Hyperbola

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

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

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

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

$ \Rightarrow k = 2$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

By (i) and (ii)

$7a = 4b$

then by (i)

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

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

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

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

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

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

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

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

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

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

Now, equation of normal to hyperbola

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

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

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

Given, equation of hyperbola

$ \begin{aligned} & x^2-2 y^2=1 \\ \Rightarrow & \frac{x^2}{1^2}-\frac{y^2}{(1 / \sqrt{2})^2}=1 \end{aligned} $

On comparing with

$ \begin{array}{ll} & \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ a^2=1 & \\ b^2=1 / 2 & \end{array} $

Now, $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}$

$ \text { ∴ } \text { Focus }=(a e, 0)=\left(\frac{\sqrt{3}}{\sqrt{2}}, 0\right) $

Now, equation of line $P S$

$ y-0=\frac{0-1}{\frac{\sqrt{3}}{\sqrt{2}}+1}\left(x-\frac{\sqrt{3}}{\sqrt{2}}\right) $

$ \begin{aligned} &\begin{aligned} & y=-\frac{\sqrt{2}}{\sqrt{3}+\sqrt{2}}\left(x-\frac{\sqrt{3}}{\sqrt{2}}\right) \\ & y=-\left(\frac{\sqrt{2}}{\sqrt{3}+\sqrt{2}}\right) x+\frac{\sqrt{3}}{\sqrt{3}+\sqrt{2}} \end{aligned}\\ &\text { Now coordinate of } A \text { put } x=0\\ &y=\frac{\sqrt{3}}{\sqrt{3}+\sqrt{2}}=A O\\ &\text { } \begin{aligned}Area \,\,of \,\, \triangle A O S & =\frac{1}{2} O A \times O S \\ & =\frac{1}{2} \times \frac{\sqrt{3}}{(\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}}{\sqrt{2}} \\ & =\frac{3}{2 \sqrt{2}(\sqrt{3}+\sqrt{2})} \\ & =\frac{3}{2(\sqrt{6}+2)} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, } S P+S^{\prime} P=2\left(S P-S^{\prime} P\right)\\ &\begin{array}{lc} \Rightarrow & \left(e x_1-a\right)+\left(e x_1+a\right)=2(2 a) \\ \Rightarrow & 2 e x_1=4 a \\ \Rightarrow & e x_1=2 a \\ \Rightarrow & e\left(a \sec \frac{\pi}{6}\right)=2 a \\ \Rightarrow & e \times \frac{2}{\sqrt{3}}=2 \\ \Rightarrow & e=\sqrt{3} \end{array} \end{aligned} $

Let $e_1$ eccentricity of hyperbola.

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

Distance between its focii $=2$ times distance between its directices

$ \begin{aligned} \Rightarrow & & 2 a e_1 & =2 \times \frac{2 a}{e_1} \\ \Rightarrow & & e_1 & =\sqrt{2} \end{aligned} $

and let $e_2$ be eccentricity of another hyperbola.

Length of transverse axis $=2 \times$ Length of conjugate axis

$ \begin{array}{ll} \Rightarrow & 2 a^{\prime}=2 \times 2 b^{\prime} \\ \Rightarrow & a^{\prime}=2 b^{\prime} \\ \Rightarrow & e_2^2=1+\frac{b^{\prime 2}}{a^{\prime 2}} \\ \Rightarrow & e_2^2=1+\frac{b^{\prime 2}}{4 b^{\prime 2}}=e_2=\frac{\sqrt{5}}{2} \\ \therefore & e_1 e_2=\sqrt{2} \times \frac{\sqrt{5}}{2}=\frac{\sqrt{10}}{2} \end{array} $

Assertion (A)

$ \quad \begin{aligned} &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 9 x^2+16 y^2=9 \\ & \Rightarrow \quad \frac{x^2}{1}+\frac{y^2}{\left(\frac{4}{3}\right)^2}=1 \end{aligned} $

Not the equation of hyperbola. So, Assertion (A) is wrong. Reason (R) is correct.

