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

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} $

Given, hyperbola $5 x^2-9 y^2=90$

$ \frac{x^2}{18}-\frac{y^2}{10}=1 $

Let equation of tangent to the ellipse be

$ y=m x \pm \sqrt{18 m^2-10} $

or $ (y-m x)^2=18 m^2-10 $

$ \begin{array}{ll} \therefore & y^2+m^2 x^2-2 m x y-18 m^2+10=0 \\ & m^2\left(x^2-18\right)-2 m x y+y^2+10=0 \end{array} $

Given tangent makes $\alpha$ and $\beta$ angles with transverse axis are complementary

$ \begin{aligned} & \because \quad \alpha+\beta=\frac{\pi}{2} \\ & \tan (\alpha+\beta)=\text { not defined } \\ & \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\text { not defined } \\ & \text { And } m_1=\tan \alpha \quad m_2=\tan \beta \\ & m_1+m_2=\frac{2 x y}{x^2-18} \text { and } m_1 m_2=\frac{y^2+10}{x^2-18} \\ & \because \quad 1-m_1 m_2=0 \Rightarrow m_1 m_2=+1 \\ & \quad \frac{y^2+10}{x^2-18}=+1 \\ & \Rightarrow \quad y^2+10=+x^2-18 \\ & \Rightarrow \quad y^2=x^2-28 \text { or } x^2-y^2=28 \end{aligned} $

We have hyperbola

$ \begin{aligned} & 7 x^2-9 y^2=63 \\ & \frac{x^2}{9}-\frac{y^2}{7}=1 \end{aligned} $

Now, its asymptotes is

$ \begin{aligned} & \frac{x}{3} \pm \frac{y}{\sqrt{7}}=0 \\ & r=\frac{x^2}{9}+0 \cdot x y-\frac{y^2}{7}=0 \\ & a=\frac{1}{9}, b=-\frac{1}{7} \end{aligned} $

Angle between pair of asymptotes

$ \tan \theta=\frac{2 \sqrt{0+\frac{1}{63}}}{\frac{1}{9}-\frac{1}{7}}=\left|\frac{2 \frac{1}{\sqrt{63}}}{\frac{-2}{63}}\right| $

$ \begin{array}{ll} \Rightarrow & \tan \theta=\frac{2}{\sqrt{63}} \times \frac{63}{2}=\sqrt{63} \\ \therefore & \cos \theta=\frac{1}{8} \end{array} $

Given, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $b^2=36$ Coordinate of one of $e^2=1+\frac{b^2}{a^2}=1+\frac{36}{a^2}$ The latus rectum is $L\left(a e,+\frac{b^2}{a}\right)$ and $L^{\prime}\left(a e,-\frac{b^2}{a}\right)$.

Now, slope of $O L=\frac{b^2}{a^2 e}$ Slope of $O L^{\prime}=-\frac{b^2}{a^2 e}$ Now, angle between $O L$ and $O L^{\prime}$

$ \begin{aligned} & 2 \tan ^{-1} \frac{b^2}{a^2 e}=2 \tan ^{-1} \frac{3}{2} \\ & \therefore \quad \frac{b^2}{a^2+e}=\frac{3}{2} \Rightarrow \frac{3 e}{2}=\frac{b^2}{a^2} \\ & \frac{3 e}{2}=e^2-1 \quad\left[\because e^2=1+\frac{b^2}{a^2}\right] \\ & 2 e^2-3 e-2=0 \\ & (e-2)(2 e+1)=10 \Rightarrow e=2, e \neq-\frac{1}{2} \\ & a^2=\frac{2 b^2}{3 e}=\frac{2 \times 36}{3 \times 2}=12 \,\,\,\,\left[\because b^2=36\right]\\ & \therefore \sqrt{a^2+e^2}=\sqrt{12+4}=4 \end{aligned} $

Given, foci $(8,3)$ and $(0,3)$ and $e=\frac{4}{3}$ centre $=(4,3)$

Distance between $\mathrm{foci}=2 a e=8$

$ \begin{aligned} a e & =4 \\ a \cdot \frac{4}{3} & =4 \\ a & =3 \Rightarrow a^2=9 \text { and } b^2=7 \end{aligned} $

