Hyperbola

The equation of the ellipse is :

$ E: \frac{x^2}{144}+\frac{y^2}{169}=1 $

Comparing this with the standard form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we have :

• $a^2=144 \Rightarrow a=12$

• $b^2=169 \Rightarrow b=13$

Since $b>a$, the ellipse is vertically oriented. The distance from the center to the foci (c) is given by :

$ \begin{aligned} & c^2=b^2-a^2 \\ & \quad c^2=169-144=25 \Longrightarrow c=5 \end{aligned} $

The foci of the ellipse are at $(0, \pm 5)$.

The equation of the hyperbola is :

$H: \frac{x^2}{16}-\frac{y^2}{\lambda^2}=-1 \quad$ which is equivalent to $\quad \frac{y^2}{\lambda^2}-\frac{x^2}{16}=1$

This is a vertical hyperbola.

Let $B^2=\lambda^2$ and $A^2=16$.

Since it shares the same foci as the ellipse, its focal distance $c$ is also 5 . For a hyperbola:

$ \begin{gathered} c^2=A^2+B^2 \\ 25=16+\lambda^2 \Longrightarrow \lambda^2=9 \Longrightarrow \lambda=3 \end{gathered} $

Calculate Eccentricity ( $e$ ) and Latus Rectum ( $L$ ) of $H$

• Eccentricity (e):

$ e=\frac{c}{B}=\frac{5}{3} $

Length of Latus Rectum ( $L$ ): For a vertical hyperbola $\frac{y^2}{B^2}-\frac{x^2}{A^2}=1$, the length of the latus rectum is:

$ L=\frac{2 A^2}{B}=\frac{2 \times 16}{3}=\frac{32}{3} $

We need to find $24(e+L)$ :

$ \begin{aligned} & e+L=\frac{5}{3}+\frac{32}{3}=\frac{37}{3} \\ & \therefore 24(e+L)=24 \times \frac{37}{3}=8 \times 37=296 \end{aligned} $

PQ is a chord perpendicular to x -axis of hyperbola $\frac{x^2}{4}-\frac{y^2}{b^2}=1$

$\begin{aligned} & \text { Eccentricity, } e=\sqrt{3} \\ & e=\sqrt{1+\frac{b^2}{4}}=\sqrt{3} \\ & 1+\frac{b^2}{4}=3\end{aligned}$

$ \frac{b^2}{4}=2 \Longrightarrow b^2=8 $

equation become $\frac{x^2}{4}-\frac{y^2}{8}=1$ - (1)

let coordinate of point $P\left(x_1, y_1\right)$

PQ is perpendicular to x -axis so $Q\left(x_1,-y_1\right)$

It is given that OPQ is equilateral triangle where O is origin

so, $\angle P O Q=60^{\circ} \Longrightarrow \angle P O X=30^{\circ}$

$\tan 30^{\circ}=$ slope of $\mathrm{OP}=\frac{y_1}{x_1}$

$ \frac{y_1}{x_1}=\frac{1}{\sqrt{3}} \Longrightarrow y_1=\frac{x_1}{\sqrt{3}} $

Point $P\left(x_1, y_1\right)$ satisfy hyperbola.

$ \begin{aligned} & \frac{x_1^2}{4}-\frac{y_1^2}{8}=1 \\ & \frac{x_1^2}{4}-\frac{x_1^2}{24}=1\left\{y_1=\frac{x_1}{\sqrt{3}}\right\} \\ & 6 x_1^2-x_1^2=24 \\ & x_1^2=\frac{24}{5} \end{aligned} $

$\begin{aligned} & \text { Now, area of } \triangle O P Q=\frac{1}{2} \times \text { base } P Q \times \text { height }\left(x_1\right) \\ & =\frac{1}{2} \times 2 y_1 \times x_1=\frac{x_1^2}{\sqrt{3}} \\ & =\frac{24}{5 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{8 \sqrt{3}}{5}\end{aligned}$

$ \begin{aligned} & \log _5\left(\log _7\left(9 x-x^2-13\right)\right)>0 \\ & \Rightarrow \quad 9 x-x^2-13>7 \\ & \Rightarrow \quad x \in(4,5)=(m, n) \\ & \Rightarrow \quad \text { eccentricity }=\frac{5}{3} \\ & L(L R)=\frac{32}{3} \\ & \frac{25}{9}=1+\frac{b^2}{a^2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)\end{aligned} $

