5}{9}\right) =\frac{5a^{2}}{9}$$ $$-(3)$$.

6. Insert $$(3)$$ into the latus-rectum condition $$(2)$$:
$$\frac{2}{a}\left(\frac{5a^{2}}{9}\right)=10 \;\;\Longrightarrow\;\; \frac{10a}{9}=10 \;\;\Longrightarrow\;\; a=\frac{10\times9}{10}=9$$.

Therefore $$a^{2}=9^{2}=81$$.

7. Find $$b^{2}$$ from $$(3)$$:
$$b^{2}=\frac{5a^{2}}{9} =\frac{5\times81}{9}=5\times9=45$$.

8. Finally,
$$a^{2}+b^{2}=81+45=126$$.

Hence $$a^{2}+b^{2}=126$$. The correct option is Option B.

For the ellipse $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$, the equation of the chord with midpoint $$(h, k)$$ is given by:

$$ \frac{xh}{9} + \frac{yk}{4} = \frac{h^2}{9} + \frac{k^2}{4} $$

Given midpoint $$\left(\frac{5}{2}, \frac{1}{2}\right)$$:

$$ \frac{x \cdot \frac{5}{2}}{9} + \frac{y \cdot \frac{1}{2}}{4} = \frac{\left(\frac{5}{2}\right)^2}{9} + \frac{\left(\frac{1}{2}\right)^2}{4} $$

$$ \frac{5x}{18} + \frac{y}{8} = \frac{25}{36} + \frac{1}{16} $$

RHS: $$\frac{25}{36} + \frac{1}{16} = \frac{400 + 36}{576} = \frac{436}{576} = \frac{109}{144}$$

LHS: $$\frac{5x}{18} + \frac{y}{8}$$

Multiplying through by 144:

$$ 40x + 18y = 109 $$

Comparing with $$\alpha x + \beta y = 109$$:

$$\alpha = 40, \quad \beta = 18$$

$$\alpha + \beta = 40 + 18 = 58$$

The correct answer is Option 1: 58.

The circle is $$x^{2}+y^{2}-8x=0$$, which can be written as $$(x-4)^{2}+y^{2}=16$$. Its centre is $$(4,0)$$ and radius is $$4$$.

The hyperbola is $$\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$$, or equivalently $$4x^{2}-9y^{2}=36$$ $$-(1)$$.

For the points of intersection A and B we must satisfy both the circle and the hyperbola. From the circle, $$y^{2}=8x-x^{2}$$ $$-(2)$$. Substituting $$y^{2}$$ from $$(2)$$ into $$(1)$$ gives

$$4x^{2}-9(8x-x^{2})=36$$ $$\Rightarrow 4x^{2}-72x+9x^{2}=36$$ $$\Rightarrow 13x^{2}-72x-36=0$$ $$-(3)$$.

The discriminant of $$(3)$$ is $$\Delta = (-72)^{2}-4\cdot 13\cdot (-36)=5184+1872=7056=84^{2}$$, so the roots are

$$x=\frac{72\pm 84}{26} \;\Rightarrow\; x_{1}=6,\; x_{2}=-\frac{6}{13}$$.

Using $$(2)$$ for each root: For $$x=6$$, $$y^{2}=8\cdot 6-6^{2}=48-36=12 \;\Rightarrow\; y=\pm 2\sqrt{3}$$. For $$x=-\dfrac{6}{13}$$, $$y^{2}=-\dfrac{660}{169}\lt 0$$, which is impossible. Hence the real intersection points are $$A(6,2\sqrt{3}),\; B(6,-2\sqrt{3})$$.

Let the moving point be $$P(x,y)$$ on the line $$2x-3y+4=0$$ $$-(4)$$.

The centroid $$G(X,Y)$$ of $$\triangle PAB$$ is $$X=\frac{x+6+6}{3}=\frac{x+12}{3}, \qquad Y=\frac{y+2\sqrt{3}-2\sqrt{3}}{3}=\frac{y}{3}$$ $$-(5)$$.

Express $$x$$ and $$y$$ from $$(5)$$: $$x=3X-12, \qquad y=3Y$$ $$-(6)$$.

Substitute $$(6)$$ into the line $$(4)$$:

$$2(3X-12)-3(3Y)+4=0$$ $$\Rightarrow 6X-24-9Y+4=0$$ $$\Rightarrow 6X-9Y-20=0$$ $$\Rightarrow 6X-9Y=20$$.

Replacing $$(X,Y)$$ by the usual $$x,y$$ gives the locus of the centroid: $$6x-9y=20$$.

Thus the required line is $$6x-9y=20$$, which corresponds to Option C.

The given ellipse is $$\dfrac{x^{2}}{4^{2}}+\dfrac{y^{2}}{3^{2}}=1$$, that is $$\dfrac{x^{2}}{16}+\dfrac{y^{2}}{9}=1$$.

Case 1: Find the values of $$p$$ for which the line $$y = x + p$$ is tangent to the ellipse.

Substitute $$y = x + p$$ in the ellipse:

$$\dfrac{x^{2}}{16} + \dfrac{(x+p)^{2}}{9} = 1$$

Multiply by $$144$$ (the LCM of $$16$$ and $$9$$):
$$9x^{2} + 16(x+p)^{2} = 144$$

Expand the square:
$$9x^{2} + 16\bigl(x^{2}+2px+p^{2}\bigr) - 144 = 0$$

Simplify:
$$25x^{2} + 32p\,x + 16p^{2} - 144 = 0$$ $$-(1)$$

For tangency the quadratic in $$x$$ must have discriminant $$0$$:

$$\bigl(32p\bigr)^{2} - 4 \cdot 25 \bigl(16p^{2}-144\bigr) = 0$$

$$1024p^{2} - 1600p^{2} + 14400 = 0$$

$$-576p^{2} + 14400 = 0 \;\;\Longrightarrow\;\; 576p^{2} = 14400$$

$$p^{2} = 25 \;\;\Longrightarrow\;\; p = \pm 5$$

Thus the two tangent lines are
$$y = x + 5 \quad\text{and}\quad y = x - 5$$.

Case 2: Coordinates of the points of contact $$A$$ and $$B$$.

For a tangent quadratic $$ax^{2}+bx+c=0$$, at tangency $$x = -\dfrac{b}{2a}$$. In equation $$-(1)$$ we have $$a = 25,\; b = 32p$$, so

$$x = -\dfrac{32p}{2\cdot25} = -\dfrac{16p}{25}$$.

• For $$p = 5$$:
$$x_A = -\dfrac{16(5)}{25} = -\dfrac{16}{5},\qquad y_A = x_A + 5 = -\dfrac{16}{5} + \dfrac{25}{5} = \dfrac{9}{5}$$

• For $$p = -5$$:
$$x_B = -\dfrac{16(-5)}{25} = \dfrac{16}{5},\qquad y_B = x_B - 5 = \dfrac{16}{5} - \dfrac{25}{5} = -\dfrac{9}{5}$$

Hence
$$A\!\left(-\dfrac{16}{5},\;\dfrac{9}{5}\right),\quad B\!\left(\;\dfrac{16}{5},\;-\dfrac{9}{5}\right).$$

Case 3: Intersection points $$C$$ and $$D$$ of the line $$y = x$$ with the ellipse.

Put $$y = x$$ in the ellipse:

$$\dfrac{x^{2}}{16} + \dfrac{x^{2}}{9} = 1 \;\;\Longrightarrow\;\; x^{2}\Bigl(\dfrac{1}{16}+\dfrac{1}{9}\Bigr) = 1$$

$$x^{2}\Bigl(\dfrac{25}{144}\Bigr)=1 \;\;\Longrightarrow\;\; x^{2}=\dfrac{144}{25}$$

$$x = \pm\dfrac{12}{5}\;\;,\;\; y = x$$

Therefore
$$C\!\left(\dfrac{12}{5},\;\dfrac{12}{5}\right),\quad D\!\left(-\dfrac{12}{5},\;-\dfrac{12}{5}\right).$$

Case 4: Area of quadrilateral $$ABCD$$ (vertices taken in order $$C \rightarrow A \rightarrow D \rightarrow B$$, which goes anticlockwise).

Using the Shoelace Theorem:

Coordinates:
$$C\left(\dfrac{12}{5},\;\dfrac{12}{5}\right),\; A\left(-\dfrac{16}{5},\;\dfrac{9}{5}\right),\; D\left(-\dfrac{12}{5},-\dfrac{12}{5}\right),\; B\left(\;\dfrac{16}{5},-\dfrac{9}{5}\right)$$

Compute $$\sum x_i\,y_{i+1}$$:
$$\dfrac{12}{5}\!\cdot\!\dfrac{9}{5} + \left(-\dfrac{16}{5}\right)\!\cdot\!\left(-\dfrac{12}{5}\right) + \left(-\dfrac{12}{5}\right)\!\cdot\!\left(-\dfrac{9}{5}\right) + \dfrac{16}{5}\!\cdot\!\dfrac{12}{5} = \dfrac{108+192+108+192}{25} = \dfrac{600}{25} = 24$$

Compute $$\sum y_i\,x_{i+1}$$:
$$\dfrac{12}{5}\!\cdot\!\left(-\dfrac{16}{5}\right) + \dfrac{9}{5}\!\cdot\!\left(-\dfrac{12}{5}\right) + \left(-\dfrac{12}{5}\right)\!\cdot\!\dfrac{16}{5} + \left(-\dfrac{9}{5}\right)\!\cdot\!\dfrac{12}{5} = \dfrac{-192-108-192-108}{25} = -\dfrac{600}{25} = -24$$

Area
$$=\dfrac{1}{2}\Bigl|\;\sum x_i y_{i+1} - \sum y_i x_{i+1}\Bigr| =\dfrac{1}{2}\bigl|24 - (-24)\bigr| =\dfrac{1}{2}\times48 = 24.$$

Hence the area of quadrilateral $$ABCD$$ is $$24$$.

Therefore, Option B is correct.

The given ellipse is $$\frac{x^{2}}{18}+\frac{y^{2}}{9}=1$$.

Comparing with $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ we get $$a^{2}=18,\;b^{2}=9$$.
Hence $$a=3\sqrt{2},\;b=3$$.

For an ellipse, $$c^{2}=a^{2}-b^{2}$$, so
$$c^{2}=18-9=9\;\Longrightarrow\;c=3$$.

The foci are $$S(-3,0)$$ and $$S'(3,0)$$.

Let an arbitrary point on the ellipse be $$P(x,y)$$. Because $$P$$ lies on the ellipse,

$$\frac{x^{2}}{18}+\frac{y^{2}}{9}=1 \quad\Longrightarrow\quad y^{2}=9\left(1-\frac{x^{2}}{18}\right)=9-\frac{x^{2}}{2}\;.$$ $$-(1)$$

Distances of $$P$$ from the two foci are

$$SP=\sqrt{(x+3)^{2}+y^{2}},\qquad S'P=\sqrt{(x-3)^{2}+y^{2}}.$$

We need the product $$SP\cdot S'P$$. First square each distance:

$$SP^{2}=(x+3)^{2}+y^{2},\qquad S'P^{2}=(x-3)^{2}+y^{2}.$$

Using $$(1)$$ to eliminate $$y^{2}$$:

$$SP^{2}=(x+3)^{2}+9-\frac{x^{2}}{2}$$ $$\phantom{SP^{2}}=\frac{x^{2}}{2}+6x+9+9 =\frac{x^{2}}{2}+6x+18,$$

$$S'P^{2}=(x-3)^{2}+9-\frac{x^{2}}{2}$$ $$\phantom{S'P^{2}}=\frac{x^{2}}{2}-6x+18.$$

Therefore

$$\bigl(SP\cdot S'P\bigr)^{2}=SP^{2}\,S'P^{2} =\left(\frac{x^{2}}{2}+6x+18\right)\left(\frac{x^{2}}{2}-6x+18\right).$$

Recognise the form $$(A+6x)(A-6x)=A^{2}-36x^{2},$$ where $$A=\dfrac{x^{2}}{2}+18$$. Hence

$$\bigl(SP\cdot S'P\bigr)^{2} =\left(\frac{x^{2}}{2}+18\right)^{2}-36x^{2}.$$ Expand: $$\bigl(SP\cdot S'P\bigr)^{2} =\frac{x^{4}}{4}+18x^{2}+324-36x^{2} =\frac{x^{4}}{4}-18x^{2}+324.$$ Let $$X=x^{2}\;(X\ge 0)$$ to work with one variable: $$\bigl(SP\cdot S'P\bigr)^{2} =\frac{X^{2}}{4}-18X+324.$$ $$-(2)$$

The point $$P$$ lies on the ellipse, so $$x^{2}\le a^{2}=18$$, that is $$0\le X\le 18$$.

Equation $$(2)$$ is a quadratic in $$X$$ with coefficient $$\frac14>0$$; it opens upward. Its vertex is at

$$X_v=\frac{-(-18)}{2\left(\frac14\right)} =\frac{18}{0.5}=36.$$

The vertex lies to the right of the allowed interval $$[0,18]$$, so within the interval the quadratic is strictly decreasing.
Hence:

• Maximum of $$(SP\cdot S'P)^{2}$$ occurs at the left end $$X=0$$.
• Minimum of $$(SP\cdot S'P)^{2}$$ occurs at the right end $$X=18$$.

From $$(2):\;\bigl(SP\cdot S'P\bigr)^{2}=324 \;\Longrightarrow\;SP\cdot S'P=18.$$

Substitute $$X=18$$ in $$(2):$$ $$\bigl(SP\cdot S'P\bigr)^{2} =\frac{(18)^{2}}{4}-18(18)+324 =81-324+324=81,$$ $$\Longrightarrow\;SP\cdot S'P=9.$$

Thus

$$\min(SP\cdot S'P)=9,\qquad \max(SP\cdot S'P)=18.$$

Required sum:

$$\min(SP\cdot S'P)+\max(SP\cdot S'P)=9+18=27.$$

Option D is correct.

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.

Let the centre of the ellipse $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$$ be the origin $$O(0,0)$$.
Because the length of the major axis is given as 17,

$$2a = 17 \; \Longrightarrow \; a = \dfrac{17}{2}$$

For an ellipse, the distance of each focus from the centre is $$c$$ where

$$c^{2}=a^{2}-b^{2} \quad -(1)$$

The foci are therefore $$F_{1}(-c,0)$$ and $$F_{2}(c,0)$$.

The circle $$C$$ is also centred at $$O$$ and passes through the foci, hence its radius is $$c$$ and its equation is

$$x^{2}+y^{2}=c^{2} \quad -(2)$$

Let $$P(x,y)$$ be one of the four common points of the ellipse and the circle.
Then $$P$$ satisfies both $$(2)$$ and the ellipse equation, giving

$$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1 \quad -(3)$$

From $$(2)$$: $$y^{2}=c^{2}-x^{2} \quad -(4)$$

Substitute $$(4)$$ into $$(3)$$:

$$\dfrac{x^{2}}{a^{2}}+\dfrac{c^{2}-x^{2}}{b^{2}}=1$$

Simplify the left side:

$$x^{2}\left(\dfrac{1}{a^{2}}-\dfrac{1}{b^{2}}\right)+\dfrac{c^{2}}{b^{2}}=1$$

Using $$(1)$$, $$b^{2}=a^{2}-c^{2}$$, so

$$\dfrac{1}{a^{2}}-\dfrac{1}{b^{2}}=\dfrac{b^{2}-a^{2}}{a^{2}b^{2}}=-\dfrac{c^{2}}{a^{2}b^{2}}$$

Hence

$$-\dfrac{c^{2}}{a^{2}b^{2}}\,x^{2}+\dfrac{c^{2}}{b^{2}}=1$$

Multiply by $$b^{2}$$:

$$c^{2}-\dfrac{c^{2}}{a^{2}}\,x^{2}=b^{2}$$

Replace $$b^{2}$$ again by $$a^{2}-c^{2}$$:

$$c^{2}-\dfrac{c^{2}}{a^{2}}\,x^{2}=a^{2}-c^{2}$$

Rearrange to solve for $$x^{2}$$:

$$\dfrac{c^{2}}{a^{2}}\,x^{2}=2c^{2}-a^{2}$$

$$x^{2}=a^{2}\dfrac{2c^{2}-a^{2}}{c^{2}} \quad -(5)$$

Now obtain $$y^{2}$$ from $$(4)$$:

$$y^{2}=c^{2}-x^{2}=c^{2}-a^{2}\dfrac{2c^{2}-a^{2}}{c^{2}} \quad -(6)$$

Next use the area condition.
For the triangle with vertices $$P, F_{1}(-c,0), F_{2}(c,0)$$:

• Base $$F_{1}F_{2}=2c$$.
• The altitude from $$P$$ to the base is $$|y|$$ (since the base lies on the $$x$$-axis).

Hence

$$\text{Area}=\dfrac{1}{2}\times 2c \times |y|=c|y|$$

The area is given as 30, so

$$c|y| = 30 \;\Longrightarrow\; y^{2}=\dfrac{900}{c^{2}} \quad -(7)$$

Equate $$(6)$$ and $$(7)$$:

$$\dfrac{900}{c^{2}} = c^{2}-a^{2}\dfrac{2c^{2}-a^{2}}{c^{2}}$$

Multiply by $$c^{2}$$ to clear denominators:

$$900 = c^{4}-a^{2}(2c^{2}-a^{2}) \quad -(8)$$

Insert $$a^{2}=\left(\dfrac{17}{2}\right)^{2}=\dfrac{289}{4}$$ into $$(8)$$:

$$900 = c^{4}-\dfrac{289}{4}\left(2c^{2}-\dfrac{289}{4}\right)$$

Multiply by 16 to avoid fractions:

$$16c^{4}-8\!\cdot\!289\,c^{2}+289^{2}-14400=0$$

That is

$$16c^{4}-2312c^{2}+69121=0$$

Let $$k=c^{2}$$. The quadratic in $$k$$ becomes

$$16k^{2}-2312k+69121=0$$

Using the quadratic formula,

$$k=\dfrac{2312\pm\sqrt{2312^{2}-4\!\cdot\!16\!\cdot\!69121}}{32}$$

The discriminant is

$$2312^{2}-64\!\cdot\!69121 = 5345344-4423744 = 921600 = 960^{2}$$

Hence

$$k=\dfrac{2312\pm960}{32}$$

This gives two values: $$k=102.25$$ and $$k=42.25$$.
But $$k=c^{2}$$ must satisfy $$c^{2}\lt a^{2}=72.25$$, so we take

$$c^{2}=42.25 \;=\;\left(6.5\right)^{2}$$

Therefore

$$c = 6.5, \quad 2c = 13$$

The distance between the foci of the ellipse is 13.
Hence the correct option is Option B (13).

The two foci are $$(2,5)$$ and $$(2,-3)$$. Their common $$x$$-coordinate is $$2$$, so the major axis is the vertical line $$x = 2$$.