Let the equation of hyperbola whose centre is at origin and transverse axis is along $X$-axis is

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

As the length of transverse axis is 8,

$ \begin{aligned} & 2 a=8 \\ & \Rightarrow \end{aligned} \quad a=4 $

The equation of hyperbola is $\frac{x^2}{16}-\frac{y^2}{b^2}=1$

The hyperbola passes through $(5,2)$

$ \begin{aligned} \frac{25}{16} & -\frac{4}{b^2}=1 \\ \Rightarrow \frac{4}{b^2} & =\frac{25}{16}-1 \Rightarrow b^2=\frac{64}{9} \end{aligned} $

∴ The equation of hyperbola is

$ \frac{x^2}{16}-\frac{y^2}{64 / 9}=1 $

$ \begin{aligned} &\text { Equation of conjugate hyperbola is }\\ &\frac{x^2}{16}-\frac{y^2}{64 / 9}=-1\\ &\begin{aligned} \text { Eccentricity } & =\sqrt{1+\frac{a^2}{b^2}} \cdot=\sqrt{1+\frac{16}{64 / 9}} \\ & =\sqrt{1+\frac{16 \times 9}{64}} \\ & =\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, two hyperbola i.e. } x=\sec \theta \text {, }\\ &\begin{aligned} & y=\sqrt{2} \tan \theta \text { and } x=\sqrt{2} \sec \theta, y=\tan \theta \\ & \text { H1: } \frac{x^2}{1}-\frac{y^2}{2}=1 \\ & \text { and H2: } \frac{x^2}{2}-\frac{y^2}{1}=1 \\ & e_1=\sqrt{1+\frac{2}{1}}=\sqrt{3} \\ & \text { and } e_2=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}} \\ & \therefore \quad \frac{e_2^2}{e_1^2}=\frac{3}{2 \times 3}=\frac{1}{2} \end{aligned} \end{aligned} $

Let the equation of hyperbola be

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

Here, $e=\sqrt{1+\frac{b^2}{a^2}}$ and centre is $(0,0)$.

End points of latusrectum i.e. $L\left(a e, \frac{b^2}{a}\right)$ and $\left(a e, \frac{-b^2}{a}\right)$.

$\angle L C L^{\prime}$ is $120^{\circ}$,

$\left(m_1\right)$ slope of $L C=\frac{b^2}{a \times a e}=\frac{b^2}{a^2 e}=\frac{e^2-1}{e}$

$\left(m_2\right)$ slope of $C L^{\prime}=\frac{-b^2}{a^2 e}=\frac{1-e^2}{e}$

$ \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| $

$ \begin{aligned} \text { Slope of } L C & =\frac{b^2}{a^2 e} \\ \tan 60^{\circ} & =\frac{\left(e^2-1\right)}{e} \\ \Rightarrow \quad \sqrt{3} & =\frac{e^2-1}{e} \Rightarrow e^2-\sqrt{3} e-1=0 \\ \Rightarrow \quad e & =\frac{\sqrt{3} \pm \sqrt{3+4}}{2}=\frac{\sqrt{3}+\sqrt{7}}{2} \end{aligned} $

Given, $x^2-4 y^2=4$

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

Here, $a^2=4, b^2=1$

$ a=2 \text { and } b=1 $

Parametric form of hyperbola is

$ \dot{x}=2 \sec \theta, y=\tan \theta $

$ \begin{aligned} & \text { For point } P\left(\frac{\pi}{4}\right), x=2 \sec \frac{\pi}{4}=2 \sqrt{2} \\ & \qquad \begin{array}{l} y=\tan \frac{\pi}{4}=1 \\ \therefore \quad P(2 \sqrt{2}, 1) \end{array} \\ & \text { For point } Q, x=2 \sec \frac{5 \pi}{4}=-2 \sqrt{2} \\ & \qquad \begin{array}{l} y=\tan \frac{5 \pi}{4}=1 \\\therefore Q-2 \sqrt{2}, 1) \end{array} \\ & \text { For point } R, x=2 \sec \frac{3 \pi}{4}=-2 \sqrt{2} \\ & \qquad y=\tan \frac{3 \pi}{4}=-1 \\ & \therefore \quad R(-2 \sqrt{2},-1) \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \text { For point } T, x=2 \sec \frac{7 \pi}{4}=2 \sqrt{2} \\ & \qquad \begin{array}{l}∴\,\,\, y=\tan \frac{7 \pi}{4}=-1 \\\,\,\,\,\,\, T(2 \sqrt{2},-1) \end{array} \end{aligned}\\ &\text { ∴ Area of quadrilateral }\\ &\begin{aligned} & \left.=\frac{1}{2}| | \begin{array}{cc} 2 \sqrt{2}+2 \sqrt{2} & -2 \sqrt{2}-2 \sqrt{2} \\ 1+1 & 1+1 \end{array} \right\rvert\, \\ & =\frac{1}{2}| | \begin{array}{cc} 4 \sqrt{2} & -4 \sqrt{2} \\ 2 & 2 \end{array}| | \\ & =\frac{1}{2}|8 \sqrt{2}+8 \sqrt{2}|=8 \sqrt{2} \text { sq units } \end{aligned} \end{aligned} $