$ \begin{aligned} &\frac{2 b^2}{a}=\frac{32}{3} \Rightarrow b^2=\frac{16 a}{3}\\ &\text { Stultify in (1) }\\ &\begin{aligned} & \frac{16}{9}=\frac{16 a}{3 a^2} \Rightarrow a=3 \\ & b^2=16 \\ & \therefore b^2-a^2=16-9 \\ & =7 \end{aligned} \end{aligned} $

$ \begin{gathered} P(10,2 \sqrt{15}) \text { lies on } \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ \therefore \frac{100}{a^2}-\frac{60}{b^2}=1 \ldots \ldots \ldots \ldots . .(1) \\ \because \text { Length of latus rectum }=8 \\ \frac{2 \cdot b^2}{a}=8 \Rightarrow \frac{b^2}{a}=4 \ldots \ldots \ldots .(2) \end{gathered} $

From (1) \& (2)

$ \begin{aligned} & \frac{100}{a^2}-\frac{60}{4 a}=1 \\ & 400-60 a=4 a^2 \\ & 4 a^2+60 a-400=0 \\ & a^2+15 a-100=0 \\ & a=5 \&-20(\text { rejected }) \\ & \Rightarrow b=\sqrt{20} \end{aligned} $

$ \begin{aligned} & \therefore \text { Hyperbola is } \frac{x^2}{25}-\frac{y^2}{20}=1 \\ & \therefore \text { Focal length } S S^1=2 a e=2 \cdot 5 \cdot\left(\sqrt{1+\frac{4}{5}}\right)=6 \sqrt{5} \\ & \therefore \text { Area of } \triangle P S S^1=\frac{1}{2} \cdot 6 \sqrt{5} \cdot 2 \sqrt{15}=30 \sqrt{3}=A \\ & \therefore A^2=2700 \end{aligned} $

$ \begin{aligned} & \alpha x+2 y=1, \alpha \in R \\ & x^2-9 y^2=9 \\ & y=\frac{1-\alpha x}{2} \\ & 4 x^2-9\left(1+\alpha^2 x^2-2 \alpha x\right)=36 \end{aligned} $

$ \begin{aligned} & \Delta<0 \\ & 1620 \alpha^2-720>0 \\ & \alpha^2>0.55 \\ & \alpha=0.8 \end{aligned} $

$ \begin{aligned} & E: \frac{x^2}{A^2}+\frac{y^2}{B^2}=1 \\ & H: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \end{aligned} $

For con-focal $E$ and $H$

$ \because \quad A^2-B^2=a^2+b^2 $

(Given that $A^2=36$ and $B^2=16$ )

$ \begin{aligned} & 36-16=a^2+b^2 \\ & a^2+b^2=20 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & e_H=5 \\ & \Rightarrow \quad 25=1+\frac{b^2}{a^2} \\ & \Rightarrow \quad 25 a^2=20 \Rightarrow a^2=\frac{4}{5} \Rightarrow a=\frac{2}{\sqrt{5}} \end{aligned} $

Also $a^2+b^2=20$

$ \begin{aligned} & b^2=20-\frac{4}{5} \Rightarrow \frac{96}{5} \\ & \mathrm{~L}(\mathrm{LR})==2 \times \frac{b^2}{a}=\frac{2 \times \frac{96}{5}}{\frac{2}{\sqrt{5}}}=\frac{96}{\sqrt{5}} \end{aligned} $

Let's find the eccentricities of the given ellipse and hyperbola, and then determine the eccentricity of an ellipse that passes through all four foci.

Step 1: Find $ e_1 $ for the Ellipse

The equation of the ellipse is:

$ \frac{x^2}{b^2} + \frac{y^2}{25} = 1 $

The eccentricity $ e_1 $ is given by:

$ e_1^2 = 1 - \frac{b^2}{25} $

Step 2: Find $ e_2 $ for the Hyperbola

The equation of the hyperbola is:

$ \frac{x^2}{16} - \frac{y^2}{b^2} = 1 $

The eccentricity $ e_2 $ is given by:

$ e_2^2 = 1 + \frac{b^2}{16} $

Step 3: Using the Product $ e_1 e_2 = 1 $

Given:

$ e_1 e_2 = 1 $

Thus:

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

Expanding gives:

$ 1 + \frac{b^2}{16} - \frac{b^2}{25} - \frac{b^4}{400} = 1 $

Simplifying:

$ \frac{9b^2}{400} = \frac{b^4}{400} $

Thus:

$ b^2 = 9 $

Step 4: Determine Eccentricities $ e_1 $ and $ e_2 $

Substitute $ b^2 = 9 $:

For the ellipse:

$ e_1^2 = 1 - \frac{9}{25} = \frac{16}{25} $

$ e_1 = \frac{4}{5} $

For the hyperbola:

$ e_2 = \frac{5}{4} $

Step 5: Find the Eccentricity of the New Ellipse

The new ellipse's equation is:

$ \frac{x^2}{25} + \frac{y^2}{16} = 1 $

The eccentricity $ e $ is:

$ e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} $

Thus, the eccentricity of the ellipse that passes through all four foci is $ \frac{3}{5} $.

$\begin{aligned} &\sqrt{(c+4)^2+9}+\sqrt{(c-4)^2+9}=8 \sqrt{\frac{5}{3}}\\ &\text { Solving, } c=\frac{5}{\sqrt{6}}=\mathrm{al} \Rightarrow a^2\left(1+\frac{b^2}{a^2}\right)=\frac{25}{6} \Rightarrow a^2+b^2=\frac{25}{6}\\ &\begin{aligned} & \frac{16}{a^2}-\frac{9}{b^2}=1 \\ & 16 b^2-9 d^2=a^2 b^2 \Rightarrow 16\left(\frac{25}{6}-a^2\right)-9 a^2=9 a^2 b^2 \\ & P F_1+P F_2=8 \sqrt{\frac{5}{3}} \Rightarrow a^2=\frac{5}{2}, b^2=\frac{5}{3} \\ & \left|P F_1-P F_2\right|=2 a \\ & \frac{64.5}{3}=4 a^2+4 m \Rightarrow m=\frac{80}{3}-a^2 \\ & 6 m=160-6 a^2 \end{aligned} \end{aligned}$

$\begin{aligned} & 9 \ell^2=9\left(\frac{2 b^2}{a}\right)^2=\frac{36 b^4}{a^2} \\ & 9 \ell^2+6 m=\frac{36\left(\frac{25}{9}\right)}{\frac{5}{2}}+160-6\left(\frac{5}{2}\right) \\ & =\frac{72 \times 5}{9}+160-15 \\ & =160+40-15=185 \end{aligned}$

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

Directrix : $x=\frac{9}{\sqrt{10}}=\frac{a}{e}\quad\text{..... (i)}$

Focus : $(\sqrt{10}, 0) \equiv(a e, 0)$

$a e=\sqrt{10}\quad\text{..... (ii)}$

(i) $\times$ (ii)

$\Rightarrow a^2=9 \Rightarrow a=3$

Substitute in (ii)

$e=\frac{\sqrt{10}}{3}$

$\begin{aligned} & \text { Now } e^2=1+\frac{b^2}{a^2} \\ & \frac{10}{9}=1+\frac{b^2}{a} \\ & \Rightarrow b=1 \\ & I=\frac{2 b^2}{a}=\frac{2 \times 1}{3}=\frac{2}{3} \\ & a\left[e^2+l\right]=9\left[\frac{10}{9}+\frac{2}{3}\right]=10+6 \\ & =16 \end{aligned}$

$\begin{aligned} & \mathrm{be}=13, \mathrm{~b}=5 \\ & \mathrm{a}^2=\mathrm{b}^2\left(\mathrm{e}^2-1\right) \\ & =\mathrm{b}^2 \mathrm{e}^2-\mathrm{b}^2 \\ & =169-25=144 \\ & \ell(\mathrm{LR})=\frac{2 \mathrm{a}^2}{\mathrm{~b}}=\frac{2 \times 144}{5}=\frac{288}{5}\end{aligned}$

$E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$

$\begin{aligned} \text { Eccentricity of ellipse, } e_E & =\sqrt{1-\frac{b^2}{a^2}} \\ & =\sqrt{1-\frac{75}{100}} \\ & e_E=\frac{1}{2} \end{aligned}$

$\therefore e_H=2$ [ as eccentricity of hyperbola is reciprocal of eccentricity of ellipse]

