Hyperbola Questions in English

(C) Let $e$ and $e^{\prime}$ be the eccentricities of a hyperbola and its conjugate hyperbola respectively.
We know the relation between them is given by $\frac{1}{e^2} + \frac{1}{(e^{\prime})^2} = 1$.
Given $e = \sqrt{3}$,so $e^2 = 3$.
Substituting the value,we get $\frac{1}{3} + \frac{1}{(e^{\prime})^2} = 1$.
$\frac{1}{(e^{\prime})^2} = 1 - \frac{1}{3} = \frac{2}{3}$.
Therefore,$(e^{\prime})^2 = \frac{3}{2}$,which implies $e^{\prime} = \sqrt{\frac{3}{2}}$.

(C) The given equation $9x^2 - 16y^2 = 9$ can be written as $\frac{x^2}{1^2} - \frac{y^2}{(3/4)^2} = 1$. Here $a=1, b=3/4$.
The parametric points on a hyperbola are $(a \sec \theta, b \tan \theta)$.
For $\theta = \frac{\pi}{4}$,$P_1 = (1 \cdot \sqrt{2}, \frac{3}{4} \cdot 1) = (\sqrt{2}, \frac{3}{4})$.
For $\theta = \frac{\pi}{3}$,$P_2 = (1 \cdot 2, \frac{3}{4} \cdot \sqrt{3}) = (2, \frac{3\sqrt{3}}{4})$.
The distance $D = \sqrt{(2 - \sqrt{2})^2 + (\frac{3\sqrt{3}}{4} - \frac{3}{4})^2} = \sqrt{(4 + 2 - 4\sqrt{2}) + \frac{9}{16}(3 + 1 - 2\sqrt{3})} = \sqrt{6 - 4\sqrt{2} + \frac{9}{4} - \frac{9\sqrt{3}}{8}} = \sqrt{\frac{48 - 32\sqrt{2} + 18 - 9\sqrt{3}}{8}} = \sqrt{\frac{66 - 32\sqrt{2} - 9\sqrt{3}}{8}} = \frac{1}{2\sqrt{2}} \sqrt{66 - 32\sqrt{2} - 9\sqrt{3}}$.
The Assertion is true.
The Reason states $x = a \cosh t, y = b \sinh t$ are parametric equations for $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,which is incorrect; the correct parametric equations are $x = a \sec \theta, y = b \tan \theta$.
Thus,$(A)$ is true but $(R)$ is false.

(D) Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The centre is $(0, 0)$.
The endpoints of the latus rectum are $(ae, \frac{b^2}{a})$ and $(ae, -\frac{b^2}{a})$.
The angle subtended by the latus rectum at the centre is $120^{\circ}$.
Thus,the angle made by the line joining the centre to one endpoint $(ae, \frac{b^2}{a})$ with the $x$-axis is $60^{\circ}$.
Therefore,$\tan(60^{\circ}) = \frac{b^2/a}{ae} = \frac{b^2}{a^2e}$.
Since $\sqrt{3} = \frac{b^2}{a^2e}$ and $b^2 = a^2(e^2 - 1)$,we have $\sqrt{3} = \frac{a^2(e^2 - 1)}{a^2e} = \frac{e^2 - 1}{e}$.
This gives $e^2 - \sqrt{3}e - 1 = 0$.
Using the quadratic formula,$e = \frac{\sqrt{3} \pm \sqrt{3 - 4(1)(-1)}}{2} = \frac{\sqrt{3} \pm \sqrt{7}}{2}$.
Since $e > 1$,we take $e = \frac{\sqrt{3} + \sqrt{7}}{2}$.

(B) The equation of the chord joining points $A(\theta_1)$ and $B(\theta_2)$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is given by:
$\frac{x}{a} \cos \left(\frac{\theta_1-\theta_2}{2}\right)-\frac{y}{b} \sin \left(\frac{\theta_1+\theta_2}{2}\right)=\cos \left(\frac{\theta_1+\theta_2}{2}\right)$
Since the chord passes through the focus $S(ae, 0)$,we substitute $x=ae$ and $y=0$ into the equation:
$\frac{ae}{a} \cos \left(\frac{\theta_1-\theta_2}{2}\right) = \cos \left(\frac{\theta_1+\theta_2}{2}\right)$
$e \cos \left(\frac{\theta_1-\theta_2}{2}\right) = \cos \left(\frac{\theta_1+\theta_2}{2}\right)$
Using the relation $e = \frac{\sqrt{a^2+b^2}}{a}$,we get:
$\frac{\sqrt{a^2+b^2}}{a} \cos \left(\frac{\theta_1-\theta_2}{2}\right) = \cos \left(\frac{\theta_1+\theta_2}{2}\right)$
$\sqrt{a^2+b^2} \cos \left(\frac{\theta_1-\theta_2}{2}\right) = a \cos \left(\frac{\theta_1+\theta_2}{2}\right)$
Comparing this with the given equation $a \cos \left(\frac{\theta_1+\theta_2}{2}\right) = k \cos \left(\frac{\theta_1-\theta_2}{2}\right)$,we find:
$k = \sqrt{a^2+b^2}$