$\because$ Equation of hyperbola is

$ \frac{(x-4)^2}{9}-\frac{(y-3)^2}{7}=1 $

Clearly $\alpha=4, p=9$ and $q=7$

$ \therefore p+q=16=\alpha^2 $

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

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

$\therefore e_H=2$ [ 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 $\left( \pm \frac{5 \sqrt{5}}{3}, 2\right)$

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

We begin with the hyperbola defined by the equation:

$ \frac{x^2}{a^2} - \frac{y^2}{9} = 1 $

We can determine the eccentricity $ e $ of the hyperbola using the equation $ e^2 - 1 = \frac{9}{a^2} $. Solving for $ e $:

$ e^2 = \frac{a^2 + 9}{a^2} $

$ e = \frac{\sqrt{9 + a^2}}{a} $

The focus $ S $ is on the positive x-axis, given by the coordinates $ (c, 0) $ where $ c = \sqrt{9 + a^2} $. We also have $ Q = (0, 1) $.

Given that $ SQ = \sqrt{26} $, substitute the values:

$ \sqrt{(c - 0)^2 + (0 - 1)^2} = \sqrt{26} $

$ \sqrt{c^2 + 1} = \sqrt{26} $

$ c^2 + 1 = 26 $

$ c^2 = 25 $

$ \Rightarrow c = 5 $

Thus, $ S = (5, 0) $.

Now, concerning the point $ P $ on the hyperbola, we represent it parametrically as $ P(a \sec \theta, 3 \tan \theta) = (4 \sec \theta, 3 \tan \theta) $, since $ a = 4 $.

We use the equation for the distance $ SP = 6 $:

$ SP^2 = (4 \sec \theta - 5)^2 + (3 \tan \theta)^2 = 36 $

Expanding and substituting $ \tan^2 \theta = \sec^2 \theta - 1 $:

$ (4 \sec \theta - 5)^2 + 9 \tan^2 \theta = 36 $

$ (4 \sec \theta - 5)^2 + 9(\sec^2 \theta - 1) = 36 $

$ 16 \sec^2 \theta - 40 \sec \theta + 25 + 9 \sec^2 \theta - 9 = 36 $

$ 25 \sec^2 \theta - 40 \sec \theta - 20 = 0 $

Factoring:

$ 5 \sec^2 \theta - 8 \sec \theta - 4 = 0 $

$ 5 \sec \theta (\sec \theta - 2) + 2 (\sec \theta - 2) = 0 $

$ (\sec \theta - 2)(5 \sec \theta + 2) = 0 $

Solving gives $ \sec \theta = 2 $, leading to:

$ \theta = \frac{\pi}{3} $

Given the hyperbola $ x^2 - y^2 = c^2 $, we are examining the tangent at any point $ P(t) = (c \sec \theta, c \tan \theta) $.

For the equation of the tangent at $ P(t) $:

$ c \sec \theta \cdot x - c \tan \theta \cdot y = c^2 $

Simplifying gives:

$ \sec \theta \cdot x - \tan \theta \cdot y = c $

This tangent line intersects the $ X $-axis when $ y = 0 $:

$ T \equiv \left(\frac{c}{\sec \theta}, 0\right) $

Next, consider the normal's equation. The slope of the normal is:

$ \frac{-1}{\frac{\sec \theta}{\tan \theta}} = \frac{-\tan \theta}{\sec \theta} $

The equation of the normal through $ P(t) $ is:

$ c \sec \theta \tan \theta + c \tan \theta \sec \theta = 2c \tan \theta \sec \theta $

Thus, the normal cuts the $ Y $-axis (where $ x = 0 $) at:

$ N \equiv (0, 2c \tan \theta) $

Consider the midpoint $ M(x, y) $ of segment $ TN $. The coordinates $ (x, y) $ satisfy:

$ 2x = \frac{c}{\sec \theta} $

$ 2y = 2c \tan \theta $

Therefore:

$ \sec \theta = \frac{c}{2x}, \quad \tan \theta = \frac{y}{c} $

Using the identity $ \sec^2 \theta - \tan^2 \theta = 1 $, we have:

$ \left(\frac{c}{2x}\right)^2 - \left(\frac{y}{c}\right)^2 = 1 $

This simplifies to the equation of the locus of the midpoint:

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