Transverse axis of hyperbola $=\alpha$

Conjugate axis of hyperbola $=\beta$

Also, foci of ellipse $(1 \pm a e, 1)$

$\begin{aligned} & =\left(1 \pm\left(10 \times \frac{1}{2}\right), 1\right) \\ & =(1 \pm 5,1) \\ & =(6,1) \text { and }(-4,1) \end{aligned}$

Distance between foci $=10$

$\begin{aligned} & 2 a e=10 \\ & \Rightarrow a=\frac{5}{2} \end{aligned}$

$\begin{aligned} & \text { also, } e^2=1+\frac{b^2}{a^2} \\ & \begin{aligned} 4 & =1+\frac{4 b^2}{25} \\ b^2 & =\frac{75}{4} \\ b & =\frac{\sqrt{75}}{2} \end{aligned} \\ & \Rightarrow \quad \alpha=5 \\ & \text { and } \beta=\sqrt{75} \\ & 3 \alpha^2+2 \beta^2=3(5)^2+2(75)=225 \end{aligned}$

$\begin{aligned} & H: \frac{x^2}{a^2}-\frac{y^2}{b^2}=-1 \\ & e=\sqrt{1+\frac{a^2}{b^2}}=\sqrt{3} \\ & \Rightarrow 1+\frac{a^2}{b^2}=3 \\ & \Rightarrow \frac{a^2}{b^2}=2 \quad \text{.... (1)}\\ & \frac{2 a^2}{b}=4 \sqrt{3} \end{aligned}$

Using equation (1)

$\begin{aligned} & \frac{4 b^2}{b}=4 \sqrt{3} \\ & \Rightarrow b=\sqrt{3} \\ & a=\sqrt{6} \\ & H: \frac{x^2}{6}-\frac{y^2}{3}=-1 \\ & \frac{\alpha^2}{6}-12=-1 \\ & \frac{\alpha^2}{6}=11 \\ & \begin{array}{c} \alpha^2=66 \\ \text { Focus }:(0, b c)(0,-b c) \\ (0,3),(0,-3) \end{array} \\ & \beta=\sqrt{\alpha^2+9} \times \sqrt{\alpha^2+81} \\ & \beta=105 \\ & \alpha^2+\beta=66+105 \\ & =171 \end{aligned}$

$\begin{aligned} & C_1: x^2+y^2=a^2 \Rightarrow \text { area }=\pi a^2=36 \pi \Rightarrow a=6 \\ & C_2: x^2+(y-a e)^2=(a e-a)^2 \\ & \therefore \quad \pi(a e-a)^2=4 \pi \\ & \Rightarrow 36(e-1)^2=4 \\ & \Rightarrow e-1=\frac{1}{3} \Rightarrow e=\frac{4}{3} \\ & \Rightarrow b^2=28 \\ & \therefore \quad L R=\frac{2 b^2}{a}=\frac{2 \times 28}{6}=\frac{28}{3} \text { units } \end{aligned}$

To find the value of $\theta$, we need to determine the relationship between the eccentricities of the given hyperbola and ellipse. Let's start by writing down the standard forms of ellipse and hyperbola and then relate them to the given equations.

The standard form of an ellipse is:

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

The eccentricity $e$ of an ellipse is given by:

$e = \sqrt{1 - \frac{b^2}{a^2}}\ \text{for}\ a > b$

For the given ellipse:

$x^2 \operatorname{cosec}^2 \theta + y^2 = 5$

We can compare it with the standard form by writing it as:

$\frac{x^2}{\frac{5}{\operatorname{cosec}^2 \theta}} + \frac{y^2}{5} = 1$

From this we have $a^2 = \frac{5}{\operatorname{cosec}^2 \theta}$ and $b^2 = 5$. The eccentricity of the ellipse (let's call it $e_1$) is:

$e_1 = \sqrt{1 - \frac{5}{a^2}} = \sqrt{1 - \frac{5}{\frac{5}{\operatorname{cosec}^2 \theta}}} = \sqrt{1 - \operatorname{sin}^2 \theta} = \operatorname{cos} \theta$

The standard form of a hyperbola is:

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

The eccentricity $e$ of a hyperbola is given by:

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

For the given hyperbola:

$x^2 - y^2 \operatorname{cosec}^2 \theta = 5$

We can rewrite it as:

$\frac{x^2}{5} - \frac{y^2}{\frac{5}{\operatorname{cosec}^2 \theta}} = 1$

From this we have $a^2 = 5$ and $b^2 = \frac{5}{\operatorname{cosec}^2 \theta}$. The eccentricity of the hyperbola (let's call it $e_2$) is:

$e_2 = \sqrt{1 + \frac{\frac{5}{\operatorname{cosec}^2 \theta}}{5}} = \sqrt{1 + \operatorname{sin}^2 \theta}$

According to the question the eccentricity of the hyperbola is $\sqrt{7}$ times the eccentricity of the ellipse, so we can write:

$e_2 = \sqrt{7} \cdot e_1$

Substitute $e_1$ and $e_2$ from the above:

$\sqrt{1 + \operatorname{sin}^2 \theta} = \sqrt{7} \cdot \operatorname{cos} \theta$

Square both sides to eliminate the square root:

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

$1 + \operatorname{sin}^2 \theta = 7 \cdot (1 - \operatorname{sin}^2 \theta)$

$1 + \operatorname{sin}^2 \theta = 7 - 7 \cdot \operatorname{sin}^2 \theta$

$8 \cdot \operatorname{sin}^2 \theta = 6$

$\operatorname{sin}^2 \theta = \frac{6}{8} = \frac{3}{4}$

$\operatorname{sin} \theta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$

Looking at the options provided and knowing the sine values for common angles, we can see that $\operatorname{sin} \theta = \frac{\sqrt{3}}{2}$ corresponds to $\theta = \frac{\pi}{3}$.

The correct answer is:

Option C.

$\frac{\pi}{3}$

$\begin{aligned} & \frac{\mathrm{x}^2}{9}+\frac{\mathrm{y}^2}{25}=1 \\ & \mathrm{a}=3, \mathrm{~b}=5 \\ & \mathrm{e}=\sqrt{1-\frac{9}{25}}=\frac{4}{5} \therefore \text { foci }=(0, \pm \mathrm{be})=(0, \pm 4) \\ & \quad \therefore \mathrm{e}_{\mathrm{H}}=\frac{4}{5} \times \frac{15}{8}=\frac{3}{2} \end{aligned}$

Let equation hyperbola

$\begin{aligned} & \frac{\mathrm{x}^2}{\mathrm{~A}^2}-\frac{\mathrm{y}^2}{\mathrm{~B}^2}=-1 \\ & \therefore \mathrm{B} \cdot \mathrm{e}_{\mathrm{H}}=4 \quad \therefore \mathrm{B}=\frac{8}{3} \\ & \therefore \mathrm{A}^2=\mathrm{B}^2\left(\mathrm{e}_{\mathrm{H}}^2-1\right)=\frac{64}{9}\left(\frac{9}{4}-1\right) \therefore \mathrm{A}^2=\frac{80}{9} \\ & \therefore \frac{\mathrm{x}^2}{\frac{80}{9}}-\frac{\mathrm{y}^2}{\frac{64}{9}}=-1 \\ & \text { Directrix }: \mathrm{y}= \pm \frac{\mathrm{B}}{\mathrm{e}_{\mathrm{H}}}= \pm \frac{16}{9} \\ & \mathrm{PS}=\mathrm{e} \cdot \mathrm{PM}=\frac{3}{2}\left|\frac{14}{3} \cdot \sqrt{\frac{2}{5}}-\frac{16}{9}\right| \\ & =7 \sqrt{\frac{2}{5}}-\frac{8}{3} \end{aligned}$

$\begin{aligned} & \frac{x^2}{9}-\frac{y^2}{4}=1 \\ & a^2=9, b^2=4 \\ & b^2=a^2\left(e^2-1\right) \Rightarrow e^2=1+\frac{b^2}{a^2} \\ & e^2=1+\frac{4}{9}=\frac{13}{9} \\ & e=\frac{\sqrt{13}}{3} \Rightarrow s_1 s_2=2 \mathrm{ae}=2 \times 3 \times \sqrt{\frac{13}{3}}=2 \sqrt{13} \end{aligned}$

