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} $

$ r=\frac{|3(1)+4(2)-1|}{\sqrt{3^2+4^2}}=\frac{10}{5}=2 $

$\therefore $ Equation of circle is $(x-1)^2+(y-2)^2=4$

Passes through $(3,2) \Rightarrow \mathrm{P} \rightarrow 3$

(Q) Tangent to parabola: $y=m x+\frac{2}{m}$

Radius of circle, $r=\sqrt{2}$

Distance from $(0,0)$ to $m x-y+\frac{2}{m}=0$ must be $\sqrt{2}$

$ \begin{aligned} & \therefore \frac{\frac{2}{m}}{\sqrt{m^2+1}}=\sqrt{2} \Rightarrow \frac{4}{m^2}=2\left(m^2+1\right) \\ & \Rightarrow m^4+m^2-2=0 \\ & \Rightarrow m^2=1 \end{aligned} $

For positive slope, $m=1$

$\therefore $ Equation is $y=x+2$ passes through $(7,9) \Rightarrow Q \rightarrow 2$

$ \frac{x^2}{16}+\frac{y^2}{12}=1, a^2=16, b^2=12, e=\sqrt{1-\frac{12}{16}}=\frac{1}{2} $

Latus rectum point $M\left(1^{\text {st }}\right.$ quadrant $): x=a e=4 \times \frac{1}{2}=2$

$ \begin{aligned} & y=\frac{b^2}{a}=\frac{12}{4}=3 \\ & \therefore M(2,3) \end{aligned} $

Equation of normal at $(2,3)$ is

$ \frac{a^2 x}{x_1}-\frac{b^2 y}{y_1}=a^2 e^2 \Rightarrow 2 x-y=1 $

which passes through $(1,1)$

$ \begin{aligned} & a e=5, \frac{a}{e}=\frac{16}{5} \\ & a^2=16 \Rightarrow a=4 \\ & \therefore e=\frac{5}{4} \\ & b^2=a^2\left(e^2-1\right)=9 \\ & \therefore \frac{x^2}{16}-\frac{y^2}{9}=1 \text { passes through }(8,3 \sqrt{3}) \Rightarrow S \rightarrow 5 \\ & (P) \rightarrow(3),(Q) \rightarrow(2),(R) \rightarrow(1),(S) \rightarrow(5) \end{aligned} $

Option (B) is correct

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)

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

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

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

Let the slope of common tangent be $m$.

Equation of tangent of slope $m$ to

$ \frac{x^2}{16}-\frac{y^2}{9}=1 \text { is } y=m x \pm \sqrt{16 m^2-9} $

Equation of tangent of slope $m$ to

$ \begin{aligned} & x^2+y^2=16 \text { is } y=m x \pm 4 \sqrt{1+m^2} \\ & \Rightarrow 16 m^2-9=16+16 m^2 \\ & \Rightarrow 0 m^2+0 m-25=0 \end{aligned} $

⇒ Both roots are at infinity

⇒ Slope of common tangent is undefined which is parallel to $Y$-axis. So, number of common tangent is 2 .

$ \begin{aligned} &\text { Let } P \equiv(x, y)\\ &\begin{aligned} & \text { Area of } \triangle P A B=\left|\left|\begin{array}{lll} x & y & 1 \\ 0 & 1 & 1 \\ 1 & 2 & 1 \end{array}\right| \times \frac{1}{2}\right| \\ & =|-x+y-1| \times \frac{1}{2} \\ & \text { Area of } \triangle P A C=1\left|\begin{array}{ccc} x & y & 1 \\ 0 & 1 & 1 \\ -2 & 1 & 1 \end{array}\right| 1 \times \frac{1}{2} \\ & =\frac{1}{2}|-2 y+2| \\ & \Rightarrow \quad|y-x-1|=|2-2 y| \\ & \Rightarrow \quad y^2+x^2+1-2 x y-2 y+2 x \\ & =4+4 y^2-8 y \\ & \Rightarrow \quad x^2-2 x y-3 y^2+2 x+6 y-3=0 \end{aligned} \end{aligned} $

