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} $

For the ellipse $E$, let its equation be

$ \frac{x^2}{A^2}+\frac{y^2}{B^2}=1, \quad A>B $

since its major axis is along the $x$-axis because the directrices are of the form $x=\text{constant}$.

Given eccentricity,

$ e=\frac{\sqrt{3}}{2} $

For an ellipse with major axis along the $x$-axis, the directrices are

$ x=\pm \frac{A}{e} $

We are given

$ x=\pm \frac{4\sqrt{6}}{3} $

So,

$ \frac{A}{e}=\frac{4\sqrt{6}}{3} $

Substituting $e=\frac{\sqrt{3}}{2}$,

$ A=\frac{\sqrt{3}}{2}\cdot \frac{4\sqrt{6}}{3} $

$ A=\frac{4\sqrt{18}}{6} =\frac{4\cdot 3\sqrt{2}}{6} =2\sqrt{2} $

Thus the semi-major axis of ellipse $E$ is

$ A=2\sqrt{2} $

Now for the ellipse,

$ B^2=A^2(1-e^2) $

Since

$ A^2=(2\sqrt{2})^2=8 $

and

$ e^2=\frac{3}{4} $

we get

$ B^2=8\left(1-\frac{3}{4}\right)=8\cdot \frac{1}{4}=2 $

Hence

$ B=\sqrt{2} $

So the length of minor axis of $E$ is

$ 2B=2\sqrt{2} $

Now consider the hyperbola

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

Its eccentricity is equal to the length of semi-major axis of $E$, so

$ e_H=2\sqrt{2} $

For a hyperbola,

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

Thus,

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

$ 8=1+\frac{b^2}{a^2} $

$ \frac{b^2}{a^2}=7 \quad \Rightarrow \quad b^2=7a^2 $

Also, the length of latus rectum of the hyperbola is equal to the length of minor axis of $E$.

For the hyperbola, length of latus rectum is

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

So,

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

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

Using $b^2=7a^2$,

$ \frac{7a^2}{a}=\sqrt{2} $

$ 7a=\sqrt{2} $

$ a=\frac{\sqrt{2}}{7} $

Now for the hyperbola,

$ c=ae_H $

So,

$ c=\frac{\sqrt{2}}{7}\cdot 2\sqrt{2} =\frac{4}{7} $

Distance between the foci $=2c$,

$ 2c=\frac{8}{7} $

Hence, the correct answer is

$ \boxed{\frac{8}{7}} $

So, Option D is correct.

For the quadratic equation

$ x^2-ax+2=0 $

let its two distinct roots be $e_1$ and $e_2$.

We are given that these roots are to be interpreted as eccentricities.

Recall:

• For a hyperbola, eccentricity $e>1$

• For an ellipse, eccentricity satisfies $0

Also, since $e_1,e_2$ are roots of the quadratic, by Vieta's formulas,

$ e_1+e_2=a,\qquad e_1e_2=2 $

and since the roots are distinct real roots,

$ a^2-8>0 \quad \Rightarrow \quad |a|>2\sqrt2 $

Now we analyse the two sets.

1. Both $e_1$ and $e_2$ are eccentricities of hyperbolas

This means both roots must satisfy

$ e_1>1,\qquad e_2>1 $

Since

$ e_1e_2=2>0 $

the two roots have the same sign.

If both are hyperbola eccentricities, both must be positive and greater than $1$.

Now, for both roots of

$ x^2-ax+2=0 $

to be greater than $1$, we use the NCERT-style condition:

For both roots to be greater than $1$, we need

1. roots real and distinct:

   $ a^2-8>0 $

2. sum positive:

   $ a>0 $

3. value of polynomial at $x=1$ positive, because $1$ must lie to the left of both roots:

   $ f(1)=1-a+2=3-a>0 \text{ or } <0? $

Let us be careful.