$ \text { According to the question, } $

Given, $A B+B C+A C=20$

$ \begin{aligned} & \Rightarrow A B+A C=9 \\ & \Rightarrow \sqrt{(h+5)^2+k^2}+\sqrt{(h-6)^2+k^2}=9 \\ & \Rightarrow \sqrt{(h+5)^2+k^2}=9-\sqrt{(h-6)^2+k^2} \end{aligned} $

On squaring both sides,

$ \begin{gathered} h^2+10 h+25+k^2=81+\left(h^2-12 h\right. \\ \left.+36+k^2\right)-2 \cdot 9 \sqrt{(h-6)^2+k^2} \\ \Rightarrow 2 \times 9 \times \sqrt{h^2-12 h+36+k^2} \\ =-22 h+92=2(-11 h+46) \\ \Rightarrow 9 \sqrt{h^2-12 h+36+k^2}=-11 h+46 \end{gathered} $

Squaring on both sides,

$ \begin{aligned} & 81\left(h^2-12 h+36+k^2\right) \\ & \quad=121 h^2-1012 h+2116 \\ & \Rightarrow 40 h^2-40 h-81 k^2-800=0 \end{aligned} $

∴ Locus of the third vertex is

$ 40 x^2-81 y^2-40 x-800=0 $

$ \begin{aligned} &\\ &\begin{aligned} a & =4 \text { and } b=3 \\ e & =\sqrt{1+\frac{9}{16}}=\frac{5}{4} \\ S P & =e x_1 \pm a=\frac{5}{4} \times 5 \pm 4=\frac{41}{4}, \frac{9}{4} \end{aligned} \end{aligned} $

Hyperbola is $\frac{x^2}{25}-\frac{y^2}{9}=1$

$ \therefore \quad P(\theta) \equiv(5 \sec \theta, 3 \tan \theta) $

As, $P(\theta) \equiv\left(x_1, \frac{3 \sqrt{5}}{2}\right)\,\,\,\,\,\,\,\,\,\,\,\,(Given)$

$ \begin{aligned} & \therefore \quad 3 \tan \theta=3 \sqrt{5} / 2 \Rightarrow \tan \theta=\sqrt{5} / 2 \\ & \sec \theta=\sqrt{1+\left(\frac{\sqrt{5}}{2}\right)^2}=\frac{3}{2} \\ & \sin ^2 \theta=1-\frac{1}{\sec ^2 \theta}=1-\frac{1}{(9 / 4)}=\frac{5}{9} \\ & x_1=5 \times \frac{3}{2}=\frac{15}{2} \\ & \therefore \quad 2 x_1+9 \sin ^2 \theta=15+5=20 \end{aligned} $

Given, $\frac{x^2}{k-\frac{5}{2}}+\frac{y^2}{\frac{7}{3}-k}=1$

( $k$ is a real number)

$ \begin{aligned} & a=\sqrt{k-\frac{5}{2}}, b=\sqrt{\frac{7}{3}-k} \\ & k-\frac{5}{2}>0, \frac{7}{3}-k>0 \\ & k>\frac{5}{2}, k<\frac{7}{3} \Rightarrow k \in\left(-\infty, \frac{7}{3}\right) \cup\left(\frac{5}{2}, \infty\right) \end{aligned} $

Given, hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and focus $S(a e, 0) . A\left(\theta_1\right)$ and $B\left(\theta_2\right)$ be two points on hyperbola, equation of chord joining $A\left(\theta_1\right)$ and $B\left(\theta_2\right)$ is