$\begin{aligned} & \text { Area of } \triangle \mathrm{PS}_1 \mathrm{~S}_2=\frac{1}{2} \times \beta \times \mathrm{S}_1 \mathrm{~S}_2=2 \sqrt{13} \\ & \Rightarrow \frac{1}{2} \times \beta \times(2 \sqrt{13})=2 \sqrt{13} \Rightarrow \beta=2 \\ & \begin{aligned} & \frac{\alpha^2}{9}-\frac{\beta^2}{4}=1 \Rightarrow \frac{\alpha^2}{9}-1=1 \Rightarrow \alpha^2=18 \Rightarrow \alpha=3 \sqrt{2} \\ & \text { Distance of P from origin }=\sqrt{\alpha^2+\beta^2} \\ &=\sqrt{18+4}=\sqrt{22} \end{aligned} \end{aligned}$

$\begin{aligned} & H: \frac{x^2}{16}-\frac{y^2}{9}=1 \qquad e_1=\frac{5}{4} \\ & \therefore e_1 e_2=1 \Rightarrow e_2=\frac{4}{5} \end{aligned}$

Also, ellipse is passing through $( \pm 5,0)$

$\begin{aligned} & \therefore a=5 \text { and } b=3 \\ & E: \frac{x^2}{25}+\frac{y^2}{9}=1 \end{aligned}$

End point of chord are $\left( \pm \frac{5 \sqrt{5}}{3}, 2\right)$

$\therefore \mathrm{L}_{\mathrm{PQ}}=\frac{10 \sqrt{5}}{3}$

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

Focus of parabola : $(ae,\,0)$

Directrix : $x = - ae$.

Equation of parabola $ \equiv {y^2} = 4aex$

Length of latus rectum of parabola $ = 4ae$

Length of latus rectum of hyperbola $ = {{2.{b^2}} \over a}$

as given, $4ae = {{2{b^2}} \over a}\,.\,e$

$2 = {{{b^2}} \over {{a^2}}}$ ...... (i)

$\because$ H passes through $\left( {2\sqrt 2 , - 2\sqrt 2 } \right) \Rightarrow {8 \over {{a^2}}} - {8 \over {{b^2}}} = 1$ ........ (ii)

From (i) and (ii) ${a^2} = 4$ and ${b^2} = 8 \Rightarrow e = \sqrt 3 $

$\Rightarrow$ Equation of parabola is ${y^2} = 8\sqrt 3 x$.

Given hyperbola : ${{{x^2}} \over {6/k}} - {{{y^2}} \over 6} = 1$

Eccentricity $ = e = \sqrt {1 + {6 \over {6/k}}} = \sqrt {1 + k} $

Directrices : $x = \, \pm \,{a \over e} \Rightarrow x = \, \pm \,{{\sqrt 6 } \over {\sqrt k \sqrt {k + 1} }}$

As given : ${{\sqrt 6 } \over {\sqrt k \sqrt {k + 1} }} = 1$

$ \Rightarrow k = 2$

Here hyperbola is ${{{x^2}} \over 3} - {{{y^2}} \over 6} = 1$

Checking the option gives $\left( {\sqrt 5 , - 2} \right)$ satisfies it.

Any tangent to ${y^2} = 24x$ at ($\alpha$, $\beta$)

$\beta y = 12(x + \alpha )$

Slope $ = {{12} \over \beta }$ and perpendicular to $2x + 2y = 5$

$ \Rightarrow {{12} \over \beta } = 1 \Rightarrow \beta = 12,\,\alpha = 6$

Hence hyperbola is ${{{x^2}} \over {36}} - {{{y^2}} \over {144}} = 1$ and normal is drawn at (10, 16)

Equation of normal ${{36\,.\,x} \over {10}} + {{144\,.\,y} \over {16}} = 36 + 144$

$ \Rightarrow {x \over {50}} + {y \over {20}} = 1$

This does not pass though (15, 13) out of given option.

Ellipse : ${{{x^2}} \over {16}} + {{{y^2}} \over 7} = 1$

Eccentricity $ = \sqrt {1 - {7 \over {16}}} = {3 \over 4}$

Foci $ \equiv ( \pm \,a\,e,0) \equiv ( \pm \,3,0)$

Hyperbola : ${{{x^2}} \over {\left( {{{144} \over {25}}} \right)}} - {{{y^2}} \over {\left( {{\alpha \over {25}}} \right)}} = 1$

Eccentricity $ = \sqrt {1 + {\alpha \over {144}}} = {1 \over {12}}\sqrt {144 + \alpha } $

Foci $ \equiv ( \pm \,a\,e,0) \equiv \left( { \pm \,{{12} \over 5}\,.\,{1 \over {12}}\sqrt {144 + \alpha } ,\,0} \right)$

