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)

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