For an ellipse with vertical major axis and centre $$(h,k)$$, the standard form is
$$\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1$$ with $$a \gt b$$, foci $$(h,\,k \pm c)$$, and relations $$c^2 = a^2 - b^2$$, $$e = \frac{c}{a}$$.

Step 1: Find $$c$$
Distance between the given foci:
$$2c = |5 - (-3)| = 8$$
$$\Rightarrow c = 4$$.

Step 2: Find $$a$$ using eccentricity
Given $$e = \frac{4}{5}$$ and $$e = \frac{c}{a}$$,
$$\frac{4}{5} = \frac{4}{a} \;\Rightarrow\; a = 5$$.

Step 3: Find $$b$$
Using $$c^2 = a^2 - b^2$$:
$$4^2 = 5^2 - b^2$$
$$16 = 25 - b^2$$
$$b^2 = 25 - 16 = 9$$
$$\Rightarrow b = 3$$.

Step 4: Length of the latus-rectum
For any ellipse, length of latus-rectum $$L = \frac{2b^2}{a}$$.
Substituting $$b^2 = 9$$ and $$a = 5$$:
$$L = \frac{2 \times 9}{5} = \frac{18}{5}$$.

Hence, the length of the latus-rectum is $$\frac{18}{5}$$.

Option D is correct.

We are given the ellipse $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$ with midpoint of a chord at $$(\sqrt{2}, \frac{4}{3})$$.

We start by finding the equation of the chord with midpoint $$(h, k)$$ in the standard form of an ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, which is given by $$\frac{xh}{a^2} + \frac{yk}{b^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2}$$.

Substituting $$a^2 = 9$$, $$b^2 = 4$$, $$h = \sqrt{2}$$, and $$k = \frac{4}{3}$$ into this formula yields $$\frac{\sqrt{2}x}{9} + \frac{4y}{3 \cdot 4} = \frac{2}{9} + \frac{16}{36}$$, which simplifies to $$\frac{\sqrt{2}x}{9} + \frac{y}{3} = \frac{2}{3}$$. Multiplying through by 9 gives $$\sqrt{2}x + 3y = 6$$, so $$y = \frac{6 - \sqrt{2}x}{3}$$.

Next, we find the endpoints of the chord by substituting this expression for $$y$$ into the ellipse equation: $$\frac{x^2}{9} + \frac{1}{4}\left(\frac{6 - \sqrt{2}x}{3}\right)^2 = 1$$, which is equivalent to $$\frac{x^2}{9} + \frac{(6 - \sqrt{2}x)^2}{36} = 1$$. Multiplying both sides by 36 leads to $$4x^2 + (6 - \sqrt{2}x)^2 = 36$$, and expanding gives $$4x^2 + 36 - 12\sqrt{2}x + 2x^2 = 36$$. This simplifies to $$6x^2 - 12\sqrt{2}x = 0$$ or $$6x(x - 2\sqrt{2}) = 0$$, so the solutions are $$x_1 = 0$$ and $$x_2 = 2\sqrt{2}$$.

When $$x = 0$$, we have $$y = \frac{6}{3} = 2$$, giving the point $$(0, 2)$$, and when $$x = 2\sqrt{2}$$, we get $$y = \frac{6 - \sqrt{2} \cdot 2\sqrt{2}}{3} = \frac{6 - 4}{3} = \frac{2}{3}$$, giving the point $$(2\sqrt{2}, \frac{2}{3})$$.

To verify the midpoint, we compute $$\left(\frac{0 + 2\sqrt{2}}{2}, \frac{2 + \frac{2}{3}}{2}\right) = \left(\sqrt{2}, \frac{4}{3}\right)$$, which matches the given midpoint.

Then the length of the chord is $$\sqrt{(2\sqrt{2} - 0)^2 + \left(\frac{2}{3} - 2\right)^2} = \sqrt{8 + \frac{16}{9}} = \sqrt{\frac{72 + 16}{9}} = \frac{\sqrt{88}}{3} = \frac{2\sqrt{22}}{3}$$.

Comparing this result with the given expression $$\frac{2\sqrt{\alpha}}{3}$$ shows that $$\alpha = 22$$, so the correct answer is 22.

For the first ellipse $$\dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{25}=1$$ the larger denominator is $$25$$, so the semi-major axis length is $$a_1 = 5$$ and the semi-minor axis length is $$b_1 = b$$.

Eccentricity of an ellipse is given by $$e=\sqrt{1-\dfrac{b^{2}}{a^{2}}}$$.
Hence $$e_1 = \sqrt{1-\dfrac{b^{2}}{25}} \quad -(1)$$

For the hyperbola $$\dfrac{x^{2}}{16}-\dfrac{y^{2}}{b^{2}}=1$$ the semi-transverse axis is $$a_2 = 4$$ and the semi-conjugate axis is $$b_2 = b$$.

Eccentricity of a hyperbola is given by $$e=\sqrt{1+\dfrac{b^{2}}{a^{2}}}$$.
Hence $$e_2 = \sqrt{1+\dfrac{b^{2}}{16}} \quad -(2)$$

Given $$e_1\,e_2 = 1$$. Substituting from $$(1)$$ and $$(2)$$:

$$\sqrt{1-\dfrac{b^{2}}{25}}\;\sqrt{1+\dfrac{b^{2}}{16}} = 1$$

Squaring both sides:

$$\left(1-\dfrac{b^{2}}{25}\right)\!\left(1+\dfrac{b^{2}}{16}\right)=1$$

Expanding and simplifying:

$$1+\dfrac{b^{2}}{16}-\dfrac{b^{2}}{25}-\dfrac{b^{4}}{400}=1$$
$$\dfrac{b^{2}}{16}-\dfrac{b^{2}}{25}-\dfrac{b^{4}}{400}=0$$
$$b^{2}\left(\dfrac{1}{16}-\dfrac{1}{25}\right)-\dfrac{b^{4}}{400}=0$$
$$b^{2}\left(\dfrac{9}{400}\right)-\dfrac{b^{4}}{400}=0$$
$$\dfrac{b^{2}}{400}\Bigl(9-b^{2}\Bigr)=0$$

Since $$b\neq 0$$, we get $$b^{2}=9 \Rightarrow b=3 \;(\text{given } b\lt 5).$$

Now compute the four focal points:

Eccentricity $$e_1 = \sqrt{1-\dfrac{9}{25}}=\dfrac{4}{5}$$.
Distance of each focus from the centre: $$c_1=a_1e_1=5\left(\dfrac{4}{5}\right)=4$$.
Ellipse foci: $$(0,\pm4).$$

Eccentricity $$e_2 = \sqrt{1+\dfrac{9}{16}}=\dfrac{5}{4}$$.
Distance of each focus from the centre: $$c_2=a_2e_2=4\left(\dfrac{5}{4}\right)=5$$.
Hyperbola foci: $$(\pm5,0).$$

The required ellipse has its axes along the coordinate axes and must pass through all four foci $$(\pm5,0),\,(0,\pm4)$$. Take its equation as

$$\dfrac{x^{2}}{A^{2}}+\dfrac{y^{2}}{B^{2}}=1.$$

Substituting point $$(5,0):\; \dfrac{25}{A^{2}} = 1 \Rightarrow A^{2}=25 \Rightarrow A=5.$$

Substituting point $$(0,4):\; \dfrac{16}{B^{2}} = 1 \Rightarrow B^{2}=16 \Rightarrow B=4.$$

Thus the ellipse is $$\dfrac{x^{2}}{25}+\dfrac{y^{2}}{16}=1$$ with semi-major axis $$a=5$$ and semi-minor axis $$b=4$$ (since $$25\gt16$$).

Eccentricity of this ellipse is

$$e = \sqrt{1-\dfrac{b^{2}}{a^{2}}} = \sqrt{1-\dfrac{16}{25}} = \sqrt{\dfrac{9}{25}} = \dfrac{3}{5}.$$

Therefore, the required eccentricity is $$\dfrac{3}{5}$$.

Option B is correct.

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.

Find the area of the largest rectangle with sides parallel to axes inscribed in region $$R = \{(x,y): x \leq y \leq 9 - \frac{11}{3}x^2, x \geq 0\}$$.

We begin by considering a rectangle whose vertices are $$(0,y_1),\,(x,y_1),\,(x,y_2),\,(0,y_2)$$ subject to $$x\le y_1$$ and $$y_2\le 9-\frac{11}{3}x^2\,. $$

To maximize the area, it is natural to take the lower bound at $$y_1=x$$ and the upper bound at $$y_2=9-\frac{11}{3}x^2\,. $$

Area = $$x\bigl(9 - \frac{11}{3}x^2 - x\bigr)$$

$$= 9x - \frac{11}{3}x^3 - x^2$$

Next, we set the derivative of the area with respect to $$x$$ equal to zero:

$$\frac{dA}{dx} = 9 - 11x^2 - 2x = 0$$

$$11x^2 + 2x - 9 = 0$$

$$x = \frac{-2 \pm \sqrt{4+396}}{22} = \frac{-2 \pm 20}{22}$$

$$x = \frac{18}{22} = \frac{9}{11}$$ (taking the positive root)

Substituting $$x=\frac{9}{11}$$ back into the area formula gives

$$A = 9 \cdot \frac{9}{11} - \frac{11}{3}\left(\frac{9}{11}\right)^3 - \left(\frac{9}{11}\right)^2$$

$$= \frac{81}{11} - \frac{11}{3} \cdot \frac{729}{1331} - \frac{81}{121}$$

$$= \frac{81}{11} - \frac{729}{363} - \frac{81}{121}$$

$$= \frac{81}{11} - \frac{243}{121} - \frac{81}{121}$$

$$= \frac{81 \times 11}{121} - \frac{324}{121}$$

$$= \frac{891 - 324}{121} = \frac{567}{121}$$

The correct answer is Option 4: $$\frac{567}{121}$$.

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.

We are given the hyperbola $$H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ with one focus at $$(\sqrt{10}, 0)$$ and the corresponding directrix $$x = \frac{9}{\sqrt{10}}$$.

For a hyperbola, the focus is at $$(ae, 0)$$ and the directrix is $$x = \frac{a}{e}$$. So we have:

$$ae = \sqrt{10} \quad \text{...(1)}$$

$$\frac{a}{e} = \frac{9}{\sqrt{10}} \quad \text{...(2)}$$

Multiplying equations (1) and (2): $$ae \cdot \frac{a}{e} = \sqrt{10} \cdot \frac{9}{\sqrt{10}}$$, which gives $$a^2 = 9$$, so $$a = 3$$.

From equation (1): $$3e = \sqrt{10}$$, so $$e = \frac{\sqrt{10}}{3}$$.

Therefore $$e^2 = \frac{10}{9}$$.

Now, $$b^2 = a^2(e^2 - 1) = 9\left(\frac{10}{9} - 1\right) = 9 \cdot \frac{1}{9} = 1$$.

The length of the latus rectum is $$l = \frac{2b^2}{a} = \frac{2 \cdot 1}{3} = \frac{2}{3}$$.

Finally, $$9(e^2 + l) = 9\left(\frac{10}{9} + \frac{2}{3}\right) = 9\left(\frac{10}{9} + \frac{6}{9}\right) = 9 \cdot \frac{16}{9} = 16$$.

Hence, the correct answer is Option C.

We need to find the sum of lengths of latus rectums of ellipse $$E$$ and hyperbola $$H$$.

The ellipse is given by $$E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a > b$$ and the hyperbola by $$H: \frac{x^2}{A^2}-\frac{y^2}{B^2}=1$$. The distance between the foci of $$E$$ is $$2ae$$, while that of $$H$$ is $$2Ae_H$$, and it is given that $$2ae = 2Ae_H = 2\sqrt{3}$$, so that $$ae = Ae_H = \sqrt{3}$$. Additionally, $$a - A = 2$$ and $$\frac{e}{e_H} = \frac{1}{3}$$, i.e., $$e_H = 3e$$.

From $$ae = \sqrt{3}$$ it follows that $$e = \frac{\sqrt{3}}{a}$$. Since $$e_H = 3e = \frac{3\sqrt{3}}{a}$$ and $$Ae_H = \sqrt{3}$$, we have $$A \cdot \frac{3\sqrt{3}}{a} = \sqrt{3} \Rightarrow A = \frac{a}{3}$$. Substituting into $$a - A = 2$$ yields $$a - \frac{a}{3} = 2 \Rightarrow \frac{2a}{3} = 2 \Rightarrow a = 3$$, hence $$A = 1$$, $$e = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$$, and $$e_H = \sqrt{3}$$.

For the ellipse, $$b^2 = a^2(1-e^2) = 9(1-\frac{1}{3}) = 6$$, and for the hyperbola, $$B^2 = A^2(e_H^2-1) = 1(3-1) = 2$$.

The length of the latus rectum of the ellipse is $$\frac{2b^2}{a} = \frac{12}{3} = 4$$, and that of the hyperbola is $$\frac{2B^2}{A} = \frac{4}{1} = 4$$, so that the sum is $$4 + 4 = 8$$.

The sum of lengths of latus rectums is 8, which matches Option C. Therefore, the answer is Option C.

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 circle touches the line $$x - y + 1 = 0$$ and has its center on the positive x-axis. Let the center be $$(h, 0)$$ with $$h > 0$$. The distance from the center to the line $$x - y + 1 = 0$$ equals the radius $$r$$. The distance formula gives:

$$\frac{|h - 0 + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|h + 1|}{\sqrt{2}} = r \quad \text{(1)}$$

Since $$h > 0$$, $$|h + 1| = h + 1$$, so $$r = \frac{h + 1}{\sqrt{2}}$$.

The circle cuts a chord of length $$\frac{4}{\sqrt{13}}$$ along the line $$-3x + 2y - 1 = 0$$. The perpendicular distance $$d$$ from the center $$(h, 0)$$ to this line is:

$$d = \frac{|-3h + 2 \cdot 0 - 1|}{\sqrt{(-3)^2 + 2^2}} = \frac{|-3h - 1|}{\sqrt{13}} = \frac{|3h + 1|}{\sqrt{13}} \quad \text{(2)}$$

Since $$h > 0$$, $$|3h + 1| = 3h + 1$$, so $$d = \frac{3h + 1}{\sqrt{13}}$$.

For a chord of length $$l$$, the relationship between $$d$$, $$r$$, and $$l$$ is:

$$\left( \frac{l}{2} \right)^2 + d^2 = r^2$$

Given $$l = \frac{4}{\sqrt{13}}$$, so $$\frac{l}{2} = \frac{2}{\sqrt{13}}$$. Substituting:

$$\left( \frac{2}{\sqrt{13}} \right)^2 + d^2 = r^2 \implies \frac{4}{13} + d^2 = r^2 \quad \text{(3)}$$

Substituting $$d$$ from (2):

$$\frac{4}{13} + \left( \frac{3h + 1}{\sqrt{13}} \right)^2 = r^2 \implies \frac{4}{13} + \frac{(3h + 1)^2}{13} = r^2 \implies r^2 = \frac{4 + (3h + 1)^2}{13} \quad \text{(4)}$$

From (1), $$r^2 = \left( \frac{h + 1}{\sqrt{2}} \right)^2 = \frac{(h + 1)^2}{2}$$. Equating to (4):

$$\frac{(h + 1)^2}{2} = \frac{4 + (3h + 1)^2}{13}$$

Cross-multiplying by 26:

$$13(h + 1)^2 = 2 \left[ 4 + (3h + 1)^2 \right]$$

Expanding:

$$13(h^2 + 2h + 1) = 2[4 + 9h^2 + 6h + 1] \implies 13h^2 + 26h + 13 = 2(9h^2 + 6h + 5) \implies 13h^2 + 26h + 13 = 18h^2 + 12h + 10$$

Rearranging:

$$13h^2 + 26h + 13 - 18h^2 - 12h - 10 = 0 \implies -5h^2 + 14h + 3 = 0$$

Multiplying by -1:

$$5h^2 - 14h - 3 = 0$$

Solving the quadratic equation:

$$h = \frac{14 \pm \sqrt{(-14)^2 - 4 \cdot 5 \cdot (-3)}}{10} = \frac{14 \pm \sqrt{196 + 60}}{10} = \frac{14 \pm \sqrt{256}}{10} = \frac{14 \pm 16}{10}$$

So $$h = 3$$ or $$h = -0.2$$. Since $$h > 0$$, $$h = 3$$.

Radius $$r = \frac{3 + 1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$$. The diameter is $$2r = 4\sqrt{2}$$.

The hyperbola is $$\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$$. One focus is the center of the circle, $$(3, 0)$$. For a hyperbola, foci are at $$(\pm c, 0)$$ where $$c = \sqrt{\alpha^2 + \beta^2}$$. Thus:

$$\sqrt{\alpha^2 + \beta^2} = 3 \quad \text{(5)}$$

The transverse axis length is $$2\alpha$$, equal to the diameter $$4\sqrt{2}$$:

$$2\alpha = 4\sqrt{2} \implies \alpha = 2\sqrt{2} \quad \text{(6)}$$

Substituting $$\alpha = 2\sqrt{2}$$ into (5):

$$\sqrt{(2\sqrt{2})^2 + \beta^2} = 3 \implies \sqrt{8 + \beta^2} = 3 \implies 8 + \beta^2 = 9 \implies \beta^2 = 1$$

Now compute $$2\alpha^2 + 3\beta^2$$:

$$\alpha^2 = (2\sqrt{2})^2 = 8, \quad \beta^2 = 1$$

$$2 \times 8 + 3 \times 1 = 16 + 3 = 19$$

The answer is 19.

The standard form of a rectangular-axis hyperbola with centre at the origin and transverse axis along the x-axis is $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1$$.

The foci are $$(\pm c,0)$$ where $$c^{2}=a^{2}+b^{2} \; -(1)$$.

The given point $$P(4,\,2\sqrt{3})$$ lies on the hyperbola. Substituting in the equation of the curve,

$$\frac{4^{2}}{a^{2}}-\frac{(2\sqrt3)^{2}}{b^{2}} = 1 \;\Longrightarrow\; \frac{16}{a^{2}}-\frac{12}{b^{2}} = 1 \; -(2)$$

Let the focal distances of $$P$$ be $$PF_{1}$$ and $$PF_{2}$$. Because the foci are $$(\pm c,0)$$,

$$PF_{1}^{\,2} = (4-c)^{2} + (2\sqrt3)^{2} = (4-c)^{2} + 12 = c^{2}-8c+28 \; -(3)$$
$$PF_{2}^{\,2} = (4+c)^{2} + 12 = c^{2}+8c+28 \; -(4)$$

The product of the focal distances is given to be 32:

$$PF_{1}\,PF_{2}=32 \;\Longrightarrow\; PF_{1}^{\,2}\,PF_{2}^{\,2}=32^{2}=1024$$

Using $$(3)$$ and $$(4)$$:

$$\bigl(c^{2}-8c+28\bigr)\bigl(c^{2}+8c+28\bigr)=1024$$

Employ $$(A-B)(A+B)=A^{2}-B^{2}$$ with $$A=c^{2}+28,\;B=8c$$:

$$\bigl(c^{2}+28\bigr)^{2}-(8c)^{2}=1024$$
$$c^{4}+56c^{2}+784-64c^{2}=1024$$
$$c^{4}-8c^{2}-240=0$$

Put $$z=c^{2}$$ to obtain $$z^{2}-8z-240=0$$. Solving the quadratic,

$$z = \frac{8\pm\sqrt{8^{2}+4\cdot240}}{2} = \frac{8\pm32}{2}$$

The positive root is $$z=20$$, so

$$c^{2}=20,\qquad c = 2\sqrt5 \; -(5)$$

From $$(1)$$, $$a^{2}+b^{2}=20 \; -(6)$$. From $$(2)$$, $$\dfrac{16}{a^{2}}-\dfrac{12}{b^{2}}=1 \; -(7)$$.