$ \begin{aligned} & \quad \frac{x^2}{9}-\frac{y^2}{b^2}=1 \\ & \quad e^2=1+\frac{b^2}{9} \\ & \text { Slope of } A L_1=\frac{\frac{b^2}{3}}{a e+3} \\ & \text { Slope of } A L_2=\frac{-b^2}{3(a e+3)} \\ & \Rightarrow \quad \frac{b^2}{3(a e+3)} \times \frac{-b^2}{3(a e+3)}=-1 \\ & \Rightarrow \quad b^4=9(a e+3)^2 \\ & \Rightarrow \quad b^2=3(a e+3) \\ & \Rightarrow \quad e^2=1+\frac{b^2}{9}=1+\frac{b^2}{a^2} \\ & \Rightarrow \quad a^2 e^2=b^2+9 \\ & \Rightarrow \quad b^2=3\left(3+\sqrt{b^2+9}\right) \\ & \Rightarrow \quad b^2=9+3 \sqrt{b^2+9} \\ & \Rightarrow \quad\left(b^2-9\right)^2=9\left(b^2+9\right) \\ & \Rightarrow \quad b^4+81-18 b^2=9 b^2+81 \\ & \Rightarrow \quad b^4-27 b^2=0 \\ & \Rightarrow \quad b^2\left(b^2-27\right)=0 \\ & \therefore \quad b^2=27 \end{aligned} $

$x^2+y^2=4$

$ x y=\sqrt{3} $

Point of intersections

$ \begin{aligned} & x^2+\frac{3}{x^2}=4 \\ \Rightarrow \quad & x^4-4 x^2+3=0 \\ \Rightarrow \quad & \left(x^2-3\right)\left(x^2-1\right)=0 \\ \Rightarrow \quad & x= \pm \sqrt{3}, x= \pm 1 \\ & y= \pm 1, y= \pm \sqrt{3} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \because P(\alpha, \beta) \quad \alpha>\beta>0 \\ & \therefore P(\sqrt{3}, 1) \end{aligned}\\ &\text { ∴ Equation of tangent at } P(\sqrt{3}, 1) \text { to }\\ &\begin{aligned} & x y=\sqrt{3} \\ & \quad \frac{x(1)+y(\sqrt{3})}{2}=\sqrt{3} \\ & \Rightarrow \quad x+\sqrt{3} y=2 \sqrt{3} \end{aligned} \end{aligned} $

Tangent at $P(3 \sqrt{2}, 4)$ on hyperbola

$ \begin{array}{ll} & \frac{x^2}{9}-\frac{y^2}{16}=1 \text { is } T=0 \\ & \frac{x(3 \sqrt{2})}{9}-\frac{y(4)}{16}=1 \\ \Rightarrow & \frac{\sqrt{2} x}{3}-\frac{y}{4}=1 \\ \Rightarrow & 4 \sqrt{2} x-3 y-12=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \because & a=3, b=4 \\ \Rightarrow & e=\sqrt{1+\frac{16}{9}}=\frac{5}{3} \end{array} $

Equation of directrix $\Rightarrow x= \pm \frac{a}{e}$.

$ \begin{aligned} & \quad x= \pm \frac{9}{5} \Rightarrow \quad y=\frac{4 \sqrt{2} x-12}{3} \\ & \text { at } \quad x=\frac{9}{5} \\ & \therefore \quad y=\frac{12 \sqrt{2}-20}{5}=\beta \end{aligned} $

Given, $l$ is the maximum value of $-3 x^2+4 x+1$ and $m$ is the minimum value of $3 x^2+4 x+1$

For $-3 x^2+4 x+1, \frac{-b}{2 a}=\frac{-4}{2(-3)}=\frac{2}{3}$

$ \begin{aligned} \therefore \text { Maximum value } & = & -3\left(\frac{2}{3}\right)^2+4\left(\frac{2}{3}\right)+1 \\ & & =-3 \times \frac{4}{9}+\frac{8}{3}+1 \\ \Rightarrow & & l=\frac{-4}{3}+\frac{8}{3}+1=\frac{7}{3} \end{aligned} $

For $3 x^2+4 x+1, \frac{-b}{2 a}=\frac{-4}{2 \times 3}=\frac{-2}{3}$

$\therefore$ Minimum value

$ \begin{aligned} & =3\left(\frac{-2}{3}\right)^2+4\left(\frac{-2}{3}\right)+1 \\ & =3 \times \frac{4}{9}-\frac{8}{3}+1 \\ \Rightarrow \quad m & =\frac{4}{3}-\frac{8}{3}+1=\frac{-1}{3} \end{aligned} $