(A) The equation of the hyperbola is $4x^2 - y^2 = 4a^2$,which can be written as $\frac{x^2}{a^2} - \frac{y^2}{4a^2} = 1$.
Since the chord $x \cos \phi + y \sin \phi = P$ subtends a right angle at the origin $(0,0)$,we homogenize the equation of the hyperbola using the line equation: $\frac{x^2}{a^2} - \frac{y^2}{4a^2} = \left(\frac{x \cos \phi + y \sin \phi}{P}\right)^2$.
Expanding this,we get $x^2 \left(\frac{1}{a^2} - \frac{\cos^2 \phi}{P^2}\right) - y^2 \left(\frac{1}{4a^2} + \frac{\sin^2 \phi}{P^2}\right) - \frac{2xy \cos \phi \sin \phi}{P^2} = 0$.
For the chord to subtend a right angle at the origin,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$\left(\frac{1}{a^2} - \frac{\cos^2 \phi}{P^2}\right) - \left(\frac{1}{4a^2} + \frac{\sin^2 \phi}{P^2}\right) = 0$.
$\frac{1}{a^2} - \frac{1}{4a^2} - \frac{1}{P^2} (\cos^2 \phi + \sin^2 \phi) = 0$.
$\frac{3}{4a^2} - \frac{1}{P^2} = 0$ $\Rightarrow P^2 = \frac{4a^2}{3}$ $\Rightarrow P = \frac{2a}{\sqrt{3}}$.
Since $P$ is the perpendicular distance from the origin to the chord,it represents the radius of the circle touched by these chords. Thus,the radius is $\frac{2a}{\sqrt{3}}$.

(B) The equation of the chord of the hyperbola $S: 2x^2 - 3y^2 - 12 = 0$ with midpoint $M(h, k)$ is given by $T = S_1$,where $T = 2xh - 3yk - 12$ and $S_1 = 2h^2 - 3k^2 - 12$.
Thus,the equation of the chord is $2xh - 3yk = 2h^2 - 3k^2$.
Comparing this with the given chord equation $4x - 3y = 5$,we get:
$\frac{2h}{4} = \frac{-3k}{-3} = \frac{2h^2 - 3k^2}{5}$.
From $\frac{2h}{4} = k$,we have $k = \frac{h}{2}$.
Substituting $k = \frac{h}{2}$ into $\frac{2h}{4} = \frac{2h^2 - 3(h/2)^2}{5}$:
$\frac{h}{2} = \frac{2h^2 - \frac{3h^2}{4}}{5} = \frac{5h^2}{20} = \frac{h^2}{4}$.
So,$\frac{h}{2} = \frac{h^2}{4}$ $\Rightarrow 2h = h^2$ $\Rightarrow h(h - 2) = 0$.
Since the chord exists,$h \neq 0$ (as $h=0$ gives $k=0$,which does not satisfy the chord equation $4(0)-3(0)=5$).
Thus,$h = 2$.
Then $k = \frac{2}{2} = 1$.
The midpoint is $(2, 1)$.

(C) The equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{16} = 1$. Here $a^2 = 9$ and $b^2 = 16$,so $a = 3$ and $b = 4$.
The equation of the tangent at $P(x_1, y_1) = (3 \sqrt{2}, 4)$ is given by $\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1$.
Substituting the values,we get $\frac{x(3 \sqrt{2})}{9} - \frac{y(4)}{16} = 1$,which simplifies to $\frac{x \sqrt{2}}{3} - \frac{y}{4} = 1$.
The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$.
The directrix of the hyperbola is $x = \frac{a}{e} = \frac{3}{5/3} = \frac{9}{5}$.
Since the point $Q(\alpha, \beta)$ lies on the directrix,$\alpha = \frac{9}{5}$.
Substituting $x = \frac{9}{5}$ into the tangent equation: $\frac{(9/5) \sqrt{2}}{3} - \frac{y}{4} = 1$.
$\frac{3 \sqrt{2}}{5} - \frac{y}{4} = 1 \implies \frac{y}{4} = \frac{3 \sqrt{2}}{5} - 1 = \frac{3 \sqrt{2} - 5}{5}$.
Thus,$y = \beta = \frac{12 \sqrt{2} - 20}{5}$.
Since $Q$ is in the fourth quadrant,we check the sign. $\sqrt{2} \approx 1.414$,so $12(1.414) \approx 16.968$. $16.968 - 20 < 0$,which is consistent with the fourth quadrant.

(C) The focus $S$ is $(ae, 0)$. Given $Q = (0, 1)$,$S Q^2 = (ae)^2 + 1^2 = 26$,so $a^2 e^2 = 25$,which means $ae = 5$.
For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{9} = 1$,we have $b^2 = 9$,so $e^2 = 1 + \frac{9}{a^2}$.
Substituting $e^2 = \frac{25}{a^2}$,we get $\frac{25}{a^2} = 1 + \frac{9}{a^2}$,which implies $\frac{16}{a^2} = 1$,so $a = 4$.
Then $e = \frac{5}{4}$. The point $P$ is $(4 \sec \theta, 3 \tan \theta)$.
The distance $SP = 6$. Using the focal distance formula $SP = |ae \sec \theta - a|$ is not directly applicable as $S$ is $(ae, 0)$,so $SP^2 = (4 \sec \theta - 5)^2 + (3 \tan \theta - 0)^2 = 36$.
$16 \sec^2 \theta - 40 \sec \theta + 25 + 9 \tan^2 \theta = 36$.
Using $\tan^2 \theta = \sec^2 \theta - 1$,we get $16 \sec^2 \theta - 40 \sec \theta + 25 + 9 \sec^2 \theta - 9 = 36$.
$25 \sec^2 \theta - 40 \sec \theta - 20 = 0$,which simplifies to $5 \sec^2 \theta - 8 \sec \theta - 4 = 0$.
$(5 \sec \theta + 2)(\sec \theta - 2) = 0$.
Since $|\sec \theta| \geq 1$,we have $\sec \theta = 2$,which gives $\theta = \frac{\pi}{3}$.