$ \begin{array}{r} \Rightarrow \frac{x}{a} \cos \left(\frac{\theta_1-\theta_2}{2}\right)-\frac{y}{b} \sin \left(\frac{\theta_1+\theta_2}{2}\right) \\ =\cos \left(\frac{\theta_1+\theta_2}{2}\right) \end{array} $

$A, S$ and $B$ are collinear so, $S$ lies on line $A B$

$ \begin{array}{rlrl} & \frac{a e}{a} \cos \left(\frac{\theta_1-\theta_2}{2}\right)=\cos \left(\frac{\theta_1+\theta_2}{2}\right) \\ \Rightarrow & a \cos \left(\frac{\theta_1+\theta_2}{2}\right)= a e \cos \left(\frac{\theta_1-\theta_2}{2}\right) \end{array} $

On comparing with

$ \begin{aligned} & a \cos \left(\frac{\theta_1+\theta_2}{2}\right)=k \cos \left(\frac{\theta_1-\theta_2}{2}\right) \\ & k=a e \Rightarrow k=a \sqrt{1+\frac{b^2}{a^2}}=\sqrt{a^2+b^2} \end{aligned} $

Let $p(h, k)$ be the point of intersection of tangents at the ends of a normal chord of the hyperbola, $x^2-y^2=a^2$, then the equation of the chord is

$h x-k y=a^2 \quad \text{... (i)}$

But it is normal chord. So, its equation must be of the form $x\cos\theta+y\cot\theta=2a\quad \text{... (ii)}$

Eqs. (i) and (ii) represents the same line

$\begin{array}{ll} \therefore & \frac{\cos \theta}{h}=\frac{\cot \theta}{-k}=\frac{2 a}{a^2} \\ & \sec \theta=\frac{a}{2 h^{\prime}} \cdot \tan \theta=\frac{-a}{2 k} \\ \therefore & \sec ^2 \theta-\tan ^2 \theta=1 \\ \Rightarrow & \frac{a^2}{4 h^2}-\frac{b^2}{4 k^2}=1 \\ \Rightarrow & a^2\left(k^2-h^2\right)=4 h^2 k^2 \end{array}$

Therefore, the locus of $p(h, k)$ is $a^2\left(y^2-x^2\right)=4 x^2 y^2$

Given, hyperbola is $16 x^2-9 y^2=1$

$\begin{array}{ll} & \frac{x^2}{\frac{1}{16}}-\frac{y^2}{\frac{1}{9}}=1 \\ \Rightarrow \quad & \frac{x^2}{\left(\frac{1}{4}\right)^2}-\frac{y^2}{\left(\frac{1}{3}\right)^2}=1 \end{array}$

We know that Eccentricity of a hyperbola, $e_1^2=1+\frac{b^2}{a^2}$

$\begin{aligned} & \Rightarrow e_1^2=1+\frac{1 \times 16}{9 \times 1} \\ & \Rightarrow e_1^2=\frac{25}{9} \\ & \Rightarrow e_1=\frac{5}{3} \quad \text{.... (i)} \end{aligned}$

Now, eccentricity of its conjugate

$ \begin{array}{rlrl} & e_2^2 =1+\frac{a^2}{b^2} \\ \Rightarrow & e_2^2 =1+\frac{1 \times 9}{16 \times 1} \\ \Rightarrow & e_2^2 =\frac{25}{16} \\ \Rightarrow & e_2=\frac{5}{4} \end{array}$

From Eqs. (i) and (ii),

$\begin{aligned} 3 e_1 & =3 \times \frac{5}{3}=\frac{5}{4} \times 4 \\ \Rightarrow \quad 3 e_1 & =4 e_2 \quad \left[\because e_2=\frac{5}{4}\right] \end{aligned}$

Let $P(\sec \theta, \tan \theta)$ be the point where the normal to the rectangular hyperbola $x^2-y^2=1$ Meet equation of normal at $\frac{\pi}{4}$ is

$\begin{aligned} & \frac{x}{\sec \frac{\pi}{4}}+\frac{y}{\tan \frac{\pi}{4}}=2 \\ \Rightarrow & \frac{x}{\sqrt{2}}+y=2 \\ \Rightarrow \quad & x+\sqrt{2} y=2 \sqrt{2} \end{aligned}$