Now, foci of hyperbola $=(l, 0)$

$ =\left(\frac{7}{3}, 0\right) $

And $(7 m, 0)=\left(\frac{-7}{3}, 0\right)$

$ \begin{aligned} & \therefore \text { Distance between foci }= \\ & \Rightarrow 2 a \times 2=2 \times \frac{7}{3} \Rightarrow a=\frac{7}{6} \\ & \text { Also, } b^2=a^2\left(e^2-1\right) \\ & \Rightarrow \quad b^2=\frac{49}{36}(4-1) \\ & \Rightarrow \quad b^2=\frac{49}{36} \times 3=\frac{49}{12} \end{aligned} $

$ \begin{aligned} & \therefore \text { Equation of hyperbola, } \frac{x^2}{\frac{49}{36}}-\frac{y^2}{\frac{49}{12}}=1 \\ & \Rightarrow \quad 36 x^2-12 y^2=49 \end{aligned} $

Given, equation of curve

$ \frac{x^2}{12-\alpha}+\frac{y^2}{\alpha-10}=1 $

for circle, $12-\alpha=\alpha-10$

$ \Rightarrow \quad 22=2 \alpha \Rightarrow \alpha=11 $

$ \begin{aligned}and\,\, & 12-\alpha>0 \Rightarrow \alpha<12 \\ & \alpha-10>0 \Rightarrow \alpha>10 \end{aligned} $

$\Rightarrow$ Curve represents a circle for some value of $\alpha$ in (10,12).

Option (c) is correct.

Similarly, for ellipse $12-\alpha \neq \alpha-10$

$ \Rightarrow \quad 22 \neq 2 \alpha \Rightarrow \alpha \neq 11 $

Option (a) is wrong. for hyperbola one denominator must be positive and other negative

$ \therefore \quad 12-\alpha>0 \text { and } \alpha-10<0 $

$\Rightarrow \quad \alpha<12$ and $\alpha<10$

$\Rightarrow \quad$ Means $\alpha<10$

or $\quad 12-\alpha<0$ and $\alpha-10>0$

$\Rightarrow \quad \alpha>12$ and $\alpha>10$

$\Rightarrow \quad$ Means $\alpha>12$

Options (b) and (d) is wrong.

Let equation of hyperbola

$ \frac{X^2}{a^2}-\frac{Y^2}{b^2}=1 $

Given, $2 a=2(2 b)$

$ \begin{aligned} \Rightarrow & & 2 a & =4 b \\ \Rightarrow & & a & =2 b \end{aligned} $

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

$ \begin{aligned} & =1+\frac{b^2}{4 b^2} \\ & =1+\frac{1}{4}=\frac{5}{4} \end{aligned} $

Also, $2 a y=3\left(\frac{2 a}{y}\right)$

(for another hyperbola)

$ \begin{aligned} \Rightarrow & & y^2 & =3 \\ \therefore & & y^2-x^2 & =3-\frac{5}{4}=\frac{12-5}{4}=\frac{7}{4} \end{aligned} $

Since, the product of perpendicular distance from any point on the hyperbola to its asymptotes $=\frac{36}{13}$

So, $\quad \frac{a^2 b^2}{a^2+b^2}=\frac{36}{13}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

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

$\Rightarrow \frac{\sqrt{13}}{3}=\sqrt{1+\frac{b^2}{a^2}}$

$\Rightarrow \quad \frac{13}{9}=1+\frac{b^2}{a^2}$

(Squaring on both sides)

$ \begin{aligned} & \Rightarrow \quad \frac{b^2}{a^2}=\frac{13}{9}-1=\frac{4}{9} \\ & \Rightarrow \quad b^2=\frac{4}{9} a^2 \end{aligned} $

$ \Rightarrow \quad b=\frac{2}{3} a \quad(\because a, b>0) $

$ \begin{aligned} &\text { Substitute this into Eq. (i), we get }\\ &\begin{aligned} & \frac{a^2 \cdot\left(\frac{2}{3} a\right)^2}{a^2+\frac{4}{9} a^2}=\frac{36}{13} \\ \Rightarrow & \frac{\frac{4}{9} a^4}{\frac{13}{9} a^2}=\frac{36}{13} \Rightarrow \frac{4}{13} a^2=\frac{36}{13} \\ \Rightarrow & 4 a^2=36 \\ \Rightarrow & a^2=9 \end{aligned} \end{aligned} $

$ \Rightarrow \quad a=3 \quad(\because a>0) $

And $b=\frac{2}{3} a=\frac{2}{3} \times 3=2$

Now, $a-b=3-2=1$

The latus rectum of a hyperbola is a line segment that is perpendicular to the main axis (transverse axis) and passes through a focus.