(B) The given line is $2x + \sqrt{6}y = 2$,which can be written as $2x + \sqrt{6}y - 2 = 0$.
The equation of the hyperbola is $x^2 - 2y^2 = 4$.
Let the point of contact be $(x_1, y_1)$.
The equation of the tangent to the hyperbola $x^2 - 2y^2 = 4$ at $(x_1, y_1)$ is given by $xx_1 - 2yy_1 = 4$,or $xx_1 - 2yy_1 - 4 = 0$.
Since this line is the same as the given line $2x + \sqrt{6}y - 2 = 0$,the coefficients must be proportional:
$\frac{x_1}{2} = \frac{-2y_1}{\sqrt{6}} = \frac{-4}{-2}$
$\frac{x_1}{2} = 2 \Rightarrow x_1 = 4$
$\frac{-2y_1}{\sqrt{6}} = 2$ $\Rightarrow -2y_1 = 2\sqrt{6}$ $\Rightarrow y_1 = -\sqrt{6}$
Thus,the point of contact is $(4, -\sqrt{6})$.

(B) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is given by $\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$.
For the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$,we have $a^2 = 9$ and $b^2 = 4$.
The equation of the normal becomes $\frac{9x}{x_1} + \frac{4y}{y_1} = 9 + 4 = 13$.
Comparing this with the given line $x + y = -k$,we have the ratios:
$\frac{9/x_1}{1} = \frac{4/y_1}{1} = \frac{13}{-k}$.
This gives $x_1 = -\frac{9k}{13}$ and $y_1 = -\frac{4k}{13}$.
Since $(x_1, y_1)$ lies on the hyperbola,we substitute these values:
$\frac{(-9k/13)^2}{9} - \frac{(-4k/13)^2}{4} = 1$.
$\frac{81k^2}{169 \times 9} - \frac{16k^2}{169 \times 4} = 1$.
$\frac{9k^2}{169} - \frac{4k^2}{169} = 1$.
$\frac{5k^2}{169} = 1$.
$k^2 = \frac{169}{5}$.
$k = \pm \frac{13}{\sqrt{5}}$.

(B) Given the equations:
$x^2+y^2=4$
$xy=\sqrt{3}$
Substitute $y=\frac{\sqrt{3}}{x}$ into the circle equation:
$x^2 + \frac{3}{x^2} = 4$
$x^4 - 4x^2 + 3 = 0$
$(x^2-3)(x^2-1) = 0$
So,$x^2=3$ or $x^2=1$.
Since $\alpha > \beta > 0$,we have $x^2=3$ and $y^2=1$.
Thus,$\alpha = \sqrt{3}$ and $\beta = 1$.
Point $P = (\sqrt{3}, 1)$.
The equation of the tangent to the hyperbola $xy=c^2$ at $(x_1, y_1)$ is $xy_1 + yx_1 = 2c^2$.
Here $c^2 = \sqrt{3}$,$x_1 = \sqrt{3}$,$y_1 = 1$.
So,$x(1) + y(\sqrt{3}) = 2\sqrt{3}$.
$x + \sqrt{3}y = 2\sqrt{3}$.

(D) The locus of the point of intersection of two perpendicular tangents to a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is its director circle,given by the equation $x^2 + y^2 = a^2 - b^2$.
Given the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{4} = 1$,we have $b^2 = 4$.
The director circle is $x^2 + y^2 = a^2 - 4$.
Comparing this with the given circle $x^2 + y^2 = 5$,we get $a^2 - 4 = 5$.
$a^2 = 9 \Rightarrow a = 3$ (since $a > 0$ for a hyperbola).

(C) The equation of the hyperbola is $2x^2-y^2=4$,which can be written as $\frac{x^2}{2}-\frac{y^2}{4}=1$.
Here,$a^2=2$ and $b^2=4$.
The equation of a tangent with slope $m$ to the hyperbola is $y=mx \pm \sqrt{a^2m^2-b^2}$,which becomes $y=mx \pm \sqrt{2m^2-4}$.
Since the tangent passes through the point $(1,1)$,we substitute $x=1$ and $y=1$:
$1 = m(1) \pm \sqrt{2m^2-4}$
$1-m = \pm \sqrt{2m^2-4}$
Squaring both sides,we get:
$(1-m)^2 = 2m^2-4$
$1+m^2-2m = 2m^2-4$
$m^2+2m-5 = 0$
Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$m = \frac{-2 \pm \sqrt{2^2-4(1)(-5)}}{2(1)} = \frac{-2 \pm \sqrt{4+20}}{2} = \frac{-2 \pm \sqrt{24}}{2} = \frac{-2 \pm 2\sqrt{6}}{2} = -1 \pm \sqrt{6}$.