Let $$a^{2}=A,\;b^{2}=B$$. Then $$(6)$$ gives $$A+B=20 \; -(8)$$, and $$(7)$$ gives $$\dfrac{16}{A}-\dfrac{12}{B}=1 \; -(9)$$.

Re-write $$(9)$$: $$\dfrac{16}{A} = 1+\dfrac{12}{B} = \dfrac{B+12}{B}$$,
so $$A = \frac{16B}{B+12} \; -(10)$$.

Substitute $$(10)$$ into $$(8)$$:

$$\frac{16B}{B+12}+B = 20$$
$$16B + B(B+12) = 20(B+12)$$
$$B^{2}+28B = 20B + 240$$
$$B^{2}+8B-240 = 0$$

Solving, $$B = \frac{-8\pm\sqrt{8^{2}+4\cdot240}}{2} = \frac{-8\pm32}{2}$$.

Select the positive value: $$B = 12 \;\Longrightarrow\; b^{2}=12$$.

Using $$(8)$$, $$a^{2}=20-12 = 8$$, hence $$a = \sqrt8 = 2\sqrt2$$.

Conjugate axis length
The conjugate axis has length $$p = 2b = 2\sqrt{b^{2}} = 2\sqrt{12}=4\sqrt3$$,
so $$p^{2} = (4\sqrt3)^{2}=48$$.

Latus rectum length
For the hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$, the length of a latus rectum is $$q = \frac{2b^{2}}{a}$$.

$$q = \frac{2\cdot12}{2\sqrt2} = \frac{24}{2\sqrt2}= \frac{12}{\sqrt2}=6\sqrt2$$,
so $$q^{2} = (6\sqrt2)^{2}=72$$.

Finally,

$$p^{2}+q^{2}=48+72=120$$.

Hence, the required value is $$\boxed{120}$$.

The given circle is $$x^{2}+(y-1)^{2}=2$$.
Its centre is $$O(0,1)$$ and its radius is $$\sqrt{2}$$.

The straight line is $$x+y=3 \;\; \Longleftrightarrow \;\; x+y-3=0$$.
Distance of the centre $$O(0,1)$$ from this line is $$\frac{|0+1-3|}{\sqrt{1^{2}+1^{2}}}= \frac{2}{\sqrt{2}}=\sqrt{2}$$, that is exactly the radius, hence the line is tangent to the circle.

For a circle, the point of contact is the foot of the perpendicular drawn from the centre to the tangent. For the line $$x+y-3=0$$ the foot of the perpendicular from $$(x_{0},y_{0})$$ is $$\left(x_{0}-a\frac{ax_{0}+by_{0}+c}{a^{2}+b^{2}},\;y_{0}-b\frac{ax_{0}+by_{0}+c}{a^{2}+b^{2}}\right)$$ with $$a=1,\;b=1,\;c=-3,\;(x_{0},y_{0})=(0,1).$$ Thus

$$x_{1}=0-\frac{1(-2)}{2}=1, \qquad y_{1}=1-\frac{1(-2)}{2}=2.$$

Therefore $$P(1,2)$$ and $$x_{1}y_{1}=2.$$

Since the same straight line touches the two ellipses $$E_{1},E_{2}$$ at $$Q(x_{2},y_{2})$$ and $$R(x_{3},y_{3})$$, both points lie on the line: $$x_{2}+y_{2}=3, \qquad x_{3}+y_{3}=3 \; -(1)$$

The midpoint condition $$P$$ is the midpoint of $$QR$$:

$$\frac{x_{2}+x_{3}}{2}=1, \qquad \frac{y_{2}+y_{3}}{2}=2 \; -(2)$$

Let $$x_{2}=u \;\; (\Rightarrow y_{2}=3-u)$$. From $$(2)$$, $$x_{3}=2-u$$ and with $$(1)$$, $$y_{3}=3-(2-u)=1+u.$$ Hence

$$Q(u,\,3-u), \qquad R(2-u,\,1+u).$$

The distance $$PQ$$ is given to be $$\dfrac{2\sqrt{2}}{3}$$. Compute $$PQ^{2}:$$

$$PQ^{2}=(u-1)^{2}+\bigl((3-u)-2\bigr)^{2} =(u-1)^{2}+(1-u)^{2}=2(u-1)^{2}.$$

Therefore $$\sqrt{2}\,|u-1|=\frac{2\sqrt{2}}{3}\;\Longrightarrow\;|u-1|=\frac{2}{3}.$$ So

$$u=1+\frac{2}{3}=\frac{5}{3}\quad\text{or}\quad u=1-\frac{2}{3}=\frac{1}{3}.$$

This merely swaps $$Q$$ and $$R$$; either choice gives the same final value. Choose $$u=\dfrac{5}{3}$$ (the other works identically):

$$Q\!\left(\frac{5}{3},\,\frac{4}{3}\right), \qquad R\!\left(\frac{1}{3},\,\frac{8}{3}\right).$$

Now compute the required sum:

$$x_{1}y_{1}=1\cdot2=2,$$ $$x_{2}y_{2}=\frac{5}{3}\cdot\frac{4}{3}=\frac{20}{9},$$ $$x_{3}y_{3}=\frac{1}{3}\cdot\frac{8}{3}=\frac{8}{9}.$$

Hence $$x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}=2+\frac{20}{9}+\frac{8}{9}=2+\frac{28}{9}=\frac{46}{9}.$$

Finally, $$9\bigl(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}\bigr)=9\cdot\frac{46}{9}=46.$$

The desired value is $$46$$.

We have ellipse $$E_1: \frac{x^2}{9} + \frac{y^2}{4} = 1$$ with semi-major axis $$a_1 = 3$$, semi-minor axis $$b_1 = 2$$.

Eccentricity: $$e = \sqrt{1 - \frac{b_1^2}{a_1^2}} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$$.

For each ellipse $$E_i$$, the length of the minor axis of $$E_i$$ equals the length of the major axis of $$E_{i+1}$$.

All ellipses share the same center and eccentricity $$e = \frac{\sqrt{5}}{3}$$.

For any ellipse with eccentricity $$e$$: $$b^2 = a^2(1-e^2) = a^2 \cdot \frac{4}{9}$$, so $$b = \frac{2a}{3}$$.

Since the minor axis of $$E_i$$ (length $$2b_i$$) = major axis of $$E_{i+1}$$ (length $$2a_{i+1}$$):

$$a_{i+1} = b_i = \frac{2}{3}a_i$$

So $$a_i = 3 \cdot \left(\frac{2}{3}\right)^{i-1}$$ and $$b_i = \frac{2}{3}a_i = 2 \cdot \left(\frac{2}{3}\right)^{i-1}$$.

$$ A_i = \pi a_i b_i = \pi \cdot 3\left(\frac{2}{3}\right)^{i-1} \cdot 2\left(\frac{2}{3}\right)^{i-1} = 6\pi\left(\frac{2}{3}\right)^{2(i-1)} = 6\pi\left(\frac{4}{9}\right)^{i-1} $$

$$ \sum_{i=1}^{\infty} A_i = 6\pi \sum_{i=1}^{\infty}\left(\frac{4}{9}\right)^{i-1} = 6\pi \cdot \frac{1}{1 - 4/9} = 6\pi \cdot \frac{9}{5} = \frac{54\pi}{5} $$

$$ \frac{5}{\pi}\left(\sum_{i=1}^{\infty} A_i\right) = \frac{5}{\pi} \cdot \frac{54\pi}{5} = 54 $$

Hence the answer is 54.

The standard horizontal hyperbola with centre at the origin is written as $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$.
Its basic facts are:

• Foci $$\left(\pm c,\,0\right)$$ where $$c^{2}=a^{2}+b^{2}$$.
• Eccentricity $$e=\frac{c}{a}\ (e\gt 1)$$.
• Directrices $$x=\pm\frac{a}{e}$$ (each focus has the nearer directrix).

We are told that one focus is $$(-5,0)$$ and the corresponding directrix is $$5x+9=0$$, i.e. $$x=-\frac{9}{5}$$.
Hence

$$c=5,\qquad -\frac{a}{e}=-\frac{9}{5}\; \Longrightarrow\; \frac{a}{e}=\frac{9}{5}\; -(1)$$

Using $$e=\frac{c}{a}$$, substitute $$c=5$$ into $$e=\frac{5}{a}$$ and then into $$(1)$$:

$$\frac{a}{e}=a\left(\frac{a}{5}\right)=\frac{a^{2}}{5}=\frac{9}{5}\; \Longrightarrow\; a^{2}=9\; \Longrightarrow\; a=3.$$

Now compute $$b^{2}$$ using $$c^{2}=a^{2}+b^{2}$$:

$$25=9+b^{2}\; \Longrightarrow\; b^{2}=16\; \Longrightarrow\; b=4.$$

Therefore the hyperbola is $$\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$$.

The given point $$\left(\alpha,\,2\sqrt{5}\right)$$ lies on the curve, so substitute it:

$$\frac{\alpha^{2}}{9}-\frac{(2\sqrt{5})^{2}}{16}=1 \quad\Longrightarrow\quad \frac{\alpha^{2}}{9}-\frac{20}{16}=1.$$

Simplify:

$$\frac{\alpha^{2}}{9}-\frac{5}{4}=1\; \Longrightarrow\; \frac{\alpha^{2}}{9}=\frac{9}{4}\; \Longrightarrow\; \alpha^{2}=\frac{81}{4}\; \Longrightarrow\; \alpha=\pm\frac{9}{2}.$$

Let $$P(\alpha,2\sqrt{5})$$ be the point; its distances to the two foci $$F_{1}(5,0)$$ and $$F_{2}(-5,0)$$ are

$$PF_{1}=\sqrt{(\alpha-5)^{2}+(2\sqrt{5})^{2}},\qquad PF_{2}=\sqrt{(\alpha+5)^{2}+(2\sqrt{5})^{2}}.$$

The required product is $$p=PF_{1}\,PF_{2}=\sqrt{\bigl[(\alpha-5)^{2}+20\bigr]\bigl[(\alpha+5)^{2}+20\bigr]}.$$

$$\alpha=\frac{9}{2}=4.5$$

$$\bigl[(\alpha-5)^{2}+20\bigr]=(4.5-5)^{2}+20=0.25+20=20.25,$$ $$\bigl[(\alpha+5)^{2}+20\bigr]=(4.5+5)^{2}+20=90.25+20=110.25.$$

Product inside the root: $$20.25\times110.25=2232.5625,$$ hence

$$p=\sqrt{2232.5625}=47.25.$$

$$\alpha=-\frac{9}{2}=-4.5$$

This merely interchanges the two factors, giving the same product $$p=47.25$$.

Finally, $$4p=4\times47.25=189.$$

Hence, $$\boxed{4p=189}$$.

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

We need to find $$12\lambda_1 + 29\lambda_2$$ where $$\lambda_1, \lambda_2$$ are the values of $$\lambda$$ for which the circle passes through certain intersection points.

Find the focus of the parabola

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

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

Circle equation

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

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

Find intersection of lines 3x - y = 0 and x + λy = 4

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

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

$$(4+3k)^2 + 144 = 25k^2$$

$$16 + 24k + 9k^2 + 144 = 25k^2$$

$$16k^2 - 24k - 160 = 0$$

$$2k^2 - 3k - 20 = 0$$

$$(2k + 5)(k - 4) = 0$$

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

Find $$\lambda$$ values

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

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

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

Calculate

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

The answer is 15.

The two foci are $$(4,2)$$ and $$(8,2)$$, so the transverse axis is horizontal and its midpoint is the centre of the hyperbola.

Centre $$\,(h,k)=\left(\frac{4+8}{2},\,2\right)=(6,2)$$.

Let the hyperbola in standard form be$$(x-h)^2/a^2-\,(y-k)^2/b^2=1.$$ Here $$c$$ is the focal distance from the centre, given by $$c=\lvert8-6\rvert=2\quad\Rightarrow\quad c^2=4.$$ For a rectangular hyperbola,$$c^2=a^2+b^2\quad-(1).$$

After shifting the origin to $$(6,2)$$ and clearing denominators we obtain $$b^2\,(x-6)^2-a^2\,(y-2)^2=a^2b^2\quad-(2).$$ Expanding $$(2)$$ gives $$b^2(x^2-12x+36)-a^2(y^2-4y+4)=a^2b^2.$$ Collect the terms on the left: $$b^2x^2-12b^2x+36b^2-a^2y^2+4a^2y-4a^2-a^2b^2=0.$$ The required general form is $$3x^2-y^2-\alpha x+\beta y+\gamma=0,$$ so we match coefficients.

Coefficient of $$x^2:$$ $$b^2=3.$$ Coefficient of $$y^2:$$ $$-a^2=-1\;\Rightarrow\;a^2=1.$$ With $$a^2=1,\;b^2=3,$$ equation $$-(1)$$ gives $$c^2=a^2+b^2=1+3=4,$$ agreeing with $$c=2,$$ so the values are consistent.

Substituting $$a^2=1,\;b^2=3$$ in $$b^2(x-6)^2-a^2(y-2)^2=a^2b^2$$: $$3(x-6)^2-(y-2)^2=3.$$ Expand: $$3(x^2-12x+36)-(y^2-4y+4)=3,$$ $$3x^2-36x+108-y^2+4y-4=3.$$ Move everything to one side: $$3x^2-y^2-36x+4y+108-4-3=0,$$ $$3x^2-y^2-36x+4y+101=0.$$ Therefore $$-\alpha=-36\;\Rightarrow\;\alpha=36,$$ $$\beta=4,$$ $$\gamma=101.$$

Finally, $$\alpha+\beta+\gamma=36+4+101=141.$$

The required value is $$141.$$

We are given two hyperbolas $$H_1: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ and $$H_2: -\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$$ with the latus rectum of $$H_1$$ equal to $$15\sqrt{2}$$, the latus rectum of $$H_2$$ equal to $$12\sqrt{5}$$, the eccentricity $$e_1 = \sqrt{\frac{5}{2}}$$, and the product of their transverse axes equal to $$100\sqrt{10}$$. Our goal is to find $$25e_2^2$$.

For the first hyperbola, the transverse axis is $$2a$$, the latus rectum is $$\frac{2b^2}{a}$$, and we have $$b^2 = a^2(e_1^2 - 1)$$. For the second hyperbola (in its conjugate form), the transverse axis is $$2B$$, the latus rectum is $$\frac{2A^2}{B}$$, and $$A^2 = B^2(e_2^2 - 1)$$.

Since $$e_1 = \sqrt{\frac{5}{2}}$$, it follows that $$e_1^2 = \frac{5}{2}$$ and hence $$b^2 = a^2\left(\frac{5}{2} - 1\right) = \frac{3a^2}{2}$$. Substituting into the expression for the latus rectum of $$H_1$$ gives $$\frac{2b^2}{a} = \frac{2\cdot\frac{3a^2}{2}}{a} = 3a = 15\sqrt{2}$$, which yields $$a = 5\sqrt{2}$$.

Therefore the transverse axis of $$H_1$$ is $$2a = 10\sqrt{2}$$. Since the product of the transverse axes is $$(2a)(2B) = 100\sqrt{10}$$, we have $$10\sqrt{2}\cdot 2B = 100\sqrt{10}$$. This gives $$2B = \frac{100\sqrt{10}}{10\sqrt{2}} = 10\sqrt{5}$$ and hence $$B = 5\sqrt{5}$$.

Turning to the second hyperbola, its latus rectum satisfies $$\frac{2A^2}{B} = 12\sqrt{5}$$, so $$A^2 = \frac{12\sqrt{5}\cdot B}{2} = \frac{12\sqrt{5}\cdot 5\sqrt{5}}{2} = \frac{300}{2} = 150$$.

Using the relation $$A^2 = B^2(e_2^2 - 1)$$ gives $$150 = 125(e_2^2 - 1)$$, so $$e_2^2 - 1 = \frac{150}{125} = \frac{6}{5}$$ and therefore $$e_2^2 = 1 + \frac{6}{5} = \frac{11}{5}$$.

Finally, we compute $$25e_2^2 = 25 \times \frac{11}{5} = 55$$.

The answer is 55

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.

Focal distances of $$(x,y)$$ are $$a \pm ex$$. Product $$= a^2 - e^2x^2 = 7/4$$.

Point $$(\sqrt{3}, 1/2)$$ is on ellipse: $$\frac{3}{a^2} + \frac{1}{4b^2} = 1$$. Use $$b^2 = a^2(1-e^2)$$.

Solving the system of equations gives two values for $$e$$. 

Their Difference $$\frac{3-2\sqrt{2}}{2\sqrt{3}}$$.

For an ellipse centred at the origin with the major axis along the $$x$$-axis, we use the standard form

$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\qquad a \gt b\,(\,\text{major}\, \gt \text{minor}\,)$$

Eccentricity: $$e=\sqrt{1-\frac{b^{2}}{a^{2}}}$$
Distance between the two foci: $$2ae$$
Length of the minor axis: $$2b$$

The question states: “length of the minor axis is one fourth of the distance between the foci”.
Translate this directly into an equation:

$$2b=\frac{1}{4}\times(2ae)$$

Simplify:

$$2b=\frac{ae}{2}\qquad\Longrightarrow\qquad4b=ae$$ $$-(1)$$

Replace $$b$$ by its expression in terms of $$a$$ and $$e$$. From the definition of eccentricity,

$$b=a\sqrt{1-e^{2}}$$ $$-(2)$$

Substitute $$(2)$$ into $$(1)$$:

$$4\bigl(a\sqrt{1-e^{2}}\bigr)=ae$$

Assuming $$a\neq0$$, divide both sides by $$a$$:

$$4\sqrt{1-e^{2}}=e$$

Square both sides to eliminate the square root:

$$16(1-e^{2})=e^{2}$$

Expand and collect like terms:

$$16-16e^{2}=e^{2}\qquad\Longrightarrow\qquad17e^{2}=16$$

Solve for $$e$$:

$$e^{2}=\frac{16}{17}\qquad\Longrightarrow\qquad e=\frac{4}{\sqrt{17}}$$

Since $$e$$ must be positive and less than 1, this value is acceptable.
Hence, the eccentricity of the ellipse is $$\frac{4}{\sqrt{17}}$$.