The angle $\theta$ is the angle at the center of the hyperbola made by one latus rectum.

For a hyperbola, the angle $\theta$ subtended at the center by the latus rectum always follows this rule:

$ \sin \theta = \frac{1}{2} \tan \left(\frac{\theta}{2}\right) $

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

So, $a=2, b=\sqrt{5}$ and $e=\sqrt{1+\frac{5}{4}}=3 / 2$

The latus rectum for the focus in the positive $x$-direction is at $x=a c=3$

$ \therefore \quad \frac{3^2}{4}-\frac{y^2}{5}=1 \Rightarrow y= \pm 5 / 2 $

in first quadrant, the extremity is ( $3,5 / 2$ ) and the equation of tangental ( $3,5 / 2$ ) is

$ \begin{aligned} & \frac{x \cdot 3}{4}-\frac{y \cdot \frac{5}{2}}{5}=1 \Rightarrow \frac{3 x}{4}-\frac{y}{2}=1 \\ \Rightarrow & 3 x-2 y=4 \end{aligned} $

the tangent meet $X$-axis at $y=0$

$ \begin{aligned} & \Rightarrow 3 x=4 \Rightarrow x=4 / 3 \\ & \therefore(O A)^2=\left(\frac{4}{3}-0\right)^2+(0-0)^2=16 / 9 \\ & \text { and when } x=0 \Rightarrow-2 y=4 \Rightarrow y=-3 \end{aligned} $

So, $O B^2=(0-0)^2+(0+2)^2=4$

Hence, $(O A)^2-(O B)^2=\frac{16}{9}-4=-20/9$

$ \begin{aligned} & \text { } H: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \text { passes through }(4,6) \\ & \therefore \quad \frac{16}{a^2}-\frac{36}{b^2}=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { and } \quad e=2 \end{aligned} $

$ \sqrt{1+\frac{b^2}{a^2}}=2 \Rightarrow \frac{b^2}{a^2}=3 \Rightarrow b^2=3 a^2 $

From Eq. (i), we get

$ \frac{16}{a^2}-\frac{36}{3 a^2}=1 \Rightarrow \frac{4}{a^2}=1 \Rightarrow a^2=4 $

and   $ b^2=12 $

$ \therefore \quad H: \frac{x^2}{4}-\frac{y^2}{12}=1 $

Equation of tangent at (4,6)

$ \begin{aligned} T & =0 \\ \Rightarrow \frac{x(4)}{4}-\frac{y(6)}{12} & =1 \Rightarrow 2 x-y=2 \\ \Rightarrow 2 x-y-2 & =0 \end{aligned} $

For Hyperbola

We have, focii $=( \pm 2,0)$

$ \begin{aligned} & \therefore \quad a e=2 \Rightarrow a^2 e^2=4 \\ & \Rightarrow a^2+b^2=4 \Rightarrow b^2=4-a^2 \end{aligned} $

Let equation of hyperbola

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

∵ It passes through $(\sqrt{2}, \sqrt{3})$

$ \therefore \quad \frac{2}{a^2}-\frac{3}{b^2}=1 $

$ \begin{aligned} & \Rightarrow \quad 2 b^2-3 a^2=a^2 b^2 \\ & \Rightarrow \quad 8-2 a^2-3 a^2=a^2\left(4-a^2\right) \\ & \Rightarrow \quad 8-5 a^2=4 a^2-a^4 \\ & \Rightarrow \quad 8-5 a^2=4 a^2-a^4 \\ & \Rightarrow \quad a^4-9 a^2+8=0 \\ & \Rightarrow \quad\left(a^2-8\right)\left(a^2-1\right)=0 \\ & \Rightarrow \quad a^2=8 \text { and } a^2=1 \\ & \Rightarrow \quad b^2=4-8 \quad \Rightarrow b^2=4-1 \\ & \quad=-4 \quad \Rightarrow b^2=3 \end{aligned} $

Equation of hyperbola

$ \frac{x^2}{1}-\frac{y^2}{3}=1 $

$P(2 \sqrt{2}, 3 \sqrt{3})$ satisfies equation of hyperbola.