(A) Let the point $P$ on the hyperbola $x^2-y^2=c^2$ be $(c \sec \theta, c \tan \theta)$.
Equation of tangent at $P$ is $x \sec \theta - y \tan \theta = c$.
$X$-intercept $T$ is $(c \cos \theta, 0)$.
Equation of normal at $P$ is $x \cos \theta + y \cot \theta = 2c \sec \theta$.
$Y$-intercept $N$ is $(0, 2c \csc \theta)$.
Let the midpoint of $TN$ be $(h, k) = (\frac{c}{2 \cos \theta}, c \csc \theta)$.
Then $\cos \theta = \frac{c}{2h}$ and $\sin \theta = \frac{c}{k}$.
Using $\cos^2 \theta - \sin^2 \theta = 1$ is not applicable here,but we use $\sec^2 \theta - \tan^2 \theta = 1$ or simply $\frac{1}{\cos^2 \theta} - \frac{1}{\sin^2 \theta} = \frac{1}{\cos^2 \theta} - \frac{1}{\sin^2 \theta}$.
Actually,$\frac{1}{\cos^2 \theta} - \frac{1}{\sin^2 \theta} = \frac{4h^2}{c^2} - \frac{k^2}{c^2} = 1$ is incorrect.
Correct approach: $\cos \theta = \frac{c}{2h}$ and $\sin \theta = \frac{c}{k}$.
Since $\cos^2 \theta + \sin^2 \theta = 1$ is not the identity,we use $\frac{1}{\cos^2 \theta} - \frac{1}{\sin^2 \theta} = \sec^2 \theta - \csc^2 \theta$.
Wait,the locus is $\frac{c^2}{4x^2} - \frac{y^2}{c^2} = 1$.

(B) The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$.
Given the tangent $y = mx + 4$,we have $c = 4$.
Comparing $c^2 = a^2m^2 - b^2$,we get $16 = 25m^2 - 9$.
$25m^2 = 25 \Rightarrow m^2 = 1$. Since $m$ is usually positive for such forms,$m = 1$.
The point of contact for a tangent $y = mx + c$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $\left(\frac{a^2m}{c}, \frac{b^2}{c}\right)$.
Substituting $a^2 = 25, b^2 = 9, m = 1, c = 4$:
Point of contact = $\left(\frac{25 \times 1}{4}, \frac{9}{4}\right) = \left(\frac{25}{4}, \frac{9}{4}\right)$.

(D) The equation of the hyperbola is $x^2 - 4y^2 = 4$, which can be written as $\frac{x^2}{4} - \frac{y^2}{1} = 1$.
Here, $a = 2$ and $b = 1$.
The parametric coordinates are given by $(a \sec \theta, b \tan \theta) = (2 \sec \theta, \tan \theta)$.
Calculating the points:
$P(\frac{\pi}{4}) = (2 \sec \frac{\pi}{4}, \tan \frac{\pi}{4}) = (2 \sqrt{2}, 1)$
$Q(\frac{5 \pi}{4}) = (2 \sec \frac{5 \pi}{4}, \tan \frac{5 \pi}{4}) = (-2 \sqrt{2}, 1)$
$R(\frac{3 \pi}{4}) = (2 \sec \frac{3 \pi}{4}, \tan \frac{3 \pi}{4}) = (-2 \sqrt{2}, -1)$
$T(\frac{7 \pi}{4}) = (2 \sec \frac{7 \pi}{4}, \tan \frac{7 \pi}{4}) = (2 \sqrt{2}, -1)$
These points form a rectangle with vertices $(2 \sqrt{2}, 1), (-2 \sqrt{2}, 1), (-2 \sqrt{2}, -1),$ and $(2 \sqrt{2}, -1)$.
The length of the sides are:
Width $= |2 \sqrt{2} - (-2 \sqrt{2})| = 4 \sqrt{2}$
Height $= |1 - (-1)| = 2$
Area $= \text{width} \times \text{height} = 4 \sqrt{2} \times 2 = 8 \sqrt{2}$ square units.
Thus, option $D$ is correct.

(A) Let the point $P$ be $(h, k)$. The equation of a line passing through $(h, k)$ with slope $m$ is $y - k = m(x - h)$,or $y = mx + (k - mh)$.
For this line to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the condition of tangency is $c^2 = a^2m^2 - b^2$,where $c = k - mh$.
Substituting $c$,we get $(k - mh)^2 = a^2m^2 - b^2$.
Expanding this,$k^2 - 2mhk + m^2h^2 = a^2m^2 - b^2$.
Rearranging as a quadratic in $m$: $m^2(h^2 - a^2) - 2mhk + (k^2 + b^2) = 0$.
Let the slopes of the two tangents be $m_1$ and $m_2$. The product of the roots is $m_1m_2 = \frac{k^2 + b^2}{h^2 - a^2}$.
Given that the product of the slopes is $k^2$,we have $\frac{k^2 + b^2}{h^2 - a^2} = k^2$.
Replacing $(h, k)$ with $(x, y)$,the locus of $P$ is $\frac{y^2 + b^2}{x^2 - a^2} = k^2$,which simplifies to $y^2 + b^2 = k^2(x^2 - a^2)$.

(B) The given equation of the hyperbola is $3x^2 - y^2 = 3$,which can be written as $\frac{x^2}{1} - \frac{y^2}{3} = 1$.
Here,$a^2 = 1$ and $b^2 = 3$.
The equation of a tangent with slope $m$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$.
Given the line is $y = 2x + 4$,the slope $m = 2$.
Substituting $m = 2, a^2 = 1, b^2 = 3$ into the tangent equation:
$y = 2x \pm \sqrt{1(2)^2 - 3} = 2x \pm \sqrt{4 - 3} = 2x \pm 1$.
The two parallel tangents are $2x - y + 1 = 0$ and $2x - y - 1 = 0$.
The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
$d = \frac{|1 - (-1)|}{\sqrt{2^2 + (-1)^2}} = \frac{2}{\sqrt{4 + 1}} = \frac{2}{\sqrt{5}}$.