Option A is correct.

The ellipse is $$\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$$. Write it in the standard form $$S = 0$$, where

$$S = \frac{x^{2}}{25} + \frac{y^{2}}{16} - 1 = 0$$.

Mid-point formula for a chord of a conic
If $$(x_1 , y_1)$$ is the mid-point of a chord of the conic $$S = 0$$, then the chord is given by $$T = S_1$$, where

$$T$$ is obtained from $$S$$ by replacing $$x^{2}$$ with $$x x_1$$ and $$y^{2}$$ with $$y y_1$$, and $$S_1$$ is obtained by replacing $$x, y$$ in $$S$$ with $$x_1 , y_1$$.

For our ellipse and mid-point $$(3,1)$$:

$$T = \frac{x\,x_1}{25} + \frac{y\,y_1}{16} - 1 = \frac{3x}{25} + \frac{1\;y}{16} - 1$$.

$$S_1 = \frac{x_1^{2}}{25} + \frac{y_1^{2}}{16} - 1 = \frac{3^{2}}{25} + \frac{1^{2}}{16} - 1 = \frac{9}{25} + \frac{1}{16} - 1$$.

Take the common denominator $$400$$: $$\frac{9}{25} = \frac{144}{400}, \quad \frac{1}{16} = \frac{25}{400}.$$ Hence

$$S_1 = \frac{144 + 25}{400} - 1 = \frac{169}{400} - 1 = -\frac{231}{400}.$$

Equating $$T$$ and $$S_1$$:

$$\frac{3x}{25} + \frac{y}{16} - 1 = -\frac{231}{400}.$$

Multiply every term by $$400$$ to remove fractions:

$$400 \left(\frac{3x}{25}\right) + 400 \left(\frac{y}{16}\right) - 400 = -231.$$ $$48x + 25y - 400 = -231.$$

Bring all terms to the left:

$$48x + 25y - 400 + 231 = 0$$ $$48x + 25y - 169 = 0.$$

Therefore, the chord with mid-point $$(3,1)$$ is

$$48x + 25y = 169.$$ Hence, Option A is correct.

We have two parabolas with the same focus $$(4,3)$$, where the directrix of the first parabola is the x-axis and the directrix of the second parabola is the y-axis. We need to find $$(AB)^2$$ where $$A$$ and $$B$$ are the intersection points.

For the first parabola with focus $$(4,3)$$ and directrix $$y=0$$, the distance from any point $$(x,y)$$ to the focus equals its distance to the directrix:

$$ \sqrt{(x-4)^2 + (y-3)^2} = |y| $$

Squaring both sides gives

$$ (x-4)^2 + (y-3)^2 = y^2 $$

which expands to

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

and hence

$$ (x-4)^2 = 6y - 9 \quad \cdots (1) $$

For the second parabola with focus $$(4,3)$$ and directrix $$x=0$$, we similarly have

$$ \sqrt{(x-4)^2 + (y-3)^2} = |x| $$

Squaring both sides gives

$$ (x-4)^2 + (y-3)^2 = x^2 $$

which expands to

$$ x^2 - 8x + 16 + (y-3)^2 = x^2 $$

so

$$ (y-3)^2 = 8x - 16 $$

and

$$ (y-3)^2 = 8(x - 2) \quad \cdots (2) $$

At the intersection points the distances to the focus and the two directrices are equal, implying $$|y| = |x|$$. Thus the intersection points lie on $$y = x$$ or $$y = -x$$.

In the case $$y = x$$, substituting into (1) yields

$$ (x-4)^2 = 6x - 9 $$

so

$$ x^2 - 8x + 16 = 6x - 9 $$

and

$$ x^2 - 14x + 25 = 0 $$

giving

$$ x = \frac{14 \pm \sqrt{196 - 100}}{2} = \frac{14 \pm \sqrt{96}}{2} = 7 \pm 2\sqrt{6} $$

Hence the points are $$(7 + 2\sqrt{6},\;7 + 2\sqrt{6})$$ and $$(7 - 2\sqrt{6},\;7 - 2\sqrt{6})$$.

In the case $$y = -x$$, substituting into (1) yields

$$ (x-4)^2 = -6x - 9 $$

so

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

The discriminant $$4 - 100 = -96 < 0$$ shows there are no real solutions.

Letting $$A = (7 + 2\sqrt{6},\;7 + 2\sqrt{6})$$ and $$B = (7 - 2\sqrt{6},\;7 - 2\sqrt{6})$$, we compute

$$ (AB)^2 = (4\sqrt{6})^2 + (4\sqrt{6})^2 = 96 + 96 = 192 $$

The correct answer is Option 3: 192.

The ellipse is $$\frac{x^{2}}{36} + \frac{y^{2}}{25} = 1$$ and the external point is $$P(\sqrt{5},\sqrt{5})$$.
Let a variable line through P have slope $$m$$. Its equation can be written in parametric form as

$$x = \sqrt{5} + t ,\; y = \sqrt{5} + mt \qquad (t\in\mathbb{R})$$
The point $$P$$ corresponds to $$t = 0$$. The intersections with the ellipse (points $$A,\,B$$) occur for those values of $$t$$ that satisfy

$$\frac{(\sqrt{5}+t)^{2}}{36} + \frac{(\sqrt{5}+mt)^{2}}{25} = 1$$

Expand each square:
$$(\sqrt{5}+t)^{2}=5+2\sqrt{5}\,t+t^{2},\qquad (\sqrt{5}+mt)^{2}=5+2m\sqrt{5}\,t+m^{2}t^{2}$$

Substituting and clearing denominators by multiplying by $$900 \;(=36\!\cdot\!25)$$ gives

$$(25+36m^{2})t^{2}+\sqrt{5}(50+72m)\,t-595=0\tag{1}$$

Equation $$(1)$$ is quadratic in $$t$$, so its roots $$t_{1},t_{2}$$ represent the directed distances (along the x-axis direction of the line) from $$P$$ to $$A,\,B$$ respectively.

For a quadratic $$At^{2}+Bt+C=0$$ we have $$t_{1}t_{2}=C/A.$$
Here
$$A = 25+36m^{2},\quad C = -595 \;\Rightarrow\; t_{1}t_{2}= \frac{-595}{25+36m^{2}}$$

The actual (positive) distances are $$PA = |t_{1}|\sqrt{1+m^{2}},\; PB = |t_{2}|\sqrt{1+m^{2}}.$$
Hence

$$PA\cdot PB = |t_{1}t_{2}|(1+m^{2}) = \frac{595(1+m^{2})}{25+36m^{2}}\tag{2}$$

To maximise $$PA\cdot PB$$ we maximise the function
$$g(m)=\frac{1+m^{2}}{25+36m^{2}},\qquad m\in\mathbb{R}$$

Set $$t=m^{2}\ge 0.$$ Then $$g(t)=\frac{1+t}{25+36t}.$$ Differentiate:

$$\frac{dg}{dt}= \frac{(25+36t)\cdot1 - (1+t)\cdot36}{(25+36t)^{2}} = \frac{25+36t-36-36t}{(25+36t)^{2}} = \frac{-11}{(25+36t)^{2}} \lt 0$$

Since $$dg/dt$$ is negative for all $$t\ge 0$$, $$g(t)$$ decreases as $$t$$ increases. Therefore $$g(t)$$ (and hence $$PA\cdot PB$$) is maximum at $$t = 0$$, i.e. at $$m = 0$$.

Thus the required line is horizontal: $$y = \sqrt{5}.$$
Its intersections with the ellipse are obtained by substituting $$y=\sqrt{5}:$$

$$\frac{x^{2}}{36} + \frac{5}{25}=1 \;\Longrightarrow\; \frac{x^{2}}{36}=1-\frac{1}{5}=\frac{4}{5} \;\Longrightarrow\; x^{2}=\frac{144}{5}$$

Hence the points are $$A\bigl(\;+\frac{12}{\sqrt{5}},\,\sqrt{5}\bigr)$$ and $$B\bigl(\;-\frac{12}{\sqrt{5}},\,\sqrt{5}\bigr).$$

Compute the distances from $$P(\sqrt{5},\sqrt{5})$$:
$$PA = \left|\frac{12}{\sqrt{5}}-\sqrt{5}\right| = \left|\frac{12-5}{\sqrt{5}}\right| = \frac{7}{\sqrt{5}},$$
$$PB = \left|-\frac{12}{\sqrt{5}}-\sqrt{5}\right| = \left|\frac{-12-5}{\sqrt{5}}\right| = \frac{17}{\sqrt{5}}.$$

Now evaluate $$5(PA^{2}+PB^{2}):$$
$$PA^{2} = \frac{49}{5},\qquad PB^{2} = \frac{289}{5}$$
$$PA^{2}+PB^{2} = \frac{49+289}{5} = \frac{338}{5}$$
$$\therefore\; 5(PA^{2}+PB^{2}) = 5\cdot\frac{338}{5} = 338$$

Hence $$5(PA^{2}+PB^{2}) = 338,$$ which matches Option D.

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 ellipse is $$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$$ and the circle $$\mathcal{C}$$ is $$x^{2}+y^{2}=9$$ (radius $$3$$, centre at the origin $$O(0,0)$$).
The point $$M$$ is $$(3,0)$$, so the ray $$OM$$ is the positive $$x$$-axis.

Because $$\angle ROM=\frac{\pi}{6}$$ and $$|OR|=3$$, the point $$R$$ on the circle is obtained by standard polar-Cartesian conversion:

$$R=\bigl(3\cos\frac{\pi}{6},\,3\sin\frac{\pi}{6}\bigr)=\left(\frac{3\sqrt{3}}{2},\,\frac{3}{2}\right).$$

Similarly, $$\angle SOM=\frac{\pi}{3}$$ gives

$$S=\bigl(3\cos\frac{\pi}{3},\,3\sin\frac{\pi}{3}\bigr)=\left(\frac{3}{2},\,\frac{3\sqrt{3}}{2}\right).$$

The vertical lines through $$R$$ and $$S$$ intersect the ellipse at $$P$$ and $$Q$$, so

$$x_{1}=x_{R}=\frac{3\sqrt{3}}{2},\qquad x_{2}=x_{S}=\frac{3}{2}.$$

Point P
Insert $$x_{1}$$ in the ellipse:

$$\frac{x_{1}^{2}}{9}=\frac{\left(\dfrac{3\sqrt{3}}{2}\right)^{2}}{9}=\frac{27/4}{9}=\frac34.$$

Therefore $$\frac{y_{1}^{2}}{4}=1-\frac34=\frac14 \;\Longrightarrow\; y_{1}=1 \;(\text{positive branch}).$$

Hence $$P=\left(\frac{3\sqrt{3}}{2},\,1\right).$$

Point Q
Insert $$x_{2}$$ in the ellipse:

$$\frac{x_{2}^{2}}{9}=\frac{\left(\dfrac{3}{2}\right)^{2}}{9}=\frac{9/4}{9}=\frac14.$$

Therefore $$\frac{y_{2}^{2}}{4}=1-\frac14=\frac34 \;\Longrightarrow\; y_{2}=\sqrt{3}.$$

Hence $$Q=\left(\frac{3}{2},\,\sqrt{3}\right).$$

Equation of line $$PQ$$
Slope $$m=\dfrac{\sqrt{3}-1}{\frac{3}{2}-\frac{3\sqrt{3}}{2}} =\dfrac{\sqrt{3}-1}{\frac{3}{2}(1-\sqrt{3})} =-\frac{2}{3}.$$

Using point $$P$$:

$$y-1=-\frac{2}{3}\Bigl(x-\frac{3\sqrt{3}}{2}\Bigr) \;\Longrightarrow\;3y-3=-2x+3\sqrt{3}.$$

Rearranging,

$$2x+3y=3(1+\sqrt{3}).$$

This matches Option A. Option B (with $$2x+y$$) is therefore incorrect.

Checking Statement C
Take $$N_{2}=(x_{2},0)=\left(\dfrac32,0\right).$$
Because $$Q$$ and $$S$$ have the same $$x$$-coordinate $$x_{2},$$ the required distances are vertical:

$$|N_{2}Q|=|y_{2}|=\sqrt{3},\qquad |N_{2}S|=\left|\frac{3\sqrt{3}}{2}\right|=\frac{3\sqrt{3}}{2}.$$

Thus $$3|N_{2}Q|=3\sqrt{3},\qquad 2|N_{2}S|=2\cdot\frac{3\sqrt{3}}{2}=3\sqrt{3},$$ so $$3|N_{2}Q|=2|N_{2}S|.$$
Statement C is true.

Checking Statement D
Take $$N_{1}=(x_{1},0)=\left(\dfrac{3\sqrt{3}}{2},0\right).$$

Vertical distances give $$|N_{1}P|=1,\quad |N_{1}R|=\frac{3}{2}.$$

But $$9|N_{1}P|=9,\quad 4|N_{1}R|=4\cdot\frac32=6,$$ which are not equal. Hence Statement D is false.

Final result
Option A and Option C are the only correct statements.

Correct choices: Option A, Option C.

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

To find the points of intersection:

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:

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

$$ (x_1 + 1)^2 = 4 $$

$$ x_1 + 1 = \pm 2 $$

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

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

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

$$ (x_2 + 1)^2 = 4 $$

$$ x_2 = 1 $$ 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$$ (0, -3) $$.

To find the area of $$ \triangle PQR $$ with vertices P$$ (1, 0) $$, Q$$ (-3, 0) $$, and R$$ (0, -3) $$, use the shoelace formula.

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

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

$$= \frac{1}{2} \left| 3 + 9 \right|$$

$$= \frac{1}{2} \left| 12 \right|$$

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

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

Consider the ellipse $$\frac{x^2}{4}+\frac{y^2}{2}=1$$ and let the midpoint of a chord be $$(1, 1/2)$$. For an ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$, a chord with midpoint $$(h,k)$$ is given by the equation $$\frac{xh}{a^2}+\frac{yk}{b^2} = \frac{h^2}{a^2}+\frac{k^2}{b^2}$$. Substituting $$a^2=4, b^2=2, h=1, k=1/2$$ yields

$$\frac{x}{4}+\frac{y/2}{2} = \frac{1}{4}+\frac{1/4}{2} = \frac{1}{4}+\frac{1}{8} = \frac{3}{8}$$

which simplifies to

$$\frac{x}{4}+\frac{y}{4} = \frac{3}{8} \implies x + y = \frac{3}{2} \implies y = \frac{3}{2}-x$$.

Substituting into the ellipse equation gives

$$\frac{x^2}{4}+\frac{(3/2-x)^2}{2}=1$$

which expands to

$$\frac{x^2}{4}+\frac{9/4-3x+x^2}{2}=1$$

and simplifies as

$$\frac{x^2}{4}+\frac{9}{8}-\frac{3x}{2}+\frac{x^2}{2}=1$$

leading to

$$\frac{3x^2}{4}-\frac{3x}{2}+\frac{1}{8}=0 \implies 6x^2-12x+1=0$$.

Solving this quadratic yields

$$x = \frac{12\pm\sqrt{144-24}}{12} = \frac{12\pm\sqrt{120}}{12} = \frac{12\pm2\sqrt{30}}{12} = 1\pm\frac{\sqrt{30}}{6}$$.

The difference between the two $$x$$-values is $$\Delta x = \frac{2\sqrt{30}}{6} = \frac{\sqrt{30}}{3}$$. Since $$y = 3/2-x$$, it follows that $$\Delta y = -\Delta x$$. Therefore the length of the chord is

$$\sqrt{(\Delta x)^2+(\Delta y)^2} = |\Delta x|\sqrt{2} = \frac{\sqrt{30}}{3}\cdot\sqrt{2} = \frac{\sqrt{60}}{3} = \frac{2\sqrt{15}}{3}$$.

The correct answer is Option 3: $$\frac{2}{3}\sqrt{15}$$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Hence the answer is 12.

The ellipse is $$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$$.
Hence $$a=3,\;b=2$$ and its centre is $$O(0,0)$$. The vertex with positive $$x$$-coordinate is $$R(3,0)$$.

Let the second point of contact in the fourth quadrant be $$T(x_T,y_T)$$.
Since $$O$$ and $$R$$ both lie on the $$x$$-axis, the area of $$\triangle ORT$$ equals

$$\text{Area}=\frac12\,(OR)\,\bigl|y_T\bigr| = \frac12\,(3)\,\bigl|y_T\bigr|.$$

Given $$\text{Area}= \frac32,$$ we get
$$\frac12\,(3)\,\bigl|y_T\bigr|=\frac32 \;\Longrightarrow\; |y_T|=1.$$ Because $$T$$ is in the fourth quadrant, $$y_T=-1$$.

Substituting $$y_T=-1$$ in the ellipse,

$$\frac{x_T^{2}}{9}+\frac{(-1)^{2}}{4}=1 \;\Longrightarrow\; \frac{x_T^{2}}{9}+\frac14=1 \;\Longrightarrow\; \frac{x_T^{2}}{9}=\frac34 \;\Longrightarrow\; x_T^{2}=\frac{27}{4} \;\Longrightarrow\; x_T=\frac{3\sqrt3}{2}\;(\text{positive in quadrant IV}).$$

Thus $$T\Bigl(\dfrac{3\sqrt3}{2},\,-1\Bigr).$$

Equation of the two tangents
For a point $$(x_1,y_1)$$ on the ellipse, the tangent is $$\frac{xx_1}{9}+\frac{yy_1}{4}=1.$$

• At the upper end of the minor axis $$(0,2):$$ $$\frac{x\cdot0}{9}+\frac{y\cdot2}{4}=1\;\Longrightarrow\;y=2.$$
• At $$T\Bigl(\dfrac{3\sqrt3}{2},-1\Bigr):$$ $$\frac{x\left(\dfrac{3\sqrt3}{2}\right)}{9}+\frac{y(-1)}{4}=1 \;\Longrightarrow\; \frac{x\sqrt3}{6}-\frac{y}{4}=1 \;\Longrightarrow\; y=\frac{2\sqrt3}{3}\,x-4.$$

Both tangents meet at the external point $$S(p,q)$$, so $$S$$ is the intersection of

$$y=2 \quad\text{and}\quad y=\frac{2\sqrt3}{3}\,x-4.$$

Putting $$y=2$$ in the second equation:

$$2=\frac{2\sqrt3}{3}\,x-4 \;\Longrightarrow\; \frac{2\sqrt3}{3}\,x=6 \;\Longrightarrow\; x=\frac{6\cdot3}{2\sqrt3}=3\sqrt3.$$

Hence $$S(3\sqrt3,\,2).$$

Verification that $$S$$ lies outside the ellipse

$$\frac{p^{2}}{9}+\frac{q^{2}}{4} =\frac{(3\sqrt3)^{2}}{9}+\frac{2^{2}}{4} =\frac{27}{9}+1=3+1=4\gt1,$$ so $$S$$ is indeed outside the ellipse.