$P(a \sec \theta, b \tan \theta)$ and

$Q(a \sec \phi, b \tan \phi)$

$ \because \theta+\phi=\pi / 2 $

Equation of normal at $(a \sec \theta, b \tan \theta)$ to hyperbola

$ \begin{aligned} & \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \text { is } \\ & a x \cos \theta+b y \cot \theta=a^2+b^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & a x \cos \phi+b y \cot \phi=a^2+b^2 \\ & \Rightarrow \phi=\frac{\pi}{2}-\theta \\ & \therefore a x \sin \theta+b y \tan \theta=a^2+b^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Multiply Eq. (i) by $\sin \theta$ and Eq. (ii) $\cos \theta$ and subtract

$ \begin{aligned} & \text { by } \cot \theta \cdot \sin \theta-\text { by } \tan \theta \cos \theta \\ & =\left(a^2+b^2\right)(\sin \theta-\cos \theta) \\ & \text { by }(\cos \theta-\sin \theta)=-\left(a^2+b^2\right) \\ & \Rightarrow \quad y=-\frac{a^2+b^2}{b} \end{aligned} $

$ \begin{aligned} & \because \quad 2 \theta=2 \tan ^{-1} 2 / 3 \\ & \Rightarrow \quad \theta=\tan ^{-1} 2 / 3 \\ & \Rightarrow \tan \theta=2 / 3 \end{aligned} $

$ \begin{array}{ll} \because & e=\sec \theta \\ \Rightarrow & e^2=\sec ^2 \theta=1+\frac{4}{9}=\frac{13}{9} \\ \therefore & e^2=\frac{13}{9} \\ \Rightarrow & 1+\frac{b^2}{a^2}=\frac{13}{9} \Rightarrow \frac{b^2}{a^2}=\frac{4}{9} \\ \Rightarrow & \frac{b}{a}=\frac{2}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \Rightarrow & a^2-b^2=45 \Rightarrow a^2-\frac{4 a^2}{9}=45 \\ \Rightarrow & 5 a^2=9 \times 45 \\ \Rightarrow & a^2=81 \\ \Rightarrow & a=9 \Rightarrow b=\frac{2}{3} \times 9=6 \\ \therefore & a b=9 \times 6=54 \end{array} $

Given, equation of hyperbola is

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

Now, the equation of the tangent to be hyperbola is

$ 3 \sqrt{2} x-4 y=12 $

$ \Rightarrow \quad y=\frac{3 \sqrt{2}}{4} x-3 $

So, slope of the tangent, $m_t=\frac{3 \sqrt{2}}{4}$ and $y$-intercept is $c=-3$

Now, the line $y=m x+c$ tangent to the hyperbola is given by

$ \begin{array}{rlrl} c^2 & =a^2 m^2-b^2 \\ \Rightarrow \quad(-3)^2 & =a^2\left(\frac{3 \sqrt{2}}{4}\right)^2-b^2 \\ \Rightarrow \quad 9 & =\frac{9}{8} a^2-b^2 \\ \Rightarrow \quad 9 a^2-8 b^2 & =72 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Given, eccentricity, $e=\frac{5}{4}$

$ \text {} \begin{aligned} So, \,\,\, b^2 & =a^2\left(e^2-1\right) \\ & =a^2\left(\left(\frac{5}{4}\right)^2-1\right) \\ & =a^2\left(\frac{25}{16}-1\right)=\frac{9}{16} a^2 \\ \Rightarrow \quad 16 b^2 & =9 a^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

From Eqs. (i) and (ii), we get

$ \begin{aligned} b^2 & =9 \text { and } a^2=16 \\ \therefore a^2-b^2 & =16-9=7 \end{aligned} $

Given hyperbola equation is $x y=16$

Differentiate it w.r.t $x$, we get

$ \begin{aligned} & & y+x \cdot \frac{d y}{d x} & =0 \\ \Rightarrow & & \frac{d y}{d x} & =\frac{-y}{x} \end{aligned} $