(A) The given hyperbola is $5 x^2-y^2=5$,which can be rewritten as $\frac{x^2}{1}-\frac{y^2}{5}=1$.
Here,$a^2=1$ and $b^2=5$.
The equation of a tangent with slope $m$ is $y=m x \pm \sqrt{a^2 m^2-b^2}$,which becomes $y=m x \pm \sqrt{m^2-5}$.
Since the tangent passes through $(2,8)$,we have $8=2 m \pm \sqrt{m^2-5}$,or $(8-2 m)^2 = m^2-5$.
Expanding this,$64+4 m^2-32 m = m^2-5$,which simplifies to $3 m^2-32 m+69=0$.
Factoring the quadratic equation,$(3 m-23)(m-3)=0$,so $m=3$ or $m=\frac{23}{3}$.
For $m=3$,the tangent equation is $y=3 x \pm \sqrt{3^2-5} \Rightarrow y=3 x \pm 2$.
Thus,$3 x-y+2=0$ or $3 x-y-2=0$ are the tangents.

(D) The equation of the normal at point $\theta$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is given by $ax \cos \theta + by \cot \theta = a^2+b^2$.
For point $P(\theta)$,the normal is $ax \cos \theta + by \cot \theta = a^2+b^2$ (Equation $1$).
For point $Q(\phi)$,the normal is $ax \cos \phi + by \cot \phi = a^2+b^2$.
Since $\phi = \frac{\pi}{2} - \theta$,we have $\cos \phi = \sin \theta$ and $\cot \phi = \tan \theta$.
Thus,the normal at $Q$ is $ax \sin \theta + by \tan \theta = a^2+b^2$ (Equation $2$).
Solving for the intersection $(h, k)$ by eliminating $x$ or $y$:
From Equation $1$,$ax \cos \theta = a^2+b^2 - by \cot \theta$.
From Equation $2$,$ax \sin \theta = a^2+b^2 - by \tan \theta$.
Using Cramer's rule or substitution,we find the $y$-coordinate $k$:
$k = \frac{(a^2+b^2)a(\cos \theta - \sin \theta)}{ab(\sin \theta - \cos \theta)} = \frac{-(a^2+b^2)}{b}$.
Therefore,$k = -\left(\frac{a^2+b^2}{b}\right)$.

(B) The given equation of the line is $x+y+n=0$,which can be written as $y=-x-n$.
Comparing this with $y=mx+c$,we get $m=-1$ and $c=-n$.
The condition for the line $y=mx+c$ to be a normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $c^2 = \frac{(a^2+b^2)^2 m^2}{a^2-b^2 m^2}$.
Here,$a^2=6$ and $b^2=2$.
Substituting the values,we get $(-n)^2 = \frac{(6+2)^2 (-1)^2}{6-2(-1)^2}$.
$n^2 = \frac{8^2 \times 1}{6-2} = \frac{64}{4} = 16$.
Therefore,$n = \pm 4$.

(D) The equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.
The equation of the normal at a point $P(a \sec \theta, b \tan \theta)$ is $\frac{ax}{\sec \theta} + \frac{by}{\tan \theta} = a^2+b^2$,which simplifies to $ax \cos \theta + by \cot \theta = a^2+b^2$.
Similarly,the equation of the normal at $Q(a \sec \phi, b \tan \phi)$ is $ax \cos \phi + by \cot \phi = a^2+b^2$.
Given $\theta + \phi = \frac{\pi}{2}$,we have $\phi = \frac{\pi}{2} - \theta$. Thus,$\cos \phi = \sin \theta$ and $\cot \phi = \tan \theta$.
The two equations are:
$(1) \quad ax \cos \theta + by \cot \theta = a^2+b^2$
$(2) \quad ax \sin \theta + by \tan \theta = a^2+b^2$
To find $k$ (the $y$-coordinate),we eliminate $x$ by multiplying $(1)$ by $\sin \theta$ and $(2)$ by $\cos \theta$:
$ax \cos \theta \sin \theta + by \cot \theta \sin \theta = (a^2+b^2) \sin \theta$
$ax \sin \theta \cos \theta + by \tan \theta \cos \theta = (a^2+b^2) \cos \theta$
Subtracting the two equations:
$by(\cot \theta \sin \theta - \tan \theta \cos \theta) = (a^2+b^2)(\sin \theta - \cos \theta)$
$by(\cos \theta - \sin \theta) = (a^2+b^2)(\sin \theta - \cos \theta)$
$by = -(a^2+b^2)$
$y = -\frac{a^2+b^2}{b}$
Therefore,$k = -\frac{a^2+b^2}{b}$.