The obtained values match Option A:

Option A which is: $$q = 2,\; p = 3\sqrt{3}$$

Consider the ellipse given by $$\frac{x^2}{9} + \frac{y^2}{25} = 1.$$ Since the larger denominator is 25, the major axis runs along the y-axis, so we set $$a^2 = 25$$ and $$b^2 = 9.$$ It follows that $$c^2 = a^2 - b^2 = 25 - 9 = 16 \implies c = 4,$$ and the foci lie at $$(0,\pm4).$$ The eccentricity of this ellipse is $$e_1 = \frac{c}{a} = \frac{4}{5}.$$

The hyperbola we seek shares these foci at $$(0,\pm4)$$, so its transverse axis is also vertical and its focal parameter is $$c_h = 4.$$ We are told its eccentricity satisfies $$e_2 = \frac{15}{8}\,e_1 = \frac{15}{8}\times\frac{4}{5} = \frac{3}{2}.$$ Thus $$a_h = \frac{c_h}{e_2} = \frac{4}{3/2} = \frac{8}{3},$$ and $$b_h^2 = c_h^2 - a_h^2 = 16 - \frac{64}{9} = \frac{80}{9}.$$ Accordingly, the equation of the hyperbola is $$\frac{y^2}{64/9} \;-\;\frac{x^2}{80/9} \;=\;1.$$

To verify that the point $$P\bigl(\sqrt{2},\,\tfrac{14}{3}\sqrt{\tfrac{2}{5}}\bigr)$$ lies on this curve, note first that $$y^2 = \frac{196}{9}\times\frac{2}{5} = \frac{392}{45},$$ so $$\frac{y^2}{64/9} = \frac{392/45}{64/9} = \frac{392}{45}\times\frac{9}{64} = \frac{49}{40},$$ while $$\frac{x^2}{80/9} = \frac{2}{80/9} = \frac{9}{40}.$$ Their difference is $$\frac{49}{40} - \frac{9}{40} = 1,$$ confirming that $$P$$ indeed satisfies the hyperbola’s equation.

For any point on a hyperbola with eccentricity $$e_2$$ and transverse semi-axis $$a_h$$, the distances to the two foci satisfy $$|PF| = \bigl|e_2\,y \pm a_h\bigr|.$$ Since the y-coordinate of $$P$$ is positive, we compute $$e_2\,y = \frac{3}{2}\times\frac{14}{3}\sqrt{\frac{2}{5}} = 7\sqrt{\frac{2}{5}},$$ and hence the two focal distances are $$PF_1 = 7\sqrt{\frac{2}{5}} - \frac{8}{3},\qquad PF_2 = 7\sqrt{\frac{2}{5}} + \frac{8}{3}.$$ The smaller of these is $$7\sqrt{\frac{2}{5}} - \frac{8}{3}.$$

Therefore, the smaller focal distance is $$7\sqrt{\frac{2}{5}} - \frac{8}{3}$$.

Line $$5x + 7y = 50$$: $$A(\alpha,0) \Rightarrow 5\alpha = 50 \Rightarrow \alpha = 10$$. $$B(0,\beta) \Rightarrow 7\beta = 50 \Rightarrow \beta = 50/7$$.

P divides AB in ratio 7:3 internally: $$P = \left(\frac{7(0)+3(10)}{10}, \frac{7(50/7)+3(0)}{10}\right) = (3, 5)$$.

Directrix: $$3x - 25 = 0 \Rightarrow x = 25/3$$. For ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$: directrix $$x = a/e$$.

So $$a/e = 25/3$$. Focus $$S = (ae, 0)$$.

Since the perpendicular from S on x-axis passes through P, the x-coordinate of S = x-coordinate of P = 3. So $$ae = 3$$.

From $$a/e = 25/3$$ and $$ae = 3$$: $$a^2 = (a/e)(ae) = (25/3)(3) = 25$$, so $$a = 5$$.

$$e = ae/a = 3/5$$. $$b^2 = a^2(1-e^2) = 25(1-9/25) = 16$$.

Length of latus rectum = $$\frac{2b^2}{a} = \frac{32}{5}$$.

The answer is Option (4): $$\boxed{\frac{32}{5}}$$.

We need to find which point lies on the line through $$P$$ and $$Q$$, where $$PQ$$ is a chord of $$y^2 = 12x$$ with midpoint $$(4, 1)$$.

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.

Ellipse: $$\frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$$. $$a^2=100, b^2=75$$. $$c^2=25$$, $$c=5$$. $$e_E = 1/2$$. Foci: $$(1\pm5, 1) = (6,1)$$ and $$(-4,1)$$.

Hyperbola eccentricity $$e_H = 1/e_E = 2$$. Same foci, centre at (1,1).

For hyperbola: $$c_H = 5$$ (same foci), $$e_H = 2$$, so $$a_H = c_H/e_H = 5/2$$.

$$b_H^2 = c_H^2 - a_H^2 = 25-25/4 = 75/4$$.

Transverse axis $$\alpha = 2a_H = 5$$, conjugate axis $$\beta = 2b_H = 2\sqrt{75/4} = \sqrt{75} = 5\sqrt{3}$$.

$$3\alpha^2 + 2\beta^2 = 3(25) + 2(75) = 75+150 = 225$$.

The correct answer is Option 4: 225.

A hyperbola $$H$$ centered at the origin with foci on the x-axis is given, and a circle $$C_1$$ centered at the origin touches $$H$$ and has area $$36\pi$$ while another circle $$C_2$$ centered at one focus touches $$H$$ at its vertex and has area $$4\pi$$.

From the areas we get $$r_1^2 = 36 \implies r_1 = 6$$ and $$r_2^2 = 4 \implies r_2 = 2$$.

The circle $$C_1$$ is tangent to the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ at its closest point to the origin, namely the vertex $$(a,0)$$, so $$r_1 = a = 6$$.

Since the circle $$C_2$$ is centered at the focus $$(ae,0)$$ and touches the hyperbola at the vertex $$(a,0)$$, its radius is the distance from the focus to the vertex: $$r_2 = ae - a = a(e - 1) = 6(e - 1) = 2 \implies e - 1 = \frac{1}{3} \implies e = \frac{4}{3}$$.

Then $$b^2 = a^2(e^2 - 1) = 36\left(\frac{16}{9} - 1\right) = 36 \times \frac{7}{9} = 28$$.

Finally, the length of the latus rectum is $$\ell = \frac{2b^2}{a} = \frac{2 \times 28}{6} = \frac{56}{6} = \frac{28}{3}$$, so the correct answer is Option B: $$\frac{28}{3}$$.

For an ellipse, the semi-minor axis is $$b$$ and the distance between foci is $$2c$$ where $$c = ae$$.

Length of minor axis = $$2b$$. Half the distance between foci = $$\frac{2c}{2} = c = ae$$.

Given: $$2b = ae$$, so $$b = \frac{ae}{2}$$.

Using the relation $$b^2 = a^2(1 - e^2)$$:

$$ \frac{a^2e^2}{4} = a^2(1 - e^2) $$

$$ \frac{e^2}{4} = 1 - e^2 $$

$$ e^2 + \frac{e^2}{4} = 1 \Rightarrow \frac{5e^2}{4} = 1 \Rightarrow e^2 = \frac{4}{5} $$

$$ e = \frac{2}{\sqrt{5}} $$

The answer is Option (4): $$\boxed{\frac{2}{\sqrt{5}}}$$.

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: $$20$$.

Hyperbola: $$\frac{x^2}{16} - \frac{y^2}{9} = 1$$. Here $$a_h = 4, b_h = 3$$.

$$e_1 = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$$

Foci of hyperbola: $$(\pm 5, 0)$$.

Given $$e_1 e_2 = 1$$: $$e_2 = \frac{4}{5}$$.

Ellipse passes through $$(\pm 5, 0)$$:

$$\frac{25}{a^2} = 1 \Rightarrow a^2 = 25, a = 5$$.

$$e_2 = \sqrt{1 - \frac{b^2}{a^2}} = \frac{4}{5}$$

$$\frac{b^2}{25} = 1 - \frac{16}{25} = \frac{9}{25}$$

$$b^2 = 9$$

Ellipse: $$\frac{x^2}{25} + \frac{y^2}{9} = 1$$.

Chord at $$y = 2$$: $$\frac{x^2}{25} + \frac{4}{9} = 1 \Rightarrow \frac{x^2}{25} = \frac{5}{9} \Rightarrow x^2 = \frac{125}{9}$$

$$x = \pm\frac{5\sqrt{5}}{3}$$

Length = $$\frac{10\sqrt{5}}{3}$$.

The answer corresponds to Option (3).

Consider an ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ with $$a > b$$, eccentricity $$e_1 = \frac{1}{\sqrt{2}}$$, and latus rectum length $$= \sqrt{14}$$. Since $$e_1^2 = 1 - \frac{b^2}{a^2}$$, we have $$\frac{1}{2} = 1 - \frac{b^2}{a^2}$$, which gives $$\frac{b^2}{a^2} = \frac{1}{2}$$ and hence $$b^2 = \frac{a^2}{2}$$. The length of the latus rectum is $$\frac{2b^2}{a} = \sqrt{14}$$; substituting $$b^2 = \frac{a^2}{2}$$ yields $$\frac{2\cdot (a^2/2)}{a} = a = \sqrt{14}$$. Therefore, $$a = \sqrt{14}$$, $$a^2 = 14$$, and $$b^2 = 7$$.

Now consider the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$, which becomes $$\frac{x^2}{14} - \frac{y^2}{7} = 1$$ when we use the above values. Its eccentricity satisfies $$e_2^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{7}{14} = 1 + \frac{1}{2} = \frac{3}{2}$$. The answer is Option C: $$\frac{3}{2}$$.

$$f(x) = x^2 + 9$$, $$g(x) = \frac{x}{x-9}$$.

$$a = f(g(10)) = f\left(\frac{10}{1}\right) = f(10) = 100 + 9 = 109$$.

$$b = g(f(3)) = g(9+9) = g(18) = \frac{18}{18-9} = \frac{18}{9} = 2$$.

Ellipse: $$\frac{x^2}{109} + \frac{y^2}{2} = 1$$. Here $$a^2 = 109 > b^2 = 2$$.

Eccentricity: $$e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{2}{109} = \frac{107}{109}$$.

Latus rectum: $$l = \frac{2b^2}{a} = \frac{4}{\sqrt{109}}$$. So $$l^2 = \frac{16}{109}$$.

$$8e^2 + l^2 = 8 \cdot \frac{107}{109} + \frac{16}{109} = \frac{856 + 16}{109} = \frac{872}{109} = 8$$.

The correct answer is Option 1: 8.

Let P be a point on the ellipse $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$. Parametrize P using the parametric equations for the ellipse: $$x = 3\cos\theta$$, $$y = 2\sin\theta$$, so P is $$(3\cos\theta, 2\sin\theta)$$.

The line passing through P and parallel to the y-axis has the equation $$x = 3\cos\theta$$. This line intersects the circle $$x^2 + y^2 = 9$$. Substitute $$x = 3\cos\theta$$ into the circle equation:

$$(3\cos\theta)^2 + y^2 = 9$$

$$9\cos^2\theta + y^2 = 9$$

$$y^2 = 9 - 9\cos^2\theta = 9\sin^2\theta$$

$$y = \pm 3|\sin\theta|$$

Since P and Q are on the same side of the x-axis, the y-coordinate of Q must have the same sign as that of P. For P, the y-coordinate is $$2\sin\theta$$, so when $$\sin\theta \geq 0$$, Q has a non-negative y-coordinate, and when $$\sin\theta < 0$$, Q has a negative y-coordinate. In both cases, Q can be taken as $$(3\cos\theta, 3\sin\theta)$$.

Thus, Q is $$(3\cos\theta, 3\sin\theta)$$.

Point R is on PQ such that $$PR : RQ = 4 : 3$$. Since PQ is a vertical line segment (same x-coordinate), R divides PQ in the ratio $$PR : RQ = 4 : 3$$. Using the section formula, if R divides the line segment joining $$P(x_1, y_1)$$ and $$Q(x_2, y_2)$$ in the ratio $$m : n$$, then:

$$R = \left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right)$$

Here, $$m = 4$$, $$n = 3$$, $$P(3\cos\theta, 2\sin\theta)$$, $$Q(3\cos\theta, 3\sin\theta)$$.

The x-coordinate of R is:

$$x_R = \frac{4 \cdot (3\cos\theta) + 3 \cdot (3\cos\theta)}{7} = \frac{12\cos\theta + 9\cos\theta}{7} = \frac{21\cos\theta}{7} = 3\cos\theta$$

The y-coordinate of R is:

$$y_R = \frac{4 \cdot (3\sin\theta) + 3 \cdot (2\sin\theta)}{7} = \frac{12\sin\theta + 6\sin\theta}{7} = \frac{18\sin\theta}{7}$$

So R is $$(3\cos\theta, \frac{18}{7} \sin\theta)$$.

To find the locus of R as $$\theta$$ varies, eliminate $$\theta$$:

Set $$x = 3\cos\theta$$ and $$y = \frac{18}{7} \sin\theta$$. Then:

$$\frac{x}{3} = \cos\theta, \quad \frac{7y}{18} = \sin\theta$$

Squaring and adding:

$$\left( \frac{x}{3} \right)^2 + \left( \frac{7y}{18} \right)^2 = \cos^2\theta + \sin^2\theta = 1$$

$$\frac{x^2}{9} + \frac{49y^2}{324} = 1$$

This is the equation of an ellipse. Rewriting it in standard form $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$:

$$\frac{x^2}{9} + \frac{y^2}{\frac{324}{49}} = 1$$

So $$a^2 = 9$$ and $$b^2 = \frac{324}{49}$$. Since $$a^2 > b^2$$ (as $$9 > \frac{324}{49} \approx 6.612$$), the major axis is along the x-axis.

The eccentricity $$e$$ is given by:

$$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{\frac{324}{49}}{9}} = \sqrt{1 - \frac{324}{49 \times 9}} = \sqrt{1 - \frac{324}{441}}$$

Simplify $$\frac{324}{441}$$:

$$\frac{324 \div 9}{441 \div 9} = \frac{36}{49}$$

So:

$$e = \sqrt{1 - \frac{36}{49}} = \sqrt{\frac{49 - 36}{49}} = \sqrt{\frac{13}{49}} = \frac{\sqrt{13}}{7}$$

The eccentricity is $$\frac{\sqrt{13}}{7}$$, which corresponds to option D.

Hyperbola: $$\frac{x^2}{9} - \frac{y^2}{4} = 1$$. So $$a = 3, b = 2, c = \sqrt{13}$$. Foci: $$(\pm\sqrt{13}, 0)$$.

Let $$P = (x_0, y_0)$$ in first quadrant on hyperbola.

Area of triangle with foci $$F_1(-\sqrt{13},0)$$ and $$F_2(\sqrt{13},0)$$:

Area = $$\frac{1}{2} \times 2\sqrt{13} \times y_0 = \sqrt{13} \cdot y_0 = 2\sqrt{13}$$.

So $$y_0 = 2$$. From hyperbola: $$\frac{x_0^2}{9} - \frac{4}{4} = 1 \Rightarrow \frac{x_0^2}{9} = 2 \Rightarrow x_0^2 = 18$$.

Distance from origin squared: $$x_0^2 + y_0^2 = 18 + 4 = 22$$.

The answer is Option (3): $$\boxed{22}$$.

We need to find $$k$$ from the circle equation, then use it to find the latus rectum of the ellipse, and compute $$2m + n$$. The line is $$2x + 3y - k = 0$$ with $$k > 0$$. Its x-intercept is found by setting $$y = 0$$, which gives $$2x = k \implies x = \frac{k}{2}$$ so point $$A = \left(\frac{k}{2}, 0\right)$$, and its y-intercept by setting $$x = 0$$, which gives $$3y = k \implies y = \frac{k}{3}$$ so $$B = \left(0, \frac{k}{3}\right)$$.

The equation of the circle with AB as diameter follows from the general form for endpoints $$(x_1, y_1)$$ and $$(x_2, y_2)$$: $$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$ which becomes $$\left(x - \frac{k}{2}\right)(x - 0) + \left(y - 0\right)\left(y - \frac{k}{3}\right) = 0$$ and simplifies to $$x^2 - \frac{k}{2}x + y^2 - \frac{k}{3}y = 0$$.

Comparing this with the given circle equation $$x^2 + y^2 - 3x - 2y = 0$$ shows that $$\frac{k}{2} = 3$$ and $$\frac{k}{3} = 2$$, both yielding $$k = 6$$.

Substituting $$k = 6$$ into the ellipse equation gives $$x^2 + 9y^2 = 36$$, which may be written as $$\frac{x^2}{36} + \frac{y^2}{4} = 1$$. Here $$a^2 = 36$$ and $$b^2 = 4$$, so $$a = 6$$ and $$b = 2$$. The length of the latus rectum is given by $$\ell = \frac{2b^2}{a} = \frac{2 \times 4}{6} = \frac{8}{6} = \frac{4}{3}$$. Writing the latus rectum as $$\frac{m}{n} = \frac{4}{3}$$ with coprime integers $$m = 4$$ and $$n = 3$$, one finds $$2m + n = 2(4) + 3 = 11$$.

The correct answer is Option (1): 11.

The given hyperbola is $$x^{2}-y^{2}\csc^{2}\theta = 5$$.

Divide by $$5$$ to convert it into the standard form $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$$:

$$\frac{x^{2}}{5}-\frac{y^{2}\csc^{2}\theta}{5}=1 \;\;\Longrightarrow\;\; \frac{x^{2}}{5}-\frac{y^{2}}{\,5/\csc^{2}\theta\,}=1.$$

Hence for the hyperbola $$a^{2}=5,\qquad b^{2}= \frac{5}{\csc^{2}\theta}=5\sin^{2}\theta.$$ For a hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1,$$ the eccentricity is $$e_{h}=\sqrt{1+\frac{b^{2}}{a^{2}}}.$$

Therefore $$e_{h}= \sqrt{1+\frac{5\sin^{2}\theta}{5}} =\sqrt{1+\sin^{2}\theta}.$$

The given ellipse is $$x^{2}\csc^{2}\theta + y^{2}=5.$$

Divide by $$5$$ to write it as $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$$:

$$\frac{x^{2}\csc^{2}\theta}{5} + \frac{y^{2}}{5}=1 \;\;\Longrightarrow\;\; \frac{x^{2}}{\,5\sin^{2}\theta\,}+\frac{y^{2}}{5}=1.$$

Thus for the ellipse $$a^{2}=5\sin^{2}\theta,\qquad b^{2}=5.$$ Since $$b^{2}\gt a^{2},$$ the semi-major axis is $$b=\sqrt{5}.$$

For an ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\;(b\gt a),$$ the eccentricity is $$e_{e}= \sqrt{1-\frac{a^{2}}{b^{2}}}.$$

Hence $$e_{e}= \sqrt{1-\frac{5\sin^{2}\theta}{5}} =\sqrt{1-\sin^{2}\theta} =\cos\theta.$$

According to the question, $$e_{h}= \sqrt{7}\,e_{e}.$$ Substitute the expressions for $$e_{h}$$ and $$e_{e}$$:

$$\sqrt{1+\sin^{2}\theta}= \sqrt{7}\,\cos\theta.$$

Square both sides:

$$1+\sin^{2}\theta = 7\cos^{2}\theta.$$

Using $$\cos^{2}\theta = 1-\sin^{2}\theta,$$ get an equation in $$s=\sin^{2}\theta$$:

$$1+s = 7(1-s) \;\;\Longrightarrow\;\; 1+s = 7-7s \;\;\Longrightarrow\;\; 8s = 6 \;\;\Longrightarrow\;\; s = \frac{3}{4}.$$

Thus $$\sin^{2}\theta = \frac{3}{4} \;\;\Longrightarrow\;\; \sin\theta = \frac{\sqrt{3}}{2}.$$

For $$0 \lt \theta \lt \frac{\pi}{2},$$ this gives $$\theta = \frac{\pi}{3}.$$

Hence the correct option is Option C: $$\displaystyle \frac{\pi}{3}$$.