Now, slope of tangent at $(8,2)$ is

$ m_1=\frac{-2}{8}=\frac{-1}{4} $

And slope of normal is

$ m_n=\frac{-1}{-1 / 4}=4 $

Now, equation of normal at $(8,2)$ is

$ \begin{aligned} & y-2=4(x-8) \\ \Rightarrow \quad & y=4 x-30 \end{aligned} $

Normal intersects the hyperbola again at a point $(\alpha, \beta)$,

So, $\alpha \beta=16$ and $\beta=4 \alpha-30$

Solving these two equations, we get

$ \begin{aligned} & \alpha(4 \alpha-30)=16 \\ & \Rightarrow \quad 2 \alpha^2-15 \alpha-8=0 \end{aligned} $

$ \begin{aligned}So, \alpha & =\frac{-(-15) \pm \sqrt{(-15)^2-4 \cdot 2 \cdot(-8)}}{2 \cdot 2} \\ & =\frac{15 \pm \sqrt{289}}{4}=\frac{15 \pm 17}{4} \end{aligned} $

So, $\alpha_1=\frac{15+17}{4}=\frac{32}{4}=8$ and

$ \alpha_2=\frac{15-17}{4}=\frac{-1}{2} $

Since $(8,2)$ is one point, the other point is where $\alpha=\frac{-1}{2}$

$ \begin{aligned} & \text { So, } \beta=4 \alpha-30 \\ & =4\left(\frac{-1}{2}\right)-30=-32 \\ & \begin{aligned} \therefore|\beta| & +\frac{1}{|\alpha|}=|-32|+\frac{1}{|-1 / 2|} \\ & =32+2=34 \end{aligned} \end{aligned} $

$4 x^2-9 y^2=36 \Rightarrow \frac{x^2}{9}-\frac{y^2}{4}=0$

So, $a^2=9, b^2=4$

Thus, equation of normal

$ \frac{9 x}{x_1}+\frac{4 y}{y_1}=(9+4)=13 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and the given equation is

$ \begin{aligned} & 3 x+2 \sqrt{2} y+k=0 \\ \Rightarrow \quad & 3 x+2 \sqrt{2} y=-k \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

$\because$ Both equation are proportional

Thus, $\frac{\frac{9}{x_1}}{3}=\frac{\frac{4}{y_1}}{2 \sqrt{2}}=\frac{13}{-k}$

$\Rightarrow \quad \frac{3}{x_1}=\frac{\sqrt{2}}{y_1} \Rightarrow x_1=\frac{3}{\sqrt{2}} y_1$

Put in equation of hyperbola

$\frac{\left(\frac{3}{\sqrt{2}} y_1\right)^2}{9}-\frac{y^2}{4}=1 \Rightarrow y_1^2=4 \Rightarrow y_1= \pm 2$

at $y_1=2 \Rightarrow x_1=3 \sqrt{2}$ and at $y_1=-2$

$\Rightarrow x_1=-3 \sqrt{2}$

Also, $\frac{3}{x_1}=-\frac{13}{k} \Rightarrow-k=\frac{13}{3} \times 3 \sqrt{2}=13 \sqrt{2}$

Hence, $k=-13 \sqrt{2}$

Asymptotes are

$ \begin{aligned} & 3 x-4 y-1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & 4 x-3 y-6=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

By Eqs. (i) and (ii), we get

$ x=3, y=2 $

Now, the direction vectors of asymptotes for $3 x-4 y=1 \Rightarrow \mathrm{~d}_1=\langle 4,3\rangle$ and for $4 x-3 y=6 \Rightarrow \mathbf{d}_2=\langle 3,4\rangle$ Thus, transverse and conjugate axis directions.

$ \begin{aligned} &\begin{array}{ll} & \frac{L_1}{\sqrt{a_1^2+b_1^2}}= \pm \frac{L_2}{\sqrt{a_2^2+b_2^2}} \\ \Rightarrow & \frac{3 x-4 y-1}{\sqrt{3^2+(-4)^2}}= \pm \frac{4 x-3 y-6}{\sqrt{4^2+(-3)^2}} \\ \Rightarrow & 3 x-4 y-1= \pm 4 x-3 y-6 \\ \text { So, } & 3 x-4 y-1=4 x-3 y-6 \\ \text { or } & 3 x-4 y-1=-4 x+3 y+6 \\ \Rightarrow & x+y-5=0 \\ \text { or } & 7 x-7 y-7=0 \\ \Rightarrow & x+y-5=0 \text { or } x-y-1=0 \end{array}\\ &\text { Hence, transverse and conjugate axis are }\\ &x+y=5, x-y=1 \end{aligned} $