(A) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(a \sec \theta, b \tan \theta)$ is $ax \cos \theta + by \cot \theta = a^2 + b^2$.
For the given hyperbola $\frac{x^2}{4} - \frac{y^2}{9} = 1$,we have $a^2 = 4$ and $b^2 = 9$.
The normal at $A(2 \sec \theta, 3 \tan \theta)$ is $2x \cos \theta + 3y \cot \theta = 4 + 9 = 13$ $(i)$.
The normal at $B(2 \sec \phi, 3 \tan \phi)$ is $2x \cos \phi + 3y \cot \phi = 4 + 9 = 13$ (ii).
Since $\theta + \phi = \frac{\pi}{2}$,we have $\phi = \frac{\pi}{2} - \theta$. Thus,$\cos \phi = \sin \theta$ and $\cot \phi = \tan \theta$.
Substituting these into (ii): $2x \sin \theta + 3y \tan \theta = 13$ (iii).
To find the intersection $(\alpha, \beta)$,we solve the system of equations:
$2x \cos \theta + 3y \cot \theta = 13$
$2x \sin \theta + 3y \tan \theta = 13$
Subtracting the equations or using Cramer's rule,we find the $y$-coordinate $\beta$.
From $2x \cos \theta + 3y \frac{\cos \theta}{\sin \theta} = 13$,we get $2x \cos \theta \sin \theta + 3y \cos \theta = 13 \sin \theta$.
From $2x \sin \theta + 3y \frac{\sin \theta}{\cos \theta} = 13$,we get $2x \sin \theta \cos \theta + 3y \sin \theta = 13 \cos \theta$.
Subtracting the two resulting equations: $3y(\cos \theta - \sin \theta) = 13(\sin \theta - \cos \theta)$.
Since $\sin \theta \neq \cos \theta$ (for distinct points),we divide by $(\cos \theta - \sin \theta)$ to get $3y = -13$.
Therefore,$\beta = -\frac{13}{3}$.

(C) The equation of the line is $lx + my - 1 = 0$,so $n = -1$.
For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the condition for the line $lx + my + n = 0$ to be a normal is $\frac{a^2}{l^2} - \frac{b^2}{m^2} = \frac{(a^2 + b^2)^2}{n^2}$.
Substituting $n = -1$ into the condition,we get:
$\frac{a^2}{l^2} - \frac{b^2}{m^2} = \frac{(a^2 + b^2)^2}{(-1)^2} = (a^2 + b^2)^2$.

(D) The asymptotes of a rectangular hyperbola are parallel to the coordinate axes,so its equation is of the form $(x - h)(y - k) = c$.
Since the point of intersection of the two perpendicular tangents is the center of the hyperbola,we have $(h, k) = (1, 1)$.
Thus,the equation is $(x - 1)(y - 1) = c$.
Since the hyperbola passes through $(3, 2)$,we substitute these coordinates: $(3 - 1)(2 - 1) = c$ $\Rightarrow 2 \times 1 = c$ $\Rightarrow c = 2$.
Substituting $c = 2$ into the equation,we get $(x - 1)(y - 1) = 2$.
Expanding this,we get $xy - x - y + 1 = 2$,which simplifies to $xy = x + y + 1$.

(B) The equation of the hyperbola is $\frac{x^2}{20} - \frac{y^2}{4/3} = 1$. Here $a^2 = 20$ and $b^2 = \frac{4}{3}$.
Given the line $x + 3y = 7$,the slope $m = -\frac{1}{3}$.
The equation of the tangent to the hyperbola in slope form is $y = mx \pm \sqrt{a^2m^2 - b^2}$.
Substituting the values: $y = -\frac{1}{3}x \pm \sqrt{20(-\frac{1}{3})^2 - \frac{4}{3}} = -\frac{1}{3}x \pm \sqrt{\frac{20}{9} - \frac{12}{9}} = -\frac{1}{3}x \pm \sqrt{\frac{8}{9}} = -\frac{1}{3}x \pm \frac{2\sqrt{2}}{3}$.
This gives the two parallel tangents: $x + 3y - 2\sqrt{2} = 0$ and $x + 3y + 2\sqrt{2} = 0$.
The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
$d = \frac{|2\sqrt{2} - (-2\sqrt{2})|}{\sqrt{1^2 + 3^2}} = \frac{4\sqrt{2}}{\sqrt{10}} = \frac{4\sqrt{2}}{\sqrt{2}\sqrt{5}} = \frac{4}{\sqrt{5}}$.

(B) Given the hyperbola $x^2-y^2=9$ and the chord of contact $x=9$.
Substituting $x=9$ into the hyperbola equation:
$81-y^2=9$ $\Rightarrow y^2=72$ $\Rightarrow y = \pm 6\sqrt{2}$.
Thus,the points of contact are $P_1(9, 6\sqrt{2})$ and $P_2(9, -6\sqrt{2})$.
Differentiating $x^2-y^2=9$ with respect to $x$:
$2x - 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{y}$.
At $P_1(9, 6\sqrt{2})$,the slope $m_1 = \frac{9}{6\sqrt{2}} = \frac{3}{2\sqrt{2}}$.
The equation of the tangent at $P_1$ is $y - 6\sqrt{2} = \frac{3}{2\sqrt{2}}(x-9)$,which simplifies to $2\sqrt{2}y - 24 = 3x - 27$,or $3x - 2\sqrt{2}y - 3 = 0$.
At $P_2(9, -6\sqrt{2})$,the slope $m_2 = \frac{9}{-6\sqrt{2}} = -\frac{3}{2\sqrt{2}}$.
The equation of the tangent at $P_2$ is $y + 6\sqrt{2} = -\frac{3}{2\sqrt{2}}(x-9)$,which simplifies to $2\sqrt{2}y + 24 = -3x + 27$,or $3x + 2\sqrt{2}y - 3 = 0$.
Comparing with the options,$3x + 2\sqrt{2}y - 3 = 0$ is option $B$.