Let the given parabola $$P$$ have vertex $$V(2,3)$$ and directrix $$L:2x+y-6=0$$.

For a parabola, the distance from the vertex to the directrix equals the focal length $$p$$. Hence

Distance from $$V$$ to $$L$$ is $$ p \;=\;\frac{\bigl|2\cdot 2 + 3 -6\bigr|}{\sqrt{2^2+1^2}} \;=\;\frac{1}{\sqrt{5}}. $$

The axis of the parabola is perpendicular to the directrix, so its direction is the unit normal to $$L$$, $$ \mathbf{n}=\frac{(2,1)}{\sqrt{5}}. $$ Since the vertex lies in the half-plane $$2x+y-6>0$$, the focus lies in the same half-plane. Thus the focus is $$ F \;=\;V + p\,\mathbf{n} =\;(2,3)+\frac{1}{\sqrt{5}}\;\frac{(2,1)}{\sqrt{5}} =\Bigl(2+\tfrac{2}{5},\,3+\tfrac{1}{5}\Bigr) =\Bigl(\tfrac{12}{5},\,\tfrac{16}{5}\Bigr). $$

Next, let the ellipse $$E$$ be $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \quad a>b, \quad\text{with eccentricity }e=\frac{1}{\sqrt{2}}. $$ We have the standard formula for an ellipse, $$ e=\frac{c}{a},\quad c^2=a^2-b^2. $$ Since $$e^2=\tfrac12$$, it follows $$ c^2=a^2\,e^2=\frac{a^2}{2}, \quad b^2=a^2-c^2=a^2-\frac{a^2}{2}=\frac{a^2}{2}. $$

Thus the equation of the ellipse becomes $$ \frac{x^2}{a^2}+\frac{y^2}{\tfrac{a^2}{2}}=1 \quad\Longrightarrow\quad \frac{x^2}{a^2}+\frac{2y^2}{a^2}=1 \quad\Longrightarrow\quad x^2+2y^2=a^2. $$

Since the focus $$F\bigl(\tfrac{12}{5},\tfrac{16}{5}\bigr)$$ of the parabola lies on this ellipse, substitute into $$x^2+2y^2=a^2$$: $$ \Bigl(\tfrac{12}{5}\Bigr)^2+2\Bigl(\tfrac{16}{5}\Bigr)^2 =\frac{144}{25}+\frac{512}{25} =\frac{656}{25} \;=\;a^2. $$

The length of the latus rectum of the ellipse is given by $$ \ell=\frac{2b^2}{a} =\frac{2\bigl(\tfrac{a^2}{2}\bigr)}{a} =\frac{a^2}{a} =a, $$ so the square of its length is $$ \ell^2=a^2=\frac{656}{25}. $$

Answer: Option D. The required value is $$\displaystyle\frac{656}{25}\,. $$

For the ellipse $$\frac{x^2}{25} + \frac{y^2}{16} = 1$$, we need the chord with midpoint $$(1, \frac{2}{5})$$.

The equation of the chord with midpoint $$(h, k)$$ is given by $$T = S_1$$:

$$\frac{xh}{25} + \frac{yk}{16} = \frac{h^2}{25} + \frac{k^2}{16}$$

With $$(h, k) = (1, \frac{2}{5})$$:

$$\frac{x}{25} + \frac{2y/5}{16} = \frac{1}{25} + \frac{4/25}{16} = \frac{1}{25} + \frac{1}{100}$$

$$\frac{x}{25} + \frac{y}{40} = \frac{4 + 1}{100} = \frac{1}{20}$$

Multiply by 200:

$$8x + 5y = 10$$

So $$y = \frac{10 - 8x}{5} = 2 - \frac{8x}{5}$$.

Substituting into the ellipse equation:

$$\frac{x^2}{25} + \frac{(2 - 8x/5)^2}{16} = 1$$

$$\frac{x^2}{25} + \frac{4 - 32x/5 + 64x^2/25}{16} = 1$$

$$\frac{x^2}{25} + \frac{1}{4} - \frac{2x}{5} + \frac{4x^2}{25} = 1$$

$$\frac{5x^2}{25} - \frac{2x}{5} + \frac{1}{4} = 1$$

$$\frac{x^2}{5} - \frac{2x}{5} = \frac{3}{4}$$

$$\frac{x^2 - 2x}{5} = \frac{3}{4}$$

$$4(x^2 - 2x) = 15$$

$$4x^2 - 8x - 15 = 0$$

$$x = \frac{8 \pm \sqrt{64 + 240}}{8} = \frac{8 \pm \sqrt{304}}{8}$$

The two x-values: $$x_1 + x_2 = 2$$, $$x_1 x_2 = -\frac{15}{4}$$.

$$(x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1x_2 = 4 + 15 = 19$$

The corresponding y-values: $$y = 2 - \frac{8x}{5}$$, so $$y_1 - y_2 = -\frac{8}{5}(x_1 - x_2)$$.

Length of chord:

$$L^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 = (x_1 - x_2)^2\left(1 + \frac{64}{25}\right) = 19 \times \frac{89}{25} = \frac{1691}{25}$$

$$L = \frac{\sqrt{1691}}{5}$$

The answer is $$\frac{\sqrt{1691}}{5}$$, which corresponds to Option (1).

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.

$$f(x) = 4\sqrt{2}x^3 - 3\sqrt{2}x - 1$$ on $$[\frac{1}{2}, 1]$$.

Statement (I): The curve intersects the x-axis exactly at one point.

Statement (II): The curve intersects the x-axis at $$x = \cos\frac{\pi}{12}$$.

Check Statement (II) by recognising the triple angle formula.

Recall that $$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$$.

If $$x = \cos\theta$$, then $$4x^3 - 3x = \cos 3\theta$$.

$$f(x) = \sqrt{2}(4x^3 - 3x) - 1 = \sqrt{2}\cos 3\theta - 1$$

Setting $$f(x) = 0$$: $$\sqrt{2}\cos 3\theta = 1$$, so $$\cos 3\theta = \frac{1}{\sqrt{2}}$$.

$$3\theta = \frac{\pi}{4} \implies \theta = \frac{\pi}{12}$$

$$x = \cos\frac{\pi}{12} \approx 0.9659$$, which lies in $$[\frac{1}{2}, 1]$$. Statement (II) is correct.

Check Statement (I) - uniqueness.

$$f'(x) = 12\sqrt{2}x^2 - 3\sqrt{2} = 3\sqrt{2}(4x^2 - 1)$$

$$f'(x) = 0 \implies x = \frac{1}{2}$$ (the only critical point in the domain).

$$f(\frac{1}{2}) = 4\sqrt{2} \cdot \frac{1}{8} - 3\sqrt{2} \cdot \frac{1}{2} - 1 = \frac{\sqrt{2}}{2} - \frac{3\sqrt{2}}{2} - 1 = -\sqrt{2} - 1 < 0$$

$$f(1) = 4\sqrt{2} - 3\sqrt{2} - 1 = \sqrt{2} - 1 > 0$$

Since $$f(\frac{1}{2}) < 0$$ and $$f(1) > 0$$, and $$f'(x) > 0$$ for $$x > \frac{1}{2}$$, the function is strictly increasing on $$(\frac{1}{2}, 1]$$. There is exactly one root. Statement (I) is correct.

Both statements are correct. The answer is Option 4.

We need to find the area enclosed by the curves $$y = 1 + 3x - 2x^2$$ and $$y = \frac{1}{x}$$, given that one intersection point is $$\left(\frac{1}{2}, 2\right)$$.

At intersection: $$1 + 3x - 2x^2 = \frac{1}{x}$$

$$x(1 + 3x - 2x^2) = 1$$

$$x + 3x^2 - 2x^3 = 1$$

$$2x^3 - 3x^2 - x + 1 = 0$$

We know $$x = \frac{1}{2}$$ is a root. Factor out $$(2x - 1)$$:

$$2x^3 - 3x^2 - x + 1 = (2x - 1)(x^2 - x - 1) = 0$$

From $$x^2 - x - 1 = 0$$: $$x = \frac{1 \pm \sqrt{5}}{2}$$

So the roots are $$x = \frac{1}{2}$$, $$x = \frac{1 + \sqrt{5}}{2}$$ (the golden ratio $$\phi$$), and $$x = \frac{1 - \sqrt{5}}{2}$$ (negative, so we consider the region for $$x > 0$$).

For $$x$$ between $$\frac{1}{2}$$ and $$\frac{1+\sqrt{5}}{2}$$, test at $$x = 1$$:

Parabola: $$1 + 3 - 2 = 2$$

Hyperbola: $$\frac{1}{1} = 1$$

So the parabola is above the hyperbola in this interval.

$$A = \int_{1/2}^{(1+\sqrt{5})/2} \left(1 + 3x - 2x^2 - \frac{1}{x}\right) dx$$

$$= \left[x + \frac{3x^2}{2} - \frac{2x^3}{3} - \ln x\right]_{1/2}^{(1+\sqrt{5})/2}$$

Let $$\phi = \frac{1+\sqrt{5}}{2}$$. Note that $$\phi^2 = \phi + 1 = \frac{3+\sqrt{5}}{2}$$ and $$\phi^3 = 2\phi + 1 = 2 + \sqrt{5}$$.

At $$x = \phi$$:

$$\phi + \frac{3\phi^2}{2} - \frac{2\phi^3}{3} - \ln\phi$$

$$= \frac{1+\sqrt{5}}{2} + \frac{3(3+\sqrt{5})}{4} - \frac{2(2+\sqrt{5})}{3} - \ln\phi$$

$$= \frac{1+\sqrt{5}}{2} + \frac{9+3\sqrt{5}}{4} - \frac{4+2\sqrt{5}}{3} - \ln\phi$$

$$= \frac{6(1+\sqrt{5}) + 3(9+3\sqrt{5}) - 4(4+2\sqrt{5})}{12} - \ln\phi$$

$$= \frac{6+6\sqrt{5}+27+9\sqrt{5}-16-8\sqrt{5}}{12} - \ln\phi$$

$$= \frac{17+7\sqrt{5}}{12} - \ln\phi$$

At $$x = \frac{1}{2}$$:

$$\frac{1}{2} + \frac{3}{8} - \frac{1}{12} - \ln\frac{1}{2} = \frac{12+9-2}{24} + \ln 2 = \frac{19}{24} + \ln 2$$

Area:

$$A = \frac{17+7\sqrt{5}}{12} - \ln\phi - \frac{19}{24} - \ln 2$$

$$= \frac{34+14\sqrt{5}-19}{24} - \ln(2\phi)$$

$$= \frac{15+14\sqrt{5}}{24} - \ln(1+\sqrt{5})$$

This can be written as $$\frac{1}{24}(14\sqrt{5} + 15) - \ln(1+\sqrt{5})$$.

Comparing with $$\frac{1}{24}(l\sqrt{5} + m) - n\log_e(1+\sqrt{5})$$:

$$l = 14, m = 15, n = 1$$

$$l + m + n = 14 + 15 + 1 = 30$$

The correct answer is Option (3): $$\boxed{30}$$.

To find the area enclosed by the curves $$xy + 4y = 16$$ and $$x + y = 6$$, first rewrite the equations in terms of $$y$$.

The first curve: $$xy + 4y = 16$$ can be factored as $$y(x + 4) = 16$$, so $$y = \frac{16}{x+4}$$ for $$x \neq -4$$.

The second curve: $$x + y = 6$$ can be written as $$y = 6 - x$$.

Next, find the points of intersection by setting the $$y$$-values equal:

$$\frac{16}{x+4} = 6 - x$$

Multiply both sides by $$x+4$$ (assuming $$x \neq -4$$):

$$16 = (6 - x)(x + 4)$$

Expand the right side:

$$16 = 6x + 24 - x^2 - 4x$$

$$16 = -x^2 + 2x + 24$$

Bring all terms to one side:

$$x^2 - 2x - 8 = 0$$

Factor the quadratic:

$$(x - 4)(x + 2) = 0$$

So, $$x = 4$$ or $$x = -2$$.

Find the corresponding $$y$$-values using $$y = 6 - x$$:

When $$x = 4$$, $$y = 6 - 4 = 2$$, so point $$(4, 2)$$.

When $$x = -2$$, $$y = 6 - (-2) = 8$$, so point $$(-2, 8)$$.

The enclosed region is between $$x = -2$$ and $$x = 4$$. To determine which curve is above, evaluate at a test point, say $$x = 0$$:

For the hyperbola: $$y = \frac{16}{0+4} = 4$$.

For the line: $$y = 6 - 0 = 6$$.

Since $$6 > 4$$, the line $$y = 6 - x$$ is above the hyperbola $$y = \frac{16}{x+4}$$ in the interval $$[-2, 4]$$.

The area enclosed is given by the integral:

$$\text{Area} = \int_{-2}^{4} \left[ (6 - x) - \frac{16}{x+4} \right] dx$$

Split the integral into two parts:

$$\text{Area} = \int_{-2}^{4} (6 - x) dx - \int_{-2}^{4} \frac{16}{x+4} dx$$

Compute the first integral:

Antiderivative of $$6 - x$$ is $$6x - \frac{x^2}{2}$$.

Evaluate from $$-2$$ to $$4$$:

At $$x = 4$$: $$6(4) - \frac{(4)^2}{2} = 24 - 8 = 16$$.

At $$x = -2$$: $$6(-2) - \frac{(-2)^2}{2} = -12 - 2 = -14$$.

So, $$\int_{-2}^{4} (6 - x) dx = 16 - (-14) = 30$$.

Now compute the second integral:

$$\int_{-2}^{4} \frac{16}{x+4} dx = 16 \int_{-2}^{4} \frac{1}{x+4} dx$$

Antiderivative of $$\frac{1}{x+4}$$ is $$\ln|x+4|$$.

Evaluate from $$-2$$ to $$4$$:

At $$x = 4$$: $$\ln|4+4| = \ln 8$$.

At $$x = -2$$: $$\ln|-2+4| = \ln 2$$.

So, $$\int_{-2}^{4} \frac{1}{x+4} dx = \ln 8 - \ln 2 = \ln \left( \frac{8}{2} \right) = \ln 4$$.

Thus, $$16 \int_{-2}^{4} \frac{1}{x+4} dx = 16 \ln 4$$.

Since $$\ln 4 = \ln (2^2) = 2 \ln 2$$, this becomes $$16 \times 2 \ln 2 = 32 \ln 2$$.

Therefore, the area is:

$$\text{Area} = 30 - 32 \ln 2$$

Since $$\log_e 2 = \ln 2$$, the area is $$30 - 32 \log_e 2$$.

Comparing with the options:

A. $$28 - 30\log_e 2$$
B. $$30 - 28\log_e 2$$
C. $$30 - 32\log_e 2$$
D. $$32 - 30\log_e 2$$

The expression matches option C.

The correct answer is 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 a hyperbola with eccentricity $$e$$, latus rectum length 9, and directrices $$x = \pm \dfrac{4}{\sqrt{13}}$$.

For a standard hyperbola $$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$$ the directrices are $$x = \pm \dfrac{a}{e}$$ and the length of the latus rectum is $$\dfrac{2b^2}{a}$$.

From the directrices $$\dfrac{a}{e} = \dfrac{4}{\sqrt{13}}$$ so $$a = \dfrac{4e}{\sqrt{13}}$$, and from the latus rectum $$\dfrac{2b^2}{a} = 9$$ so $$b^2 = \dfrac{9a}{2}$$. Using $$b^2 = a^2(e^2 - 1)$$ we obtain

$$\dfrac{9a}{2} = a^2(e^2 - 1)$$

$$\dfrac{9}{2} = a(e^2 - 1)$$

Substituting $$a = \dfrac{4e}{\sqrt{13}}$$ gives

$$\dfrac{9}{2} = \dfrac{4e}{\sqrt{13}}(e^2 - 1)$$

$$\dfrac{9\sqrt{13}}{8} = e(e^2 - 1) = e^3 - e$$

If we try $$e = \dfrac{\sqrt{13}}{2}$$ then

$$e^3 - e = \dfrac{13\sqrt{13}}{8} - \dfrac{\sqrt{13}}{2} = \dfrac{13\sqrt{13} - 4\sqrt{13}}{8} = \dfrac{9\sqrt{13}}{8}$$

This works, so $$e = \dfrac{\sqrt{13}}{2}$$. Hence $$a = \dfrac{4 \cdot \frac{\sqrt{13}}{2}}{\sqrt{13}} = 2$$ and $$b^2 = a^2(e^2 - 1) = 4\left(\dfrac{13}{4} - 1\right) = 9$$. Thus the equation of the hyperbola is $$\dfrac{x^2}{4} - \dfrac{y^2}{9} = 1$$.

The tangent line is given by $$y - \sqrt{3}x + \sqrt{3} = 0$$, or $$y = \sqrt{3}(x - 1)$$. Its slope is $$m_t = \sqrt{3}$$. The tangent to $$\dfrac{x^2}{4} - \dfrac{y^2}{9} = 1$$ at a point $$(x_0,y_0)$$ has equation

$$\dfrac{x\,x_0}{4} - \dfrac{y\,y_0}{9} = 1$$

or equivalently $$y = \dfrac{9x_0}{4y_0}x - \dfrac{9}{y_0}$$. Comparing with $$y = \sqrt{3}x - \sqrt{3}$$ gives $$\dfrac{9x_0}{4y_0} = \sqrt{3}$$ and $$-\dfrac{9}{y_0} = -\sqrt{3}$$, so $$y_0 = 3\sqrt{3}$$. Then $$x_0 = \dfrac{4\sqrt{3}\,y_0}{9} = 4$$. Hence the point of tangency is $$(4,3\sqrt{3})$$, which indeed lies on the hyperbola since $$\dfrac{16}{4} - \dfrac{27}{9} = 1$$.