Substitute $x=2 \sqrt{2}-\sqrt{2} y$ in hyperbola's equation,

$\begin{aligned} & {[\sqrt{2}(2-y)]^2-y^2=1 } \\ \Rightarrow & 2\left[4+y^2-4 y\right]-y^2=1 \\ \Rightarrow & 8+2 y^2-8 y-y^2-1=0 \\ \Rightarrow & y^2-8 y+7=0 \\ \Rightarrow & y=7,1 \\ & x+7 \sqrt{2}=2 \sqrt{2} \text { or } x+\sqrt{2}=2 \sqrt{2} \\ \Rightarrow & x=-5 \sqrt{2} \text { or } x=\sqrt{2} \end{aligned}$

The two points are $(-5 \sqrt{2}, 7),(\sqrt{2}, 1)$.

At $\left(\frac{\pi}{4}\right)$, the point is $(\sqrt{2}, 1)$ and at $\theta$, the point is $(-5 \sqrt{2}, 7)$.

$\begin{aligned} & \sec \theta=-5 \sqrt{2}, \tan \theta=7 \\ & \sec ^2 \theta+\tan \theta=(-5 \sqrt{2})^2+7=57 \end{aligned}$

Vertices and foci of the hyperola are $( \pm 3,0)$ and $( \pm 4,0)$.

Thus, $a= \pm 3$ and $a e= \pm 4$ and $e=\frac{4}{3}$

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

$\frac{16}{9}=1+\frac{b^2}{9} \Rightarrow b^2=7$

Thus, equation of hyperbola is $\frac{x^2}{9}-\frac{y^2}{7}=1$

whose parametric equation is $x=3 \sec \theta$ and $y=\sqrt{7} \tan \theta$

$\begin{aligned} & \frac{1+\tanh x}{1-\tanh x} \\ & =\frac{1+\frac{e^x-e^{-x}}{e^x+e^{-x}}}{1-\frac{e^x-e^{-x}}{e^x+e^{-x}}} \quad\left[\because \tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}\right] \\ & =\frac{e^x+e^{-x}+e^x-e^{-x}}{e^x+e^{-x}-e^x+e^{-x}} \\ & =\frac{2 e^x}{2 e^{-x}}=e^{2 x} \end{aligned}$

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

Centre : $O(0,0)$

$\begin{aligned} & \text { Foci : } S( \pm a e, o) \equiv( \pm 3,0) \\ & \Rightarrow \quad a e=3 \\ \end{aligned}$

Eccentricity, $e=\sqrt{1+\left(\frac{b}{a}\right)^2}=\frac{3}{2}$

$\begin{aligned} & \because \quad a e=3 \\ & \Rightarrow \quad a .\left(\frac{3}{2}\right)=3 \\ & \Rightarrow \quad a=2 \\ & \text { and } \sqrt{1+\left(\frac{b}{a}\right)^2}=\frac{3}{2} \\ \end{aligned}$

$\begin{aligned} & \Rightarrow 1+\left(\frac{b}{2}\right)^2=\frac{9}{4} \\ & \Rightarrow \quad \frac{b^2}{4}=\frac{5}{4} \\ & \Rightarrow \quad b^2=5 \\ \end{aligned}$

$\therefore$ Hyperbola equation is $\frac{x^2}{4}-\frac{y^2}{5}=1\quad \text{.... (i)}$

Given, line is $2 x-y=1$

Put $y=2 x-1$ into Eq. (i),

$\begin{aligned} & \frac{x^2}{4}-\frac{(2 x-1)^2}{5}=1 \\ & \Rightarrow 5 x^2-4\left(4 x^2+1-4 x\right)=20 \\ & \Rightarrow 5 x^2-16 x^2-4+16 x-20=0 \\ & \Rightarrow-11 x^2+16 x-24=0 \end{aligned}$

Here, $D=b^2-4 a c=(16)^2-4(-11)(-24)<0$

$\therefore$ Line does not intersect the hyperbola.