Since the parabola opens upward, if both roots are greater than $1$, then $x=1$ lies outside the roots on the left side, so

$ f(1)>0 $

Hence,

$ 3-a>0 \Rightarrow a<3 $

Also, for both roots to be greater than $1$, the axis of symmetry should be to the right of $1$:

$ \frac a2>1 \Rightarrow a>2 $

Combining:

$ 2

But also we need distinct real roots, and in this interval automatically $a>2\sqrt2$ is needed. Hence

$ 2\sqrt2

So,

$ (\alpha,\beta)=(2\sqrt2,\,3) $

Thus,

$ \alpha=2\sqrt2,\qquad \beta=3 $

2. $e_1$ is eccentricity of an ellipse and $e_2$ is eccentricity of a hyperbola, respectively

This means one root lies in $(0,1)$ and the other lies in $(1,\infty)$.

Since

$ e_1e_2=2>0 $

both roots are positive. So the only possibility is:

• one root in $(0,1)$

• the other root greater than $1$

Now let the roots be distinct and positive. Since their product is $2$, if one root is less than $1$, the other must automatically be greater than $2$, hence greater than $1$.

So we only need one root to be less than $1$.

For that, $x=1$ must lie between the two roots. Since the parabola opens upward, this happens when

$ f(1)<0 $

That is,

$ 1-a+2<0 $

$ 3-a<0 $

$ a>3 $

Also for real distinct roots, $a^2-8>0$, which is already true for $a>3$.

Hence the set is

$ (\gamma,\infty)=(3,\infty) $

So,

$ \gamma=3 $

3. Find $\alpha^2+\beta^2+\gamma^2$

Now,

$ \alpha^2=(2\sqrt2)^2=8 $

$ \beta^2=3^2=9 $

$ \gamma^2=3^2=9 $

Therefore,

$ \alpha^2+\beta^2+\gamma^2=8+9+9=26 $

So the correct answer is

$ \boxed{26} $

Hence, Option C.

For the hyperbola

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

we are given that it passes through the point $(6,4\sqrt{3})$.

So, substituting $x=6$ and $y=4\sqrt{3}$:

$ \frac{36}{a^2}-\frac{(4\sqrt{3})^2}{b^2}=1 $

Since

$ (4\sqrt{3})^2=48 $

we get

$ \frac{36}{a^2}-\frac{48}{b^2}=1 \qquad ...(1) $

Now for the hyperbola

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

its eccentricity is

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

So,

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

Given:

$ 15(e^2+1)=34e $

Substitute $e^2=1+\frac{b^2}{a^2}$ if needed, but first solve this equation in $e$:

$ 15e^2-34e+15=0 $

Factorising:

$ 15e^2-25e-9e+15=0 $

$ 5e(3e-5)-3(3e-5)=0 $

$ (3e-5)(5e-3)=0 $

So,

$ e=\frac{5}{3} \quad \text{or} \quad e=\frac{3}{5} $

But for a hyperbola, $e>1$, hence

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

Now,

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

So,

$ \frac{25}{9}=1+\frac{b^2}{a^2} $

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

Thus,

$ b^2=\frac{16}{9}a^2 $

Substitute this in equation $(1)$:

$ \frac{36}{a^2}-\frac{48}{\frac{16}{9}a^2}=1 $

Now,

$ \frac{48}{\frac{16}{9}a^2}=48\cdot \frac{9}{16a^2}=\frac{27}{a^2} $

So,

$ \frac{36}{a^2}-\frac{27}{a^2}=1 $

$ \frac{9}{a^2}=1 $

Hence,

$ a^2=9 $

Then,

$ b^2=\frac{16}{9}\cdot 9=16 $

Now consider the second hyperbola:

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

Substitute $a^2=9$ and $b^2=16$:

$ \frac{x^2}{16}-\frac{y^2}{2(9+1)}=1 $

$ \frac{x^2}{16}-\frac{y^2}{20}=1 $

For a hyperbola of the form

$ \frac{x^2}{A^2}-\frac{y^2}{B^2}=1 $

the length of the latus rectum is

$ \frac{2B^2}{A} $

Here,

$ A^2=16 \Rightarrow A=4, \qquad B^2=20 $

Therefore, length of latus rectum

$ =\frac{2\cdot 20}{4}=10 $

Hence, the required answer is

$ \boxed{10} $

So, the correct option is A.