If foci coincide then $3 = {1 \over 5}\sqrt {144 + \alpha } \Rightarrow \alpha = 81$

Hence, hyperbola is ${{{x^2}} \over {{{\left( {{{12} \over 5}} \right)}^2}}} - {{{y^2}} \over {{{\left( {{9 \over 5}} \right)}^2}}} = 1$

Length of latus rectum $ = 2\,.\,{{{{81} \over {25}}} \over {{{12} \over 5}}} = {{27} \over {10}}$

$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$, then

${e^2} = {{11} \over {14}}l$ (l be the length of LR)

$ \Rightarrow {a^2} + {b^2} = {{11} \over 7}{b^2}a$ ..... (i)

and $e{'^2} = {{11} \over 8}l'$ (l' be the length of LR of conjugate hyperbola)

$ \Rightarrow {a^2} + {b^2} = {{11} \over 4}{a^2}b$ ....... (ii)

By (i) and (ii)

$7a = 4b$

then by (i)

${{16} \over {49}}{b^2} + {b^2} = {{11} \over 7}{b^2}\,.\,{{4b} \over 7}$

$ \Rightarrow 44b = 65$ and $77a = 65$

$\therefore$ $77a + 44b = 130$

$1 + {{{b^2}} \over {{a^2}}} = {5 \over 2} \Rightarrow {{{b^2}} \over {{a^2}}} = {3 \over 2}$

${{2{b^2}} \over a} = 6\sqrt 2 \Rightarrow 2.\,{3 \over 2}.\,a = 6\sqrt 2 $

$ \Rightarrow a = 2\sqrt 2 ,\,{b^2} = 12$

${c^2} = {a^2}{m^2} - {b^2} = 8.4 - 12 = 20$

Given hyperbola : ${{{x^2}} \over {{a^2}}} - {{{y^2}} \over 9} = 1$

$\because$ It passes through $(8,3\sqrt 3 )$

$\because$ ${{64} \over {{a^2}}} - {{27} \over 9} = 1 \Rightarrow {a^2} = 16$

Now, equation of normal to hyperbola

${{16x} \over 8} + {{9y} \over {3\sqrt 3 }} = 16 + 9$

$ \Rightarrow 2x + \sqrt 3 y = 25$ ...... (i)

$\left( { - 1,9\sqrt 3 } \right)$ satisfies (i)

Given,

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

$ \Rightarrow {{{x^2}} \over 4} - {{{y^2}} \over 2} = 1 \Rightarrow {{{x^2}} \over {{{(2)}^2}}} - {{{y^2}} \over {{{(\sqrt 2 )}^2}}} = 1$

Here, a = 2, $b = \sqrt 2 $

$\therefore$ $e = \sqrt {1 + {{{b^2}} \over {{a^2}}}} = \sqrt {1 + {2 \over 4}} = \sqrt {1 + {1 \over 2}} = \sqrt {{3 \over 2}} $

So, Focus (F) = ($\pm$ a e, 0) = ($\pm$ $\sqrt 6 $, 0)

Now, equation of tangent at $P(4,\sqrt 6 )$ is

$x{x_1} - 2y{y_1} = 4$

$ \Rightarrow x\,.\,4 - 2y\,.\,\sqrt 6 = 4$

$ \Rightarrow 4x - 2\sqrt 6 y = 4$

$ \Rightarrow 2x - \sqrt 6 y = 2$ ....... (i)

Putting y = 0 in Eq. (i), we get x-intercept of tangent i.e. x = 1

$\therefore$ Q $\equiv$ (1, 0)

Hence, equation of corresponding latus rectum is $x = \sqrt 6 $

$\therefore$ $R \equiv \left( {\sqrt 6 ,{{2(\sqrt 6 - 1)} \over {\sqrt 6 }}} \right)$ [putting $x = \sqrt 6 $ in Eq. (i), we get $y = {{2(\sqrt 6 - 1)} \over {\sqrt 6 }}$]

$\therefore$ Area of $\Delta QFR = {1 \over 2} \times (QF) \times (RF)$

$ = {1 \over 2}(\sqrt 6 - 1) \times {{2(\sqrt 6 - 1)} \over {\sqrt 6 }} = {{{{(\sqrt 6 - 1)}^2}} \over {\sqrt 6 }} = \left( {{7 \over {\sqrt 6 }} - 2} \right)$