Given equation of asymptotes

$ x+y=-3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and $\quad 2 x-y=-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On solving Eqs. (i) and (ii), we get centre of hyperbola i.e. $x=\frac{-4}{3}, y=\frac{-5}{3}$

i.e. Centre $\equiv\left(\frac{-4}{3}, \frac{-5}{3}\right)$

Also combined equation of asymptotes is $(x+y+3)(2 x-y+1)=0$

$ \Rightarrow \quad 2 x^2-y^2+x y+7 x-2 y+3=0 $

Now, equation of hyperbola is (when equation of asymptotes are known) is

$ (x+y+3)(2 x-y+1)+c=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

where $c$ is constant.

$\because$ (iii) passes through $(1,-2)$.

So, $(1-2+3)(2+2+1)+c=0$

$ \Rightarrow \quad(2 \times 5)+c=0 \Rightarrow c=-10 $

$\therefore$ Equation of hyperbola is

$ (x+y+3)(2 x-y+1)-10=0 $

Also, equation of conjugate hyperbola is (as we know that, equation of hyperbola, conjugate hyperbola, and

pair of asymptotes are same but only difference in constant term).

$ =2 x^2+x y-y^2+7 x-2 y+13=0 $

Given equation of hyperbola can be written as

$\frac{x^2}{5}-\frac{y^2}{4}=1$ which is of the form

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

where, $a=\sqrt{5}, b=2$

As we know that, equation of any tangent to the hyperbola Eq. (i) having slope $m$ is

$ y=m x \pm \sqrt{a^2 m^2-b^2}=m x+\sqrt{5 m^2-4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) $

According to question, Eq. (ii) passes through $(1,1)$.

$ \begin{aligned} & \text { So, } 1=m \times 1 \pm \sqrt{5 m^2-4} \\ & \Rightarrow \quad 1-m= \pm \sqrt{5 m^2-4} \\ & \Rightarrow \quad(1-m)^2=5 m^2-4 \\ & \Rightarrow \quad 4 m^2+2 m-5=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

Let $m_1, m_2$ slopes of these two tangents be the roots of Eq. (iii).

So, $m_1+m_2=\frac{-2}{4}=\frac{-1}{2}$

$ \begin{aligned} & m_1 m_2=\frac{-5}{4} \\ \therefore \tan \theta & =\left|\frac{m_1-m_1}{1+m_1 m_2}\right| \\ & =\left|\frac{\sqrt{\left(m_1+m_2\right)^2-4 m_1 m_2}}{1+m_1 m_2}\right| \\ & =\left|\frac{\sqrt{\left(\frac{-1}{2}\right)^2-4 \times\left(\frac{-5}{4}\right)}}{1+\left(\frac{-5}{4}\right)}\right| \\ & =\left|\frac{\sqrt{\frac{1}{4}+\frac{20}{4}}}{1-\frac{5}{4}}\right|=\left|\frac{\sqrt{21}}{2} \times \frac{4}{-1}\right| \\ & =|-2 \sqrt{21}|=2 \sqrt{21} \end{aligned} $

The slope of tangent $m=\tan 45^{\circ}=1$

Now, $5 x^2-9 y^2-20 x-18 y-34=10$

$ \begin{aligned} & \Rightarrow 5\left(x^2-4 x\right)-9\left(y^2+2 y\right)=34 \\ & \Rightarrow 5\left(x^2-4 x+4\right)-9\left(y^2+2 y+1\right) =34+20-9 \end{aligned} $

$ \begin{aligned} & \Rightarrow 5(x-2)^2-9(y+1)^2=45 \\ & \Rightarrow \frac{(x-2)^2}{9}-\frac{(y+1)^2}{5}=1 \end{aligned} $

The equation of the tangent with slope $m$ is

$ y+1=m(x-2) \pm \sqrt{9 m^2-5} $

Since, $m=1$

$ \begin{array}{ll} \Rightarrow & y+1=(x-2) \pm \sqrt{9-5} \\ \Rightarrow & y=x-1 \text { or } y=x-5 \end{array} $