The foci are at $$(\pm ae,0) = (\pm\sqrt{13},0)$$. For the point $$(x_0,y_0)$$ on the right branch the focal distances are $$d_1 = ex_0 - a = 2\sqrt{13}-2$$ and $$d_2 = ex_0 + a = 2\sqrt{13}+2$$. Their product is

$$m = d_1\,d_2 = (2\sqrt{13})^2 - 4 = 52 - 4 = 48$$.

Finally,

$$4e^2 + m = 4 \cdot \dfrac{13}{4} + 48 = 13 + 48 = 61$$

Therefore the required value is $$\boxed{61}$$.

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 $$\boxed{192}$$.

For the hyperbola $$\frac{x^2}{3} - \frac{y^2}{5} = 1$$: $$a^2 = 3, b^2 = 5$$, so $$c^2 = a^2 + b^2 = 8$$, $$c = 2\sqrt{2}$$.

Focus on positive x-axis: $$S = (2\sqrt{2}, 0)$$.

Circle C has centre $$A(\sqrt{6}, \sqrt{5})$$ and passes through S. Radius = $$|AS|$$.

$$|AS|^2 = (2\sqrt{2}-\sqrt{6})^2 + (0-\sqrt{5})^2 = (8 - 4\sqrt{12} + 6) + 5 = 19 - 4\sqrt{12}$$

$$= 19 - 8\sqrt{3}$$

SAB is a diameter, so B is the diametrically opposite point of S through centre A.

$$B = 2A - S = (2\sqrt{6} - 2\sqrt{2}, 2\sqrt{5})$$.

Now find the area of triangle OSB, where $$O = (0,0)$$, $$S = (2\sqrt{2}, 0)$$, $$B = (2\sqrt{6}-2\sqrt{2}, 2\sqrt{5})$$.

Area = $$\frac{1}{2}|x_S \cdot y_B - x_B \cdot y_S|$$

$$= \frac{1}{2}|2\sqrt{2} \cdot 2\sqrt{5} - (2\sqrt{6}-2\sqrt{2}) \cdot 0|$$

$$= \frac{1}{2} \cdot 4\sqrt{10} = 2\sqrt{10}$$

Square of area = $$(2\sqrt{10})^2 = 40$$.

The answer is $$\boxed{40}$$.

For the ellipse $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1,\; a\gt b$$ the standard facts are

• Distance of each focus from the centre: $$c,\; c^2=a^2-b^2$$
• Length of the latus-rectum: $$L=\dfrac{2b^2}{a}$$

The question gives

$$c=5 \;\; \bigl(\text{foci } (\pm5,0)\bigr), \qquad L=\sqrt{50}$$

Step 1: Use the latus-rectum length.

$$\dfrac{2b^2}{a}= \sqrt{50}\;\; \Longrightarrow\;\; b^2=\dfrac{a\sqrt{50}}{2} \quad -(1)$$

Step 2: Use the focus relation.

$$a^2-b^2=25 \quad -(2)$$

Step 3: Substitute $$b^2$$ from $$(1)$$ into $$(2)$$.

$$a^2-\dfrac{a\sqrt{50}}{2}=25$$

Multiply by $$2$$:

$$2a^2-a\sqrt{50}-50=0$$

Step 4: Solve the quadratic in $$a$$.

Using the quadratic formula, $$a=\dfrac{\sqrt{50}\pm\sqrt{(\sqrt{50})^2-4(2)(-50)}}{4}$$

Discriminant: $$(\sqrt{50})^2-4(2)(-50)=50+400=450$$, so $$\sqrt{450}=3\sqrt{50}$$.

Hence $$a=\dfrac{\sqrt{50}\pm3\sqrt{50}}{4}$$. Since $$a\gt0$$, take the positive sign:

$$a=\sqrt{50}, \qquad\Longrightarrow\qquad a^2=50$$

Step 5: Eccentricity of the hyperbola.

The given hyperbola is $$\dfrac{x^2}{b^2}-\dfrac{y^2}{a^2b^2}=1$$

Compare with the standard form $$\dfrac{x^2}{A^2}-\dfrac{y^2}{B^2}=1$$, where

$$A^2=b^2, \qquad B^2=a^2b^2$$

For a hyperbola, $$e^2=1+\dfrac{B^2}{A^2}$$.

Therefore

$$e_{\text{hyper}}^{\,2}=1+\dfrac{a^2b^2}{b^2}=1+a^2$$

Using $$a^2=50$$, we get

$$e_{\text{hyper}}^{\,2}=1+50=51$$

Thus, the square of the eccentricity of the hyperbola equals $$51$$.

For the hyperbola $$\frac{x^2}{9} - \frac{y^2}{b^2} = 1$$, we have $$a = 3$$.

The semi-latus rectum is $$l = \frac{b^2}{a} = \frac{b^2}{3}$$. The end of the latus rectum is at $$(ae, \frac{b^2}{a})$$, where $$c = ae = \sqrt{9 + b^2}$$.

The latus rectum subtends angle $$\frac{\pi}{3}$$ at the centre. The half-angle at centre is $$\frac{\pi}{6}$$.

$$\tan\frac{\pi}{6} = \frac{b^2/3}{\sqrt{9+b^2}}$$

$$\frac{1}{\sqrt{3}} = \frac{b^2}{3\sqrt{9+b^2}}$$

$$3\sqrt{9+b^2} = \sqrt{3} b^2$$

Squaring: $$9(9+b^2) = 3b^4$$

$$81 + 9b^2 = 3b^4$$

$$3b^4 - 9b^2 - 81 = 0$$

$$b^4 - 3b^2 - 27 = 0$$

Using the quadratic formula with $$u = b^2$$:

$$u = \frac{3 + \sqrt{9 + 108}}{2} = \frac{3 + \sqrt{117}}{2} = \frac{3 + 3\sqrt{13}}{2}$$

So $$b^2 = \frac{3}{2}(1 + \sqrt{13})$$.

Here $$l = 3, m = 2, n = 13$$ (with $$l, m$$ coprime).

$$l^2 + m^2 + n^2 = 9 + 4 + 169 = 182$$.

Therefore, the answer is $$\boxed{182}$$.

We need to find the square of the area of triangle $$PQ_1Q_2$$.

First, we find the point P in the first quadrant where the circle $$x^2 + y^2 = 3$$ intersects the parabola $$x^2 = 2y$$. Substituting $$x^2 = 2y$$ into the circle gives $$2y + y^2 = 3$$, so $$y^2 + 2y - 3 = 0$$ and $$(y+3)(y-1) = 0$$. Since P is in the first quadrant, $$y = 1$$ and $$x^2 = 2$$, hence $$x = \sqrt{2}$$, which gives $$P = (\sqrt{2},1)$$.

The line $$L: \sqrt{2}x + y = \alpha$$ passes through $$P(\sqrt{2},1)$$, so $$\sqrt{2}\cdot\sqrt{2} + 1 = 2 + 1 = 3$$, giving $$\alpha = 3$$. Therefore the equation of the line is $$\sqrt{2}x + y = 3$$.

The circles $$C_1$$ and $$C_2$$ each have radius $$2\sqrt{3}$$ and centres on the y-axis, denoted $$Q_1 = (0,k_1)$$ and $$Q_2 = (0,k_2)$$. Since line $$L$$ is tangent to each circle, the distance from $$(0,k)$$ to the line $$\sqrt{2}x + y - 3 = 0$$ must equal $$2\sqrt{3}$$. This distance is $$d = \frac{|0 + k - 3|}{\sqrt{2 + 1}} = \frac{|k - 3|}{\sqrt{3}} = 2\sqrt{3}$$, which implies $$|k - 3| = 6$$. Hence $$k = 9$$ or $$k = -3$$, giving $$Q_1 = (0,9)$$ and $$Q_2 = (0,-3)$$.

With $$P = (\sqrt{2},1)$$, $$Q_1 = (0,9)$$ and $$Q_2 = (0,-3)$$, the base $$Q_1Q_2$$ has length $$|9 - (-3)| = 12$$ and the height from P to the y-axis is $$|\sqrt{2}| = \sqrt{2}$$. Therefore the area of triangle $$PQ_1Q_2$$ is $$\frac{1}{2} \times 12 \times \sqrt{2} = 6\sqrt{2}$$.

Finally, squaring this area gives $$(6\sqrt{2})^2 = 72$$. Thus, the square of the area is 72.

$$y^2 = 12x \implies a=3$$. 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 points common to the circle $$x^{2}+y^{2}=4b\quad -(1)$$ and the ellipse $$\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1\quad -(2)$$ are stated to lie on the pair of straight lines $$y^{2}=3x^{2}\; \Longrightarrow\; y=\pm\sqrt{3}\,x\quad -(3)$$.

Substitute $$y^{2}=3x^{2}$$ from $$(3)$$ in the circle $$(1)$$:

$$x^{2}+3x^{2}=4b\;\Longrightarrow\;4x^{2}=4b\;\Longrightarrow\;x^{2}=b\;\Longrightarrow\;|x|=\sqrt{b}$$.

Using $$(3)$$ again, $$y^{2}=3x^{2}=3b\;\Longrightarrow\;|y|=\sqrt{3b}$$.

Hence the four intersection points are $$\bigl(\pm\sqrt{b},\,\pm\sqrt{3b}\bigr)$$, one in each quadrant.

These points must also satisfy the ellipse $$(2)$$. Substitute $$x^{2}=b,\;y^{2}=3b$$ in $$(2)$$:

$$\frac{b}{16}+\frac{3b}{b^{2}}=1 \;\Longrightarrow\;\frac{b}{16}+\frac{3}{b}=1 \;\Longrightarrow\;b^{2}-16b+48=0 \;\Longrightarrow\;(b-12)(b-4)=0$$.

This gives $$b=12$$ or $$b=4$$. When $$b=4$$ the ellipse becomes $$\dfrac{x^{2}}{16}+\dfrac{y^{2}}{16}=1 \;\Longrightarrow\;x^{2}+y^{2}=16$$, identical to the circle. The problem states the conics are distinct, so we reject $$b=4$$ and take $$b=12$$.

Width of the rectangle formed by the four points: $$2|x|=2\sqrt{b}=2\sqrt{12}=4\sqrt{3}$$.
Height of the rectangle: $$2|y|=2\sqrt{3b}=2\sqrt{36}=12$$.

Area of the rectangle $$=4\sqrt{3}\times12=48\sqrt{3}$$.

Required expression: $$3\sqrt{3}\times(\text{area})=3\sqrt{3}\times48\sqrt{3}=3\sqrt{3}\times48\sqrt{3}=3\times48\times3=432$$.

Hence, $$3\sqrt{3}$$ times the required area equals $$432$$.

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 $$\boxed{68}$$.

We need to find point $$P$$ where lines $$ax + by + c = 0$$ are concurrent (for $$a, b, c$$ in A.P.), and point $$Q(\alpha, \beta)$$ from the system with infinitely many solutions.

Finding point P: since $$a, b, c$$ are in A.P.: $$b = a + d$$ and $$c = a + 2d$$ for some common difference $$d$$.

Substituting into $$ax + by + c = 0$$:

$$ax + (a + d)y + (a + 2d) = 0$$

$$a(x + y + 1) + d(y + 2) = 0$$

For this to hold for all values of $$a$$ and $$d$$, we need:

$$x + y + 1 = 0$$ and $$y + 2 = 0$$

So $$y = -2$$ and $$x = 1$$. Therefore $$P = (1, -2)$$.

Finding point Q: the system is: $$x + y + z = 6$$, $$2x + 5y + \alpha z = \beta$$, $$x + 2y + 3z = 4$$.

For infinitely many solutions, the determinant of the coefficient matrix must be zero:

$$\begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{vmatrix} = 1(15 - 2\alpha) - 1(6 - \alpha) + 1(4 - 5) = 0$$

$$15 - 2\alpha - 6 + \alpha + (-1) = 0 \implies 8 - \alpha = 0 \implies \alpha = 8$$

For consistency with $$\alpha = 8$$: From equations (1) and (3): $$(3) - (1)$$: $$y + 2z = -2$$.

From equations (1) and (2): $$(2) - 2(1)$$: $$3y + 6z = \beta - 12$$, i.e., $$3(y + 2z) = \beta - 12$$.

Since $$y + 2z = -2$$: $$3(-2) = \beta - 12 \implies \beta = 6$$.

So $$Q = (\alpha, \beta) = (8, 6)$$.

Computing $$(PQ)^2$$: $$ (PQ)^2 = (8 - 1)^2 + (6 - (-2))^2 = 49 + 64 = 113 $$

The answer is $$\boxed{113}$$.

We need the area of the region defined by:

1) $$x - 2y + 4 \geq 0$$ (i.e., $$x \geq 2y - 4$$)

2) $$x + 2y^2 \geq 0$$ (i.e., $$x \geq -2y^2$$)

3) $$x + 4y^2 \leq 8$$ (i.e., $$x \leq 8 - 4y^2$$)

4) $$y \geq 0$$

Let us integrate with respect to $$y$$. From condition 3: $$x \leq 8 - 4y^2$$, so $$y \leq \sqrt{2}$$ (for $$x \geq 0$$, max).

The right boundary is $$x = 8 - 4y^2$$.

The left boundary is $$x = \max(2y - 4, -2y^2)$$.

For $$y \geq 0$$: When is $$2y - 4 \geq -2y^2$$?

$$2y^2 + 2y - 4 \geq 0$$, i.e., $$y^2 + y - 2 \geq 0$$, i.e., $$(y+2)(y-1) \geq 0$$. Since $$y \geq 0$$: $$y \geq 1$$.

So left boundary = $$-2y^2$$ for $$0 \leq y \leq 1$$ and $$2y - 4$$ for $$y \geq 1$$.

Upper limit from right boundary: $$8 - 4y^2 \geq 2y - 4$$ gives $$4y^2 + 2y - 12 \leq 0$$, $$2y^2 + y - 6 \leq 0$$, $$(2y - 3)(y + 2) \leq 0$$, so $$y \leq 3/2$$.

Also $$8 - 4y^2 \geq -2y^2$$ gives $$2y^2 \leq 8$$, $$y \leq 2$$.

Area = $$\int_0^1 (8 - 4y^2 - (-2y^2)) dy + \int_1^{3/2} (8 - 4y^2 - (2y - 4)) dy$$

$$= \int_0^1 (8 - 2y^2) dy + \int_1^{3/2} (12 - 4y^2 - 2y) dy$$

First integral: $$\left[8y - \frac{2y^3}{3}\right]_0^1 = 8 - \frac{2}{3} = \frac{22}{3}$$

Second integral: $$\left[12y - \frac{4y^3}{3} - y^2\right]_1^{3/2}$$

At $$y = 3/2$$: $$18 - \frac{4 \cdot 27/8}{3} - \frac{9}{4} = 18 - \frac{9}{2} - \frac{9}{4} = 18 - \frac{27}{4} = \frac{45}{4}$$

At $$y = 1$$: $$12 - \frac{4}{3} - 1 = 11 - \frac{4}{3} = \frac{29}{3}$$

Second integral = $$\frac{45}{4} - \frac{29}{3} = \frac{135 - 116}{12} = \frac{19}{12}$$

Total area = $$\frac{22}{3} + \frac{19}{12} = \frac{88 + 19}{12} = \frac{107}{12}$$

$$m + n = 107 + 12 = 119$$.

The answer is $$\boxed{119}$$.

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, $$\frac{a}{b} + \frac{b}{a} \geq 2$$.

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

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

We need to find the area of the region $$\{(x, y) : 0 \leq x \leq 3, \, 0 \leq y \leq \min\{x^2 + 2, \, 2x + 2\}\}$$.

Find where the curves intersect: $$x^2 + 2 = 2x + 2 \implies x^2 - 2x = 0 \implies x(x - 2) = 0$$

So the curves intersect at $$x = 0$$ and $$x = 2$$.

Determine which function is smaller on each interval: for $$0 < x < 2$$: $$x^2 + 2 < 2x + 2$$ (since $$x^2 < 2x$$ for $$0 < x < 2$$).

For $$x > 2$$: $$x^2 + 2 > 2x + 2$$.

So $$\min = x^2 + 2$$ on $$[0, 2]$$ and $$\min = 2x + 2$$ on $$[2, 3]$$.

Compute the area: $$ A = \int_0^2 (x^2 + 2) \, dx + \int_2^3 (2x + 2) \, dx $$

First integral:

$$\int_0^2 (x^2 + 2) \, dx = \left[\frac{x^3}{3} + 2x\right]_0^2 = \frac{8}{3} + 4 = \frac{20}{3}$$

Second integral:

$$\int_2^3 (2x + 2) \, dx = [x^2 + 2x]_2^3 = (9 + 6) - (4 + 4) = 15 - 8 = 7$$

$$ A = \frac{20}{3} + 7 = \frac{20 + 21}{3} = \frac{41}{3} $$

Compute $$12A$$: $$ 12A = 12 \times \frac{41}{3} = 4 \times 41 = 164 $$

The answer is $$\boxed{164}$$.

The two given curves are
ellipse $$E:\dfrac{x^{2}}{6}+\dfrac{y^{2}}{3}=1$$ and parabola $$P:y^{2}=12x$$.

Write the parabola in the standard form $$y^{2}=4ax$$.
Since $$12x=4ax$$, we have $$a=3$$.

Common tangent in slope form
  • For $$P:y^{2}=4ax$$ the tangent having slope $$m$$ is $$y=mx+\dfrac{a}{m}=mx+\dfrac{3}{m}\quad -(1)$$
  • For the ellipse $$\dfrac{x^{2}}{6}+\dfrac{y^{2}}{3}=1$$ the tangent with the same slope $$m$$ is $$y=mx\pm\sqrt{6m^{2}+3}\quad -(2)$$

Because the line must be tangent to both curves, the two intercepts in (1) and (2) must coincide:
$$\dfrac{3}{m}=\pm\sqrt{6m^{2}+3}$$ Square both sides:

$$\left(\dfrac{3}{m}\right)^{2}=6m^{2}+3 \;\Longrightarrow\;9=6m^{4}+3m^{2} \;\Longrightarrow\;6m^{4}+3m^{2}-9=0$$ Divide by $$3$$, put $$n=m^{2}\ge 0$$:

$$2n^{2}+n-3=0 \;\Longrightarrow\;n=\dfrac{-1\pm5}{4}$$ The non-negative root is $$n=1\;\Rightarrow\;m^{2}=1$$, hence $$m=1\quad\text{or}\quad m=-1$$.

Equations of the two common tangents
For $$m=1$$: from (1) $$y=x+3$$   (call this $$T_{1}$$)
For $$m=-1$$: from (1) $$y=-x-3$$ (call this $$T_{2}$$)

Intersection with the x-axis
Set $$y=0$$:
For $$T_{1}$$: $$0=x+3\;\Rightarrow\;(-3,0)$$
For $$T_{2}$$: $$0=-x-3\;\Rightarrow\;(-3,0)$$
Thus both tangents meet the x-axis at the single point $$(-3,0)$$.