(A) Given equations are $x^2+y^2=a^2$ and $xy=c^2$.
From the second equation,$x = \frac{c^2}{y}$.
Substituting $x$ in the first equation:
$\left(\frac{c^2}{y}\right)^2 + y^2 = a^2$
$\frac{c^4}{y^2} + y^2 = a^2$
$c^4 + y^4 = a^2 y^2$
$y^4 - a^2 y^2 + c^4 = 0$
This is a biquadratic equation in $y$. Let the roots be $y_1, y_2, y_3, y_4$.
The equation is of the form $Ay^4 + By^3 + Cy^2 + Dy + E = 0$,where $B=0$.
By Vieta's formulas,the sum of the roots is $-\frac{B}{A} = -\frac{0}{1} = 0$.
Therefore,$y_1+y_2+y_3+y_4 = 0$.

(D) The asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are given by $\frac{x}{a} - \frac{y}{b} = 0$ and $\frac{x}{a} + \frac{y}{b} = 0$.
The product of the perpendicular distances from any point $(x_1, y_1)$ on the hyperbola to these asymptotes is given by the formula $\frac{a^2 b^2}{a^2 + b^2}$.
Given that this product is $\frac{36}{13}$,we have $\frac{a^2 b^2}{a^2 + b^2} = \frac{36}{13}$.
The eccentricity $e$ is given by $e^2 = 1 + \frac{b^2}{a^2} = \frac{a^2 + b^2}{a^2}$.
Given $e = \frac{\sqrt{13}}{3}$,so $e^2 = \frac{13}{9}$.
Thus,$\frac{a^2 + b^2}{a^2} = \frac{13}{9}$,which implies $9(a^2 + b^2) = 13a^2$,or $9b^2 = 4a^2$.
From this,$b^2 = \frac{4}{9}a^2$,so $b = \frac{2}{3}a$.
Substituting $b^2 = \frac{4}{9}a^2$ into the product equation: $\frac{a^2 (\frac{4}{9}a^2)}{a^2 + \frac{4}{9}a^2} = \frac{36}{13}$.
$\frac{\frac{4}{9}a^4}{\frac{13}{9}a^2} = \frac{36}{13} \implies \frac{4}{13}a^2 = \frac{36}{13} \implies a^2 = 9 \implies a = 3$.
Then $b = \frac{2}{3}(3) = 2$.
Therefore,$a - b = 3 - 2 = 1$.

(C) The given equation of the hyperbola is $9x^2 - 16y^2 = 144$.
Dividing by $144$,we get $\frac{x^2}{16} - \frac{y^2}{9} = 1$.
Here,$a^2 = 16$ and $b^2 = 9$,so $a = 4$ and $b = 3$.
The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.
The equation of the latus rectum is $x = ae = 4 \times \frac{5}{4} = 5$.
The equation of the asymptote is $\frac{x^2}{16} - \frac{y^2}{9} = 0$,which simplifies to $y = \pm \frac{b}{a}x = \pm \frac{3}{4}x$.
For the first quadrant,we take $y = \frac{3}{4}x$.
Substituting $x = 5$ into the asymptote equation,we get $q = \frac{3}{4}(5) = \frac{15}{4}$.
Thus,$q = \frac{15}{4}$.

(A) Given hyperbola $H: 9x^2 - 16y^2 + 72x - 32y - 16 = 0$.
Rewrite the equation by completing the squares:
$9(x^2 + 8x) - 16(y^2 + 2y) = 16$
$9(x+4)^2 - 144 - 16(y+1)^2 + 16 = 16$
$9(x+4)^2 - 16(y+1)^2 = 144$
Dividing by $144$,we get $\frac{(x+4)^2}{16} - \frac{(y+1)^2}{9} = 1$.
The equation of the conjugate hyperbola is $\frac{(x+4)^2}{16} - \frac{(y+1)^2}{9} = -1$.
Multiplying by $144$: $9(x+4)^2 - 16(y+1)^2 = -144$.
$9(x^2 + 8x + 16) - 16(y^2 + 2y + 1) = -144$
$9x^2 + 72x + 144 - 16y^2 - 32y - 16 = -144$
$9x^2 - 16y^2 + 72x - 32y + 128 = -144$
$9x^2 - 16y^2 + 72x - 32y + 272 = 0$.

(D) The angle between the asymptotes of a hyperbola is given by $2 \sec^{-1}(e)$.
Given that the angle is $30^{\circ}$,we have:
$2 \sec^{-1}(e) = 30^{\circ}$
$\sec^{-1}(e) = 15^{\circ}$
$e = \sec(15^{\circ})$
Since $\cos(15^{\circ}) = \cos(45^{\circ}-30^{\circ}) = \cos(45^{\circ}) \cos(30^{\circ}) + \sin(45^{\circ}) \sin(30^{\circ}) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$.
Therefore,$e = \frac{2\sqrt{2}}{\sqrt{3}+1} = \frac{2\sqrt{2}(\sqrt{3}-1)}{3-1} = \sqrt{6}-\sqrt{2}$.

(C) Let $(h, k)$ be any point on the hyperbola $x^2 - y^2 = 8$.
Thus,$h^2 - k^2 = 8$.
The equations of the asymptotes of the hyperbola $x^2 - y^2 = 8$ are $x + y = 0$ and $x - y = 0$.
The perpendicular distance $d_1$ from $(h, k)$ to the line $x + y = 0$ is $d_1 = \frac{|h + k|}{\sqrt{1^2 + 1^2}} = \frac{|h + k|}{\sqrt{2}}$.
The perpendicular distance $d_2$ from $(h, k)$ to the line $x - y = 0$ is $d_2 = \frac{|h - k|}{\sqrt{1^2 + (-1)^2}} = \frac{|h - k|}{\sqrt{2}}$.
The product of the lengths of the perpendiculars is $d_1 d_2 = \frac{|h + k|}{\sqrt{2}} \times \frac{|h - k|}{\sqrt{2}} = \frac{|h^2 - k^2|}{2}$.
Substituting $h^2 - k^2 = 8$,we get $d_1 d_2 = \frac{8}{2} = 4$.