$\Rightarrow$ The equation $x-y-1=0$

or  $ x-y-5=0 $

Comparing with $x+b y+c=0$

$ \begin{aligned} & \Rightarrow b=-1 \text { and } c=-1,-5 \\ & \Rightarrow b^2+c^2=(-1)^2+(-1)^2=1+1=2 \\ & \text { and } b^2+c^2=(-1)^2+(-5)^2=1+25=26 \\ & \Rightarrow b^2+c^2=2 \text { or } 26 \end{aligned} $

Distance between the foci $=2 a e$

$ \Rightarrow \quad 2 a e=26 $

distance between the directrices $=\frac{2 a}{e}$

$ \begin{aligned} & \Rightarrow \quad \frac{2 a}{e}=\frac{50}{13} \\ & \Rightarrow \quad \frac{2 a e}{\frac{2 a}{e}}=\frac{26}{\frac{50}{13}} \\ & \Rightarrow \quad e^2=\frac{13 \cdot 13}{25} \\ & \Rightarrow \quad e=\frac{13}{5} \\ & \Rightarrow \quad \frac{1}{e_H^2}+\frac{1}{e_{C H}^2}=1 \\ & \Rightarrow \quad \frac{25}{169}+\frac{1}{e_{C H}^2}=1 \\ & \Rightarrow \quad e_{C H}^2=\frac{169}{144} \Rightarrow e_{C H}=\frac{13}{12} \end{aligned} $

Given equation

$ \begin{aligned} & x^2+4 x y+y^2=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & a x^2+2 h x y+b y^2=1 \\ & x=X \cos \theta-Y \sin \theta \\ & y=X \sin \theta+Y \cos \theta \\ & a=1, h=2, b=1 \\ & h^2-a b=3>0 \Rightarrow \text { Given } \end{aligned} $

curve and hyperbola,

$ \begin{aligned} \tan 2 \theta & =\frac{2 h}{a-b}=\frac{4}{0}=\frac{\pi}{2} \\ \theta & =\frac{\pi}{4} \\ x & =\frac{X-Y}{\sqrt{2}}, y=\frac{X+Y}{\sqrt{2}} \end{aligned} $

$ \begin{aligned} &\text { From (i), we get }\\ &\begin{aligned} & \frac{X^2+Y^2-2 X Y}{2}+\frac{4\left(X^2-Y^2\right)}{2} +\frac{X^2+Y^2+2 X Y}{2}=1 \\ & \frac{1}{2}\left[6 X^2-2 Y^2\right]=1 \Rightarrow 3 X^2-Y^2=1 \\ & \frac{X^2}{1 / 3}-\frac{Y^2}{1}=1 \Rightarrow a^2=\frac{1}{3}, b^2=1 \\ & \sqrt{\frac{a^2+b^2}{a^2}}=\sqrt{\frac{\frac{1}{3}+1}{\frac{1}{3}}}=2 \end{aligned} \end{aligned} $

Given equation of hyperbola

$ x y=-1 $

Let equation of tangent to $x y=-1$ be $y=m x+c$

$ \begin{aligned} \Rightarrow \quad & x(m x+c)+1=0 \\ & m x^2+c x+1=0 \text { for tangent } Q=0 \\ & c^2-4 m=0 \Rightarrow c^2=4 m \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Equation of tangent to $y^2=8 x$

$ \begin{aligned} y & =m x+\frac{2}{m} \\ \frac{2}{m} & =c \Rightarrow \frac{4}{m^2}=4 m \\ m & =1, c=2 \\ y & =x+2 \end{aligned} $

Given equation of hyperbola

$ \begin{aligned} & 2 x^2-3 y^2=6 \\ & \frac{x^2}{3}-\frac{y^2}{2}=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Equation of tangent, $y=$ Slope of give line $m_1=\frac{1}{2}$

Slope of perpendicular line $=-2$

$y=m x+c$ is tangent to hyperbola

$ c^2=a^2 m^2-b^2 \Rightarrow c= \pm \sqrt{10} $

Equation of tangents $2 x+y+\sqrt{10}=0$,

$ 2 x+y-\sqrt{10} \text { distance }=\frac{2 \sqrt{10}}{\sqrt{5}}=2 \sqrt{2} $