For a hyperbola, the eccentricity $e>1$.

We are given:

$ 6e^2-11e+3=0 $

Solve this quadratic:

$ 6e^2-9e-2e+3=0 $

$ 3e(2e-3)-1(2e-3)=0 $

$ (2e-3)(3e-1)=0 $

So,

$ e=\frac{3}{2}\quad \text{or}\quad e=\frac{1}{3} $

Since for a hyperbola $e>1$, we take

$ e=\frac{3}{2} $

Now the foci are $(3,5)$ and $(3,-4)$.

These lie on the vertical line $x=3$, so the transverse axis is vertical.

The distance between the foci is

$ |5-(-4)|=9 $

Hence,

$ 2c=9 \Rightarrow c=\frac{9}{2} $

For a hyperbola,

$ e=\frac{c}{a} $

So,

$ a=\frac{c}{e} =\frac{\frac{9}{2}}{\frac{3}{2}} =3 $

Now use the relation

$ c^2=a^2+b^2 $

Thus,

$ b^2=c^2-a^2 =\left(\frac{9}{2}\right)^2-3^2 =\frac{81}{4}-9 =\frac{81-36}{4} =\frac{45}{4} $

Length of latus rectum of a hyperbola is

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

Therefore,

$ \text{Length of latus rectum} =\frac{2\cdot \frac{45}{4}}{3} =\frac{45}{6} =\frac{15}{2} $

Hence, the correct answer is

$ \boxed{\frac{15}{2}} $

So, Option C is correct.

The given equation of the hyperbola is:

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

Let the uniform eccentricity of the hyperbola be $e$.

According to the problem, the distance between the foci is $6$. For a standard hyperbola, the coordinates of the foci are $(\pm ae, 0)$, so the distance between them is given by $2ae$.

$ 2ae = 6 $

$ ae = 3 \quad \dots (i) $

The distance between the directrices is given as $\frac{8}{3}$. The equations of the directrices are $x = \pm \frac{a}{e}$, so the distance between them is $2\frac{a}{e}$.

$ 2\frac{a}{e} = \frac{8}{3} $

$ \frac{a}{e} = \frac{4}{3} \quad \dots (ii) $

To find the value of $a^2$, we can multiply equations (i) and (ii):

$ (ae) \cdot \left(\frac{a}{e}\right) = 3 \times \frac{4}{3} $

$ a^2 = 4 $

We know the fundamental relationship between $a, b,$ and $e$ for a hyperbola is:

$ b^2 = a^2(e^2 - 1) $

$ b^2 = a^2e^2 - a^2 $

Using the value $ae = 3$, we have $a^2e^2 = 9$. Substituting $a^2e^2$ and $a^2$ into the equation gives:

$ b^2 = 9 - 4 = 5 $

Now, substituting $a^2$ and $b^2$ back into the standard equation, the hyperbola is:

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

The line $x = \alpha$ intersects the hyperbola at points A and B. Substitute $x = \alpha$ into the equation of the hyperbola to find the y-coordinates of points A and B:

$ \frac{\alpha^2}{4} - \frac{y^2}{5} = 1 $

$ \frac{y^2}{5} = \frac{\alpha^2}{4} - 1 $

$ \frac{y^2}{5} = \frac{\alpha^2 - 4}{4} $

$ y^2 = \frac{5(\alpha^2 - 4)}{4} $

$ y = \pm \frac{\sqrt{5(\alpha^2 - 4)}}{2} $

So, the coordinates of points A and B are $\left(\alpha, \frac{\sqrt{5(\alpha^2 - 4)}}{2}\right)$ and $\left(\alpha, -\frac{\sqrt{5(\alpha^2 - 4)}}{2}\right)$.

These points lie vertically above and below each other, forming the base of $\triangle AOB$ along the line $x = \alpha$.

Length of the base $AB$ = Distance between A and B = $2 \times \frac{\sqrt{5(\alpha^2 - 4)}}{2} = \sqrt{5(\alpha^2 - 4)}$.

