Hyperbola

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

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 $

Given,

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

Let $ y=m x \pm \sqrt{a^{2} m^{2}-b^{2}} $ be the tangent

$ \begin{array}{l} \text { put } x=1, y=1 \\\\ \quad 1=m \pm \sqrt{2 m^{2}-4} \\\\ 1-m= \pm \sqrt{2 m^{2}-4} \\\\ m^{2}+1-2 m=2 m^{2}-4 \\\\ m^{2}+2 m-5=0 \\\\ m=\frac{-2 \pm \sqrt{4+20}}{2} \Rightarrow m=-1 \pm \sqrt{6} \end{array} $

Given, hyperbola, $9 x^2-16 y^2=144$

$ \Rightarrow \quad\left(\frac{x}{4}\right)^2-\left(\frac{y}{3}\right)^2=1 $

Here, $a=4$ and $b=3$

$ \begin{array}{ll} \because & a>b \\\\ \therefore & e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{9}{16}}=\frac{5}{4} \end{array} $

Asymptotes $y= \pm \frac{b}{a} x= \pm \frac{3}{4} x$

Latus rectums $x= \pm a e= \pm 4\left(\frac{5}{4}\right)= \pm 5$

$ \because \quad p>0 $

$\therefore$ Consider the latus rectum $x=5$

The points of intersection of latus rectum $x=5$ with asymptotes $y= \pm \frac{3}{4} x$ are

$ \begin{array}{ll} & \left(5, \frac{3}{4}(5)\right) \text { and }\left(5,-\frac{3}{4}(5)\right) \\\\ \because \quad & q>0 \end{array} $

$ \therefore \quad(p, q)=\left(5, \frac{15}{4}\right) $

Therefore, $q=\frac{15}{4}$

We have,

$ \begin{aligned} & H: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \quad \ldots \text { (i) } \\\\ & C: x^2+y^2=5 \quad \ldots \text { (ii) } \end{aligned} $

The equations of tangents drawn from an external point ( $x_1, y_1$ ) to hyperbola are $\left(y-y_1\right)=m_1\left(x-x_1\right)$ and $\left(y-y_1\right)=m_2\left(x-x_2\right)$

where $m_1, m_2$ are the roots of the

$ \begin{aligned} & \text { equation }\left(x_1^2-a^2\right) m^2-2 x_1 y_1 m \\\\ & +y_1^2+b^2=0 \quad \ldots \text { (iii) } \end{aligned} $

$ \Rightarrow m_1 m_2=\frac{y_1^2+b^2}{x_1^2-a^2} \quad \ldots \text { (iv) } $

$\because m_1 m_2=-1$ (tangents are perpendicular) and $\left(x_1, y_1\right)$ lies on circle

$ \Rightarrow \quad x_1^2+y_1^2=5 $

From Eq. (iv), $y_1^2+b^2=-x_1^2+a^2$

$ \begin{array}{ll} \Rightarrow & x_1^2+y_1^2=a^2-b^2 \Rightarrow a^2-4=5 \\\\ \Rightarrow & a^2=9 \Rightarrow a=3 \end{array} $

$2 x+\sqrt{6} y=2$ touches $x^2-2 y^2=4$ at

$ \begin{aligned} & P(h, k) \\ & T=0 \Rightarrow h x-2 k y=4 \end{aligned} $

On comparing, $\frac{h}{2}=\frac{-2 k}{\sqrt{6}}=\frac{4}{2}$

$\Rightarrow h=4$ and $k=-\sqrt{6}$

So, $\quad P(h, k)=(4,-\sqrt{6})$

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

∴ Asymptotes of hyperbola

$ y= \pm \frac{b}{a} x $

Then, angle between asymptotes $=30^{\circ}$

$ \begin{aligned} \Rightarrow \tan 30^{\circ} & =\frac{\frac{b}{a}-\left(-\frac{b}{a}\right)}{1+\left(\frac{b}{a}\right)\left(-\frac{b}{a}\right)} \\ & =\frac{(2 b / a)}{\left(1-b^2 / a^2\right)} \\ \Rightarrow \quad \frac{1}{\sqrt{3}} & =\frac{2(b / a)}{1-(b / a)^2} \end{aligned} $

Let $b / a=x$

$ \begin{aligned} &\begin{aligned} & \text { So, } \frac{1}{\sqrt{3}}=\frac{2 x}{1-x^2} \\ & \Rightarrow 1-x^2=2 \sqrt{3} x \\ & \Rightarrow x^2+2 \sqrt{3} x-1=0 \\ & \begin{array}{l} \therefore x=\frac{-2 \sqrt{3} \pm \sqrt{12+4}}{2}=\frac{-2 \sqrt{3} \pm 4}{2} \\ =-\sqrt{3} \pm 2 \\ =2-\sqrt{3} \quad\{\because x>0\} \end{array} \end{aligned}\\ &\text { Therefore, } \frac{b}{a}=2-\sqrt{3}\\ &\begin{aligned} \therefore e & =\sqrt{1+\left(\frac{b}{a}\right)^2}=\sqrt{1+(2-\sqrt{3})^2} \\ & =\sqrt{1+4+3-4 \sqrt{3}}=\sqrt{8-4 \sqrt{3}} \\ & =\sqrt{6}-\sqrt{2} \end{aligned} \end{aligned} $