Points of contact on the parabola
For $$y^{2}=4ax$$ the tangent $$y=mx+\dfrac{a}{m}$$ touches the curve at $$\left(\dfrac{a}{m^{2}},\dfrac{2a}{m}\right)$$.

• For $$m=1$$: $$A_{1}\bigl(\,3,\,6\bigr)$$
• For $$m=-1$$: $$A_{4}\bigl(\,3,\,-6\bigr)$$

Points of contact on the ellipse
Substitute each tangent in the ellipse and use the fact that a tangent cuts the curve at a repeated root.

For $$T_{1}:y=x+3$$:
$$\dfrac{x^{2}}{6}+\dfrac{(x+3)^{2}}{3}=1 \;\Longrightarrow\;(x+2)^{2}=0 \;\Rightarrow\;x=-2,\;y=1$$ Hence $$A_{2}(-2,1)$$.

For $$T_{2}:y=-x-3$$:
$$\dfrac{x^{2}}{6}+\dfrac{(-x-3)^{2}}{3}=1 \;\Longrightarrow\;(x+2)^{2}=0 \;\Rightarrow\;x=-2,\;y=-1$$ Hence $$A_{3}(-2,-1)$$.

Therefore the four vertices are $$A_{1}(3,6),\;A_{2}(-2,1),\;A_{3}(-2,-1),\;A_{4}(3,-6).$$

Area of quadrilateral $$A_{1}A_{2}A_{3}A_{4}$$ (shoelace formula)

$$$ \begin{array}{r|r} x & y \\ \hline 3 & 6 \\ -2 & 1 \\ -2 & -1 \\ 3 & -6 \\ \end{array} $$$ Sum_{1}=3$$\cdot$$1+(-2)(-1)+(-2)(-6)+3$$\cdot$$6=35\\ Sum_{2}=6(-2)+1(-2)+(-1)3+(-6)3=-35 \]

$$\text{Area}=\dfrac{1}{2}\,\bigl|\,\text{Sum}_{1}-\text{Sum}_{2}\bigr| =\dfrac{1}{2}\,(35-(-35))=\dfrac{70}{2}=35\text{ square units}.$$

Verification of the statements
A. Area is $$35$$ - TRUE.
B. Area is $$36$$ - FALSE.
C. Both tangents meet the x-axis at $$(-3,0)$$ - TRUE.
D. Meeting point $$(-6,0)$$ - FALSE.

Hence the correct options are:
Option A (area $$35$$ square units) and Option C (intersection point $$(-3,0)$$).

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 seek the point $$P(x_0, y_0)$$ on the hyperbola $$3x^2 - 4y^2 = 36$$ that is closest to the line $$3x + 2y = 1$$, and then compute $$\sqrt{2}(y_0 - x_0)\,.$$

First, rewrite the hyperbola in standard form as $$\frac{x^2}{12} - \frac{y^2}{9} = 1\,, $$ so that $$a^2 = 12$$ and $$b^2 = 9\,. $$ A convenient parametrization is for the right branch $$x = 2\sqrt{3}\cosh t,\quad y = 3\sinh t$$ and for the left branch $$x = -2\sqrt{3}\cosh t,\quad y = 3\sinh t\,. $$

The distance from a point $$(x,y)$$ to the line $$3x + 2y - 1 = 0$$ is given by $$d = \frac{\bigl|3x + 2y - 1\bigr|}{\sqrt{13}}\,. $$ On the right branch we set $$f(t) = 6\sqrt{3}\cosh t + 6\sinh t - 1$$ since $$3x + 2y - 1 = 6\sqrt{3}\cosh t + 6\sinh t - 1\,. $$ Differentiating yields $$f'(t) = 6\sqrt{3}\sinh t + 6\cosh t = 0 \quad\Longrightarrow\quad \tanh t = -\frac{1}{\sqrt{3}}\,. $$ Together with $$\cosh^2 t - \sinh^2 t = 1$$ one finds $$\cosh t = \sqrt{\frac{3}{2}},\qquad \sinh t = -\frac{1}{\sqrt{2}}\,. $$ Hence the critical point on the right branch is $$x_0 = 2\sqrt{3}\,\sqrt{\tfrac{3}{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2},\qquad y_0 = 3\Bigl(-\frac{1}{\sqrt{2}}\Bigr) = -\frac{3}{\sqrt{2}} = -\frac{3\sqrt{2}}{2}\,. $$

On the left branch we instead have $$f(t) = -6\sqrt{3}\cosh t + 6\sinh t - 1\,, $$ and setting $$f'(t)=0$$ gives $$\tanh t = \frac{1}{\sqrt{3}}\,, $$ leading to $$x_0 = -3\sqrt{2},\qquad y_0 = \frac{3\sqrt{2}}{2}\,. $$

To decide which point is nearer, compute the numerators of the distances. For the right‐branch point $$\bigl(3\sqrt{2},-\tfrac{3\sqrt{2}}{2}\bigr)$$ one has $$\bigl|3(3\sqrt{2}) + 2\bigl(-\tfrac{3\sqrt{2}}{2}\bigr) - 1\bigr| = \bigl|9\sqrt{2} - 3\sqrt{2} - 1\bigr| = \bigl|6\sqrt{2} - 1\bigr|\approx 7.49\,. $$ For the left‐branch point $$\bigl(-3\sqrt{2},\tfrac{3\sqrt{2}}{2}\bigr)$$ one gets $$\bigl|3(-3\sqrt{2}) + 2\bigl(\tfrac{3\sqrt{2}}{2}\bigr) - 1\bigr| = \bigl|-9\sqrt{2} + 3\sqrt{2} - 1\bigr| = \bigl|-6\sqrt{2} - 1\bigr|\approx 9.49\,. $$ Thus the right branch point is closer to the line.

Finally, at that nearest point we compute $$\sqrt{2}(y_0 - x_0) = \sqrt{2}\Bigl(-\frac{3\sqrt{2}}{2} - 3\sqrt{2}\Bigr) = \sqrt{2}\,\Bigl(-\tfrac{9\sqrt{2}}{2}\Bigr) = -\frac{9\times 2}{2} = -9\,. $$ The correct answer is Option C: $$-9\,. $$

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: $$5$$.

We need to find the area of triangle PQR. To determine point P on the parabola $$y^2 = 3x$$, note that the tangent at point $$(at^2, 2at)$$ on $$y^2 = 4ax$$ has slope $$\frac{1}{t}$$. Here $$4a = 3$$, so $$a = \frac{3}{4}$$. The line $$x + 2y = 1$$ has slope $$-\frac{1}{2}$$. Thus $$\frac{1}{t} = -\frac{1}{2}$$, giving $$t = -2$$. It follows that $$P = \left(\frac{3}{4} \times 4, 2 \times \frac{3}{4} \times (-2)\right) = (3, -3)$$.

Next, we find points Q and R on the ellipse $$\frac{x^2}{4} + y^2 = 1$$. Lines perpendicular to $$x - y = 2$$ (slope 1) have slope $$-1$$. The general tangent to the ellipse is given by $$y = mx \pm \sqrt{a^2m^2 + b^2}$$ where $$a^2 = 4, b^2 = 1$$. Hence the tangents with slope -1 are $$y = -x \pm \sqrt{4 + 1} = -x \pm \sqrt{5}$$.

For the tangent $$y = -x + \sqrt{5}$$, substituting into the ellipse equation yields

$$\frac{x^2}{4} + (-x + \sqrt{5})^2 = 1$$

$$\frac{x^2}{4} + x^2 - 2\sqrt{5}x + 5 = 1$$

$$\frac{5x^2}{4} - 2\sqrt{5}x + 4 = 0$$

$$5x^2 - 8\sqrt{5}x + 16 = 0$$

$$x = \frac{8\sqrt{5} \pm \sqrt{320 - 320}}{10} = \frac{8\sqrt{5}}{10} = \frac{4\sqrt{5}}{5}$$.

Then

$$y = -\frac{4\sqrt{5}}{5} + \sqrt{5} = \frac{\sqrt{5}}{5} = \frac{1}{\sqrt{5}}$$, so $$Q = \left(\frac{4\sqrt{5}}{5}, \frac{\sqrt{5}}{5}\right) = \left(\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)$$. Similarly, for the tangent $$y = -x - \sqrt{5}$$ one finds $$R = \left(-\frac{4}{\sqrt{5}}, -\frac{1}{\sqrt{5}}\right)$$.

Finally, the area of triangle PQR with vertices $$P(3,-3)$$, $$Q\left(\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)$$, and $$R\left(-\frac{4}{\sqrt{5}}, -\frac{1}{\sqrt{5}}\right)$$ is given by

Area $$= \frac{1}{2}|x_P(y_Q - y_R) + x_Q(y_R - y_P) + x_R(y_P - y_Q)|$$

$$= \frac{1}{2}\left|3 \cdot \frac{2}{\sqrt{5}} + \frac{4}{\sqrt{5}}\left(-\frac{1}{\sqrt{5}} - (-3)\right) + \left(-\frac{4}{\sqrt{5}}\right)\left(-3 - \frac{1}{\sqrt{5}}\right)\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{4}{\sqrt{5}} \cdot \frac{-1 + 3\sqrt{5}}{\sqrt{5}} - \frac{4}{\sqrt{5}} \cdot \frac{-3\sqrt{5} - 1}{\sqrt{5}}\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{4(-1+3\sqrt{5})}{5} + \frac{4(3\sqrt{5}+1)}{5}\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{-4+12\sqrt{5}+12\sqrt{5}+4}{5}\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{24\sqrt{5}}{5}\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{24}{\sqrt{5}}\right| = \frac{1}{2} \cdot \frac{30}{\sqrt{5}} = \frac{15}{\sqrt{5}} = 3\sqrt{5}$$.

Therefore, the area is $$\boxed{3\sqrt{5}}$$, which corresponds to Option D.

The two sides of the triangle are $$x = 0$$ (y-axis) and $$y = 3$$, and the third side is a tangent to the parabola $$y^2 = 6x$$.

Parametric tangent to $$y^2 = 6x$$.

For parabola $$y^2 = 4ax$$ where $$4a = 6$$, so $$a = \frac{3}{2}$$.

The tangent at parameter $$t$$ is:

Or equivalently: $$x = ty - \frac{3}{2}t^2$$

Find the vertices of the triangle.

Vertex A (intersection of $$x = 0$$ and $$y = 3$$): $$A = (0, 3)$$

Vertex B (intersection of tangent with $$x = 0$$):

So $$B = \left(0, \frac{3t}{2}\right)$$

Vertex C (intersection of tangent with $$y = 3$$):

So $$C = \left(3t - \frac{3}{2}t^2, 3\right)$$

Find the circumcenter.

The triangle has a right angle at $$A = (0, 3)$$ since the two sides $$x = 0$$ and $$y = 3$$ are perpendicular.

For a right triangle, the circumcenter is the midpoint of the hypotenuse (the side opposite the right angle, which is the tangent line segment $$BC$$).

Circumcenter $$= \left(\frac{0 + 3t - \frac{3}{2}t^2}{2}, \frac{\frac{3t}{2} + 3}{2}\right)$$

Eliminate parameter $$t$$.

From $$k = \frac{3(t+2)}{4}$$:

Substituting into the expression for $$h$$:

Let me compute $$t(2 - t)$$:

$$2 - t = 2 - \frac{4k-6}{3} = \frac{6 - 4k + 6}{3} = \frac{12 - 4k}{3}$$

Expanding the numerator:

Replacing $$(h, k)$$ with $$(x, y)$$:

This matches Option 3: $$4y^2 - 18y + 3x + 18 = 0$$.

The answer is $$\boxed{4y^2 - 18y + 3x + 18 = 0}$$.

The ellipse $$E_k: kx^2 + k^2y^2 = 1$$ can be written as:

$$\frac{x^2}{1/k} + \frac{y^2}{1/k^2} = 1$$

So semi-major axis $$a = \frac{1}{\sqrt{k}}$$ and semi-minor axis $$b = \frac{1}{k}$$.

The four chords join the endpoints of the major axis $$(±a, 0)$$ and the minor axis $$(0, ±b)$$, forming a rhombus.

The radius of the inscribed circle of this rhombus is:

$$r_k = \frac{ab}{\sqrt{a^2 + b^2}}$$

$$r_k = \frac{\frac{1}{\sqrt{k}} \cdot \frac{1}{k}}{\sqrt{\frac{1}{k} + \frac{1}{k^2}}} = \frac{\frac{1}{k^{3/2}}}{\sqrt{\frac{k+1}{k^2}}} = \frac{\frac{1}{k^{3/2}}}{\frac{\sqrt{k+1}}{k}} = \frac{1}{\sqrt{k}\sqrt{k+1}} = \frac{1}{\sqrt{k(k+1)}}$$

Therefore:

$$\frac{1}{r_k^2} = k(k+1)$$

$$\sum_{k=1}^{20} \frac{1}{r_k^2} = \sum_{k=1}^{20} k(k+1) = \sum_{k=1}^{20} (k^2 + k)$$

$$= \frac{20 \times 21 \times 41}{6} + \frac{20 \times 21}{2} = 2870 + 210 = 3080$$

Ellipse: $$x^2 + 9y^2 = 9$$, i.e., $$\frac{x^2}{9} + y^2 = 1$$.

Semi-major axis $$a = 3$$ (along x), semi-minor axis $$b = 1$$ (along y).

A = positive x-axis intersection = (3, 0). B = positive y-axis = (0, 1).

Circle C has major axis (length 6) as diameter: centre (0,0), radius 3. So $$C: x^2 + y^2 = 9$$.

Line through A(3,0) and B(0,1): $$\frac{x}{3} + y = 1$$ or $$x + 3y = 3$$.

Find intersection of $$x + 3y = 3$$ with $$x^2 + y^2 = 9$$:

$$x = 3 - 3y$$. Substituting: $$(3-3y)^2 + y^2 = 9$$

$$9 - 18y + 9y^2 + y^2 = 9$$

$$10y^2 - 18y = 0$$

$$y(10y - 18) = 0$$

$$y = 0$$ (point A) or $$y = 9/5$$

At $$y = 9/5$$: $$x = 3 - 27/5 = -12/5$$. So $$P = (-12/5, 9/5)$$.

Area of triangle OAP with O(0,0), A(3,0), P(-12/5, 9/5):

$$= \frac{1}{2}|x_A \cdot y_P - x_P \cdot y_A| = \frac{1}{2}|3 \cdot 9/5 - (-12/5) \cdot 0| = \frac{1}{2} \cdot \frac{27}{5} = \frac{27}{10}$$

$$m = 27, n = 10$$. Check coprime: gcd(27,10) = 1 ✓

$$m - n = 27 - 10 = 17$$

The correct answer is Option 3: 17.

1. Simplify the Ellipse Equation

The given equation is $$15x^2 + 19y^2 = 285$$. Divide the entire equation by $$285$$ to bring it into standard form:

$$\frac{x^2}{19} + \frac{y^2}{15} = 1$$

• Semi-major axis squared ($$a^2$$): $$19$$
• Semi-minor axis squared ($$b^2$$): $$15$$
• The minor axis of this ellipse lies along the y-axis.

2. Identify the Circle Equation

The circle is concentric with the ellipse (centered at $$(0,0)$$) and has a radius of $$4$$:

$$x^2 + y^2 = 16 \implies r^2 = 16$$

3. Set the Condition for Common Tangents

A line $$y = mx + c$$ is tangent to:

• The ellipse if $$c^2 = a^2m^2 + b^2$$
• The circle if $$c^2 = r^2(1 + m^2)$$

Equating the two expressions for $$c^2$$:

$$19m^2 + 15 = 16(1 + m^2)$$

$$19m^2 + 15 = 16 + 16m^2$$

$$3m^2 = 1 \implies m = \pm\frac{1}{\sqrt{3}}$$

4. Determine the Angle of Inclination

The slope $$m$$ represents the tangent of the angle $$\theta$$ that the line makes with the x-axis:

$$\tan \theta = \frac{1}{\sqrt{3}} \implies \theta = \frac{\pi}{6}$$

The question specifically asks for the angle of inclination to the minor axis (the y-axis):

$$\text{Angle with y-axis} = \frac{\pi}{2} - \theta = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$$

Correct Option: (A) $$\frac{\pi}{3}$$

·  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:

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

The ellipse $$9x^2 + 4y^2 = 36$$ can be written as $$\frac{x^2}{4} + \frac{y^2}{9} = 1$$ so that $$a^2 = 9,\ b^2 = 4$$ with semi-major axis $$a = 3$$ along the y-axis and semi-minor axis $$b = 2$$ along the x-axis. In this setting, PQ and RS denote two diameters passing through the origin that are perpendicular to each other.

First, one may parametrize the points on the ellipse by considering a line through the origin that makes an angle $$\theta$$ with the x-axis and has the equation $$y = x\tan\theta$$. Substituting this into the ellipse equation leads to $$\frac{x^2}{4} + \frac{x^2\tan^2\theta}{9} = 1$$.

Then if $$r$$ represents the distance from the origin to the intersection point, one has $$x = r\cos\theta$$ and $$y = r\sin\theta$$, which transforms the preceding relation into $$\frac{1}{r^2} = \frac{\cos^2\theta}{4} + \frac{\sin^2\theta}{9}$$.

Next, since PQ and RS are diameters, their lengths are given by $$PQ = 2r_1$$ and $$RS = 2r_2$$, where $$r_1$$ and $$r_2$$ are the corresponding radial distances for the directions $$\theta$$ and $$\theta + 90^\circ$$. Therefore, one finds

$$\frac{1}{(PQ)^2} + \frac{1}{(RS)^2} = \frac{1}{4r_1^2} + \frac{1}{4r_2^2} = \frac{1}{4}\Bigl(\frac{1}{r_1^2} + \frac{1}{r_2^2}\Bigr)\,.$$

Moreover, in the direction $$\theta$$ the relationship $$\frac{1}{r_1^2} = \frac{\cos^2\theta}{4} + \frac{\sin^2\theta}{9}$$ holds, while for the perpendicular direction $$\theta + 90^\circ$$ it becomes $$\frac{1}{r_2^2} = \frac{\sin^2\theta}{4} + \frac{\cos^2\theta}{9}$$. By adding these two expressions one obtains

$$\frac{1}{r_1^2} + \frac{1}{r_2^2} = \frac{\cos^2\theta + \sin^2\theta}{4} + \frac{\sin^2\theta + \cos^2\theta}{9} = \frac{1}{4} + \frac{1}{9} = \frac{13}{36}\,,$$

which is independent of $$\theta$$. Consequently,

$$\frac{1}{(PQ)^2} + \frac{1}{(RS)^2} = \frac{1}{4}\cdot\frac{13}{36} = \frac{13}{144}\,.$$

Since this result is of the form $$\frac{p}{q}$$ with $$p = 13$$ and $$q = 144$$ coprime, their sum is $$p + q = 13 + 144 = 157$$, giving the final answer as \boxed{157}.

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)