Equation of hyperbola

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

Let $(h, k)$ be the point whose chord of contact subtends a right angle at the centre of hyperbola. Equation of chord of contact

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

Squaring both the sides, we get

$\begin{aligned} & \left(\frac{h x}{a^2}\right)^2+\left(\frac{k y}{b^2}\right)^2-2 \frac{h x}{a^2} \cdot \frac{k y}{b^2}=(l)^2 \\ \Rightarrow \quad & \frac{h^2 x^2}{a^4}+\frac{k^2 y^2}{b^4}-\frac{2 k h}{a^2 b^2} \cdot x y=1 \quad \text{... (ii)} \end{aligned}$

Now, from Eqs. (i) and (ii), we have

$\begin{aligned} & \frac{h^2 x^2}{a^4}+\frac{k^2 y^2}{b^4}-\frac{2 h k}{a^2 b^2} \cdot x y=\frac{x^2}{a^2}-\frac{y^2}{b^2} \\ \Rightarrow \quad & \left(\frac{h^2}{a^4}-\frac{1}{a^2}\right) x^2+\left(\frac{k^2}{b^4}+\frac{1}{b^2}\right) y^2-\frac{2 h k}{a^2 b^2} \cdot(x y)=0 \end{aligned}$

The above equation represents a pair of perpendicular lines if coefficient of $x^2$ + coefficient of $y^2=0$

$\begin{aligned} & \Rightarrow \quad \frac{h^2}{a^4}-\frac{1}{a^2}+\frac{k^2}{b^4}+\frac{1}{b^2}=0 \\ & \Rightarrow \quad \frac{h^2}{a^4}+\frac{k^2}{b^4}=\frac{1}{a^2}-\frac{1}{b^2} \\ & \text { Replacing } h \rightarrow x \text { and } k \rightarrow y, \\ & \frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}-\frac{1}{b^2} \end{aligned}$

Given,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Let focal chord be latusrectum then it is given that it subtends 90$^\circ$ at its centre.

Image

Here, equation of hyperbola is

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

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

End points of latus rectum, $L=\left(a e, \frac{b^2}{a}\right)$ and $L^{\prime}\left(a e, \frac{-b^2}{a}\right) $

Centre $(0,0)$

$\because < L C L^{\prime}=90^{\circ}$

In $\triangle L C L^{\prime}$ by Pythagoras theorem

$\begin{aligned} & (L C)^2+\left(L^{\prime} C\right)^2=\left(L L^{\prime}\right)^2 \\ & (a e-0)^2+\left(\frac{b^2}{a^2}-0\right)^2+(a e-0)^2+\left(-\frac{b^2}{a}-0\right)^2=\left(\frac{2 b^2}{a}\right)^2 \\ & \Rightarrow \quad 2\left(a^2 e^2+\frac{b^4}{a^2}\right)=\frac{4 b^4}{a^2} \Rightarrow b^4=a^4 e^2 \\ & \Rightarrow \quad\left(e^2-1\right)^2=e^2 \quad \quad \quad\left[\because b^2=a^2\left(e^2-1\right)\right] \\ & \Rightarrow \quad {e^2 \pm e-1}{}=0 \\ & \Rightarrow e=\pm \frac{1\pm \sqrt{1+4}}{2}=\frac{\pm1\pm\sqrt5}{2}\end{aligned}$

$\because e > 1$ for hyperbola

$\Rightarrow e=\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2} \Rightarrow e=\frac{\sqrt{5}+1}{2}$

$\Rightarrow$ Option (c) is correct.

Distance from focus to directrix is $a\left(e-\frac{1}{e}\right)$

$a\left(\frac{5}{4}-\frac{4}{5}\right)=\frac{4 \cdot 3-3}{|5|} \Rightarrow a=4$

Slope of axis $=(-3 / 4)$

Distance from vertex to focus

$\begin{aligned} & =a e-a=a\left(\frac{5}{4}\right)-a=\frac{a}{4} \\ \text { Vertex } & =\left[3 \pm\left(\frac{-4}{5}\right), 0 \pm\left(\frac{3}{5}\right)\right] \\ & =\left(3-\frac{4}{5}, \frac{3}{5}\right)=\left(\frac{11}{5}, \frac{3}{5}\right) \end{aligned}$

Any tangent to hyperbola forms a triangle with asymptotes which has constant area $=a b$

$\begin{aligned} & \Rightarrow \quad a b=a^2 \tan \alpha \Rightarrow \frac{b}{a}=\tan \alpha \\ & \text { Eccentricity, } e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\tan ^2 \alpha}=\sec \alpha \end{aligned}$