$ \text { Area of } P A B C=\frac{1}{2}\left|\begin{array}{ll} (x-2) & (-1-1) \\ (y-1) & (-1-2) \end{array}\right| $

$ =\frac{1}{2}|-3(x-2)-2(y-1)| $

$ \begin{aligned} & =\frac{1}{2}|-3 x+6-2 y+2| \\ & =\frac{1}{2}|-3 x-2 y+8| \end{aligned} $

$2 \times$ Area of $\triangle P A B$

$ \begin{aligned} & =2 \times \frac{1}{2}\left|\begin{array}{lll} x & y & 1 \\ 1 & 2 & 1 \\ 2 & 1 & 1 \end{array}\right| \\ & =x(2-1)-y(1-2)+1(1-4) \\ & =x+y-3 \end{aligned} $

According to the question,

$ \begin{gathered} \frac{1}{2}|-3 x-2 y+8|=(x+y-3) \\ \Rightarrow \quad|-3 x-2 y+8|=2(x+y-3) \\ \Rightarrow 9 x^2+4 y^2+64+2(6 x y-16 y-24 x) \\ =4\left(x^2+y^2+9+2(x y-3 y-3 x)\right. \\ \Rightarrow 9 x^2+4 y^2+12 x y-32 y-48 x+64 \\ =4 x^2+4 y^2+36+8 x y-24 y-24 x \\ \Rightarrow 5 x^2+4 x y-8 y-24 x+28=0 \end{gathered} $

$\frac{x^2}{6}-\frac{y^2}{2}=1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For slope of hyperbola

$ \frac{2 x}{6}-\frac{2 y}{2}\left(\frac{d y}{d x}\right)=0 \Rightarrow \frac{d y}{d x}=\frac{x}{3 y} $

Then, slope of normal $=-\frac{3 y}{x}$

From the equation of normal

$ y=-x-n $

Slope $=-1$

So, $\frac{-3 y}{x}=-1 \Rightarrow x=3 y$

By using Eq. (i), we get

$ \begin{array}{rlrl} & & \frac{9 y^2}{6}-\frac{y^2}{2} & =1 \\ \Rightarrow & & y^2 & =1 \Rightarrow y= \pm 1 \\ \text { Then, } & & x & = \pm 3 \\ \text { Now, } & & x y & =-n \\ \Rightarrow & ( \pm 1)( \pm 3 y & =-n \Rightarrow n= \pm 4 \end{array} $

$y=m x+c$ is tangent to hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then point of contact is

$ \left(-\frac{a^2 m}{c},-\frac{b^2}{c}\right) $

Here, $a^2=25, b^2=9$

$ m=m, c=4 $

∴ Point of contact $\left(-\frac{25 \times m}{4},-\frac{9}{4}\right)$ and condition of tangency is

$ \begin{aligned} & \quad c^2=a^2 m^2-b^2 \\ & \Rightarrow \quad 16=25 m^2-9 \Rightarrow m=1(m>0) \\ & \therefore \text { Point of contact }\left(-\frac{25}{4},-\frac{9}{4}\right) \end{aligned} $

The given equation of hyperbola is

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

Now, normal equation at $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$ are

$a x \cos \theta+b y \cot \theta=a^2+b^2$

and $a x \cos \phi+b y \cot \phi=a^2+b^2$

Where, $\phi=\frac{\pi}{2}-\theta$ and $(h, k)$ is the point of intersection of the normals drawn at $P$ and $Q$.

So, $a h \cos \theta+b k \cot \theta=a^2+b^2$

and $a h \sin \theta+b k \tan \theta=a^2+b^2$

Now, multiply Eq. (i) by $\sin \theta$ and Eq. (ii) by $\cos \theta$ and subtract them, we get

$ \begin{aligned} & b k(\cos \theta-\sin \theta)=\left(a^2+b^2\right)(\sin \theta-\cos \theta) \\ & \Rightarrow\left(a^2+b^2+b k\right)(\sin \theta-\cos \theta)=0 \\ & \Rightarrow b k+a^2+b^2=0 \quad(\because \sin \theta-\cos \theta \neq 0) \\ & \therefore \quad \quad k=-\frac{\left(a^2+b^2\right)}{b} \quad \forall \theta \end{aligned} $

Hence, the value of $k=-\frac{\left(a^2+b^2\right)}{b}$

To find the equation of the conjugate hyperbola, we are given the equation of a hyperbola as:

$ 9x^2 - 16y^2 + 72x - 32y - 16 = 0 $

First, we'll convert this into the standard form of a hyperbola. Starting with:

$ 9(x^2 + 8x) - 16(y^2 + 2y) = 16 $

Complete the square for both $x$ and $y$:

For $x$:

$ x^2 + 8x \quad \Rightarrow \quad (x+4)^2 - 16 $

For $y$:

$ y^2 + 2y \quad \Rightarrow \quad (y+1)^2 - 1 $

Substituting these back:

$ 9((x+4)^2 - 16) - 16((y+1)^2 - 1) = 16 $

Simplify:

$ 9(x+4)^2 - 144 - 16(y+1)^2 + 16 = 16 $

$ 9(x+4)^2 - 16(y+1)^2 = 144 $

This yields the standard form of the hyperbola:

$ \frac{(x+4)^2}{16} - \frac{(y+1)^2}{9} = 1 $

Now, for the conjugate hyperbola, the equation is:

$ \frac{(x+4)^2}{16} - \frac{(y+1)^2}{9} = -1 $

This can be rewritten as:

$ 9(x+4)^2 - 16(y+1)^2 = -144 $

Expanding this:

$ 9(x^2 + 8x + 16) - 16(y^2 + 2y + 1) = -144 $

Which simplifies to:

$ 9x^2 + 72x + 144 - 16y^2 - 32y - 16 = -144 $

Combining the constants gives:

$ 9x^2 + 72x - 16y^2 - 32y + 288 - 16 = 0 $

Thus, the equation of the conjugate hyperbola is:

$ 9x^2 - 16y^2 + 72x - 32y + 272 = 0 $

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 assume equation of hyperbola

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

Given that length of transverse axis, $2 a=12$ or $a=6$

∵ Point $(8,2)$ lies on hyperbola, then it must satisfy the equation of hyperbola

$ \begin{aligned} i.e.,\frac{(8)^2}{a^2}-\frac{(2)^2}{b^2} & =1 \text { or } \frac{64}{a^2}-\frac{4}{b^2}=1 \\ \frac{64}{(6)^2}-\frac{4}{b^2} & =1 \text { or } \frac{4}{b^2}=\frac{16}{9}-1 \\ & =\frac{7}{9} \text { or } b^2=\frac{36}{7} \end{aligned} $

As we know eccentricity of parabola,

$ e=\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{(6)^2+\frac{36}{7}}}{6}=\frac{2 \sqrt{2}}{\sqrt{7}} $

$p$ and $q$ are eccentricity of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its conjugate of hyperbola.

$ \begin{align*} & \therefore p^2=\frac{a^2+b^2}{a^2} \text { and } q^2=\frac{a^2+b^2}{b^2} \\ & \quad \frac{x^2}{p^2}+\frac{y^2}{q^2}=1 \\ & \Rightarrow a^2 x^2+b^2 y^2=a^2+b^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{align*} $

and pair of line, $x^2-y^2=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

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

$ (1,1),(-1,-1),(-1,1)(1,-1) $

$ \begin{aligned} &\text { Area of square }\\ &A B C D,=A B^2=(2)^2=4 \end{aligned} $

We have,

$ \begin{aligned} x^2+y^2 & =a^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\and \,\,\,\,\,x y & =b^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Let four points

$A\left(x_1, y_1\right), B\left(x_2, y_2\right), C\left(x_3, y_3\right)$ and $D\left(x_4, y_4\right)$

$ \begin{array}{ll} \therefore & O A^2=x_1^2+y_1^2 \\ & O B^2=x_2^2+y_2^2 \\ & O C^2=x_3^2+y_3^2 \\ \text { and } & O D^2=x_4^2+y_4^2 \end{array} $

Such that $O A^2+O B^2+O C^2+O D^2$

$ =\Sigma x_1^2+\Sigma y_1^2=4 a^2 $

From Eq. (ii) $x=\frac{b^2}{y}$

From Eq. (i) $\left(\frac{b^2}{y}\right)^2+y^2=a^2$

$ \begin{aligned} & \Rightarrow & b^4+y^4 & =a^2 y^2 \\ & \Rightarrow & y^4-a^2 y^2+b^4 & =0 \end{aligned} $

This is an equation of 4 th degree in $y$ and its four roots are $y_1, y_2, y_3, y_4$, then $y_1 y_2 y_3 y_4=b^4$

Let the equation of hyperbola be

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

We have,

$ \begin{aligned} & \quad e=\sqrt{2} \text { and } 2 a e=16 \Rightarrow 2 a(\sqrt{2})=16 \\ & \Rightarrow \quad a=\frac{16}{2 \sqrt{2}}=4 \sqrt{2} \end{aligned} $

Now, $e=\sqrt{2} \Rightarrow e^2=2$

$ \begin{aligned} & \Rightarrow 1+\frac{b^2}{a^2}=2 \Rightarrow \frac{b^2}{a^2}=1 \\ & \Rightarrow \quad b^2=a^2 \Rightarrow b=a=4 \sqrt{2} \end{aligned} $

∴ Equation of hyperbola is

$ \frac{x^2}{32}-\frac{y^2}{32}=1 \Rightarrow x^2-y^2=32 $