(C) The equation of the given hyperbola is $9x^2-4y^2+36x+8y-4=0$.
Rewriting the equation by completing the square:
$9(x^2+4x+4) - 4(y^2-2y+1) = 4 + 36 - 4 = 36$.
$9(x+2)^2 - 4(y-1)^2 = 36$.
The combined equation of the asymptotes is obtained by setting the constant term to zero:
$9(x+2)^2 - 4(y-1)^2 = 0$.
This implies $3(x+2) = \pm 2(y-1)$.
Thus,the two asymptotes are $L_1: 3x - 2y + 8 = 0$ and $L_2: 3x + 2y + 4 = 0$.
The perpendicular distance from $(1,1)$ to $L_1$ is $d_1 = \frac{|3(1) - 2(1) + 8|}{\sqrt{3^2 + (-2)^2}} = \frac{9}{\sqrt{13}}$.
The perpendicular distance from $(1,1)$ to $L_2$ is $d_2 = \frac{|3(1) + 2(1) + 4|}{\sqrt{3^2 + 2^2}} = \frac{9}{\sqrt{13}}$.
The product of the distances is $d_1 \times d_2 = \frac{9}{\sqrt{13}} \times \frac{9}{\sqrt{13}} = \frac{81}{13}$.

(C) The product of the lengths of the perpendiculars from any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to its asymptotes is given by the formula $\frac{a^2 b^2}{a^2+b^2}$.
Given the equation of the hyperbola is $x^2-y^2=16$,which can be written as $\frac{x^2}{4^2}-\frac{y^2}{4^2}=1$.
Here,$a^2=16$ and $b^2=16$.
Substituting these values into the formula:
$\text{Product} = \frac{16 \times 16}{16+16} = \frac{256}{32} = 8$.

(B) The given equation of the hyperbola is $x^2 - 3y^2 = 3$,which can be written as $\frac{x^2}{3} - y^2 = 1$.
The equation of the tangent at point $(\sqrt{3}, 0)$ is given by $x x_1 - 3 y y_1 = 3$.
Substituting $(\sqrt{3}, 0)$,we get $x(\sqrt{3}) - 3y(0) = 3$,which simplifies to $x = \sqrt{3}$.
The asymptotes of the hyperbola $\frac{x^2}{3} - y^2 = 1$ are given by $\frac{x^2}{3} - y^2 = 0$,which implies $x^2 = 3y^2$,or $x = \pm \sqrt{3}y$.
Thus,the asymptotes are $x - \sqrt{3}y = 0$ and $x + \sqrt{3}y = 0$.
The vertices of the triangle are the intersection points of the tangent $x = \sqrt{3}$ with the asymptotes and the intersection of the two asymptotes:
$1$. Intersection of $x = \sqrt{3}$ and $x - \sqrt{3}y = 0$: $\sqrt{3} - \sqrt{3}y = 0 \Rightarrow y = 1$. Point is $(\sqrt{3}, 1)$.
$2$. Intersection of $x = \sqrt{3}$ and $x + \sqrt{3}y = 0$: $\sqrt{3} + \sqrt{3}y = 0 \Rightarrow y = -1$. Point is $(\sqrt{3}, -1)$.
$3$. Intersection of $x - \sqrt{3}y = 0$ and $x + \sqrt{3}y = 0$: Point is $(0, 0)$.
The area of the triangle with vertices $(0, 0)$,$(\sqrt{3}, 1)$,and $(\sqrt{3}, -1)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Area $= \frac{1}{2} |0(1 - (-1)) + \sqrt{3}(-1 - 0) + \sqrt{3}(0 - 1)| = \frac{1}{2} |0 - \sqrt{3} - \sqrt{3}| = \frac{1}{2} |-2\sqrt{3}| = \sqrt{3}$ sq units.

(C) The given equation of the hyperbola is $x^2-3y^2=3$. Dividing by $3$,we get $\frac{x^2}{3}-\frac{y^2}{1}=1$.
Here,$a^2=3$ and $b^2=1$,so $a=\sqrt{3}$ and $b=1$.
The equations of the asymptotes are $y = \pm \frac{b}{a}x$,which gives $y = \frac{1}{\sqrt{3}}x$ and $y = -\frac{1}{\sqrt{3}}x$.
Let the slopes be $m_1 = \frac{1}{\sqrt{3}}$ and $m_2 = -\frac{1}{\sqrt{3}}$.
The angle $\theta$ between the asymptotes is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values,$\tan \theta = \left| \frac{\frac{1}{\sqrt{3}} - (-\frac{1}{\sqrt{3}})}{1 + (\frac{1}{\sqrt{3}})(-\frac{1}{\sqrt{3}})} \right| = \left| \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} \right| = \left| \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} \right| = \sqrt{3}$.
Since $\tan \theta = \sqrt{3}$,we have $\theta = \frac{\pi}{3}$.

(B) Let $(a \sec \theta, b \tan \theta)$ be any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.
The equations of the asymptotes of the given hyperbola are $\frac{x}{a} + \frac{y}{b} = 0$ and $\frac{x}{a} - \frac{y}{b} = 0$.