The perpendicular height of $\triangle AOB$ from the origin $O(0,0)$ to the line $x = \alpha$ is simply $|\alpha|$.

We are given that the area of $\triangle AOB$ is $4\sqrt{15}$. Using the area of a triangle formula:

$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $

$ 4\sqrt{15} = \frac{1}{2} \cdot \sqrt{5(\alpha^2 - 4)} \cdot |\alpha| $

Squaring both sides to remove the square root and the absolute value:

$ (4\sqrt{15})^2 = \frac{1}{4} \cdot (\alpha^2) \cdot 5(\alpha^2 - 4) $

$ 240 = \frac{5}{4} \alpha^2 (\alpha^2 - 4) $

$ 240 \times \frac{4}{5} = \alpha^2(\alpha^2 - 4) $

$ 192 = \alpha^4 - 4\alpha^2 $

$ \alpha^4 - 4\alpha^2 - 192 = 0 $

Let $\alpha^2 = t$. This transforms our equation into a quadratic equation:

$ t^2 - 4t - 192 = 0 $

Splitting the middle term:

$ t^2 - 16t + 12t - 192 = 0 $

$ t(t - 16) + 12(t - 16) = 0 $

$ (t - 16)(t + 12) = 0 $

So, $t = 16$ or $t = -12$.

Since $t = \alpha^2$ and the square of a real number cannot be negative, we disregard $-12$.

Thus, we have:

$ \alpha^2 = 16 $

Option B is the correct answer.

Let

$ P(x_1,y_1), \qquad Q(x_2,y_2) $

be two points on the rectangular hyperbola

$ xy=12 $

So, both points satisfy

$ x_1y_1=12 \quad \text{and} \quad x_2y_2=12 $

Also, the midpoint of $PQ$ is given as

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

Using the midpoint formula,

$ \frac{x_1+x_2}{2}=\frac{1}{2}, \qquad \frac{y_1+y_2}{2}=-\frac{1}{2} $

Hence,

$ x_1+x_2=1, \qquad y_1+y_2=-1 $

Now we need the area of triangle $OPQ$, where $O=(0,0)$.

The area of triangle formed by points $O(0,0)$, $P(x_1,y_1)$, $Q(x_2,y_2)$ is

$ \text{Area}=\frac{1}{2}\left|x_1y_2-x_2y_1\right| $

So our aim is to find $\left|x_1y_2-x_2y_1\right|$.

Since $x_1$ and $x_2$ are the $x$-coordinates of points on the hyperbola, and their sum is $1$, let us use

$ y_1=\frac{12}{x_1}, \qquad y_2=\frac{12}{x_2} $

Now using

$ y_1+y_2=-1 $

we get

$ \frac{12}{x_1}+\frac{12}{x_2}=-1 $

$ 12\left(\frac{x_1+x_2}{x_1x_2}\right)=-1 $

Since $x_1+x_2=1$,

$ 12\left(\frac{1}{x_1x_2}\right)=-1 $

So,

$ x_1x_2=-12 $

Thus $x_1$ and $x_2$ are roots of the quadratic equation

$ t^2-(x_1+x_2)t+x_1x_2=0 $

$ t^2-t-12=0 $

$ (t-4)(t+3)=0 $

Hence,

$ x_1=4,\quad x_2=-3 $

or vice versa.

Then corresponding $y$-coordinates are

$ y_1=\frac{12}{4}=3, \qquad y_2=\frac{12}{-3}=-4 $

So the points are

$ P(4,3), \qquad Q(-3,-4) $

Now area of triangle $OPQ$ is

$ \text{Area}=\frac{1}{2}\left|4(-4)-(-3)(3)\right| $

$ =\frac{1}{2}\left|-16+9\right| $

$ =\frac{1}{2}\cdot 7 $

$ =\frac{7}{2} $

Therefore, the required area is

$ \boxed{\frac{7}{2}} $

So, the correct option is:

$ \boxed{\text{Option C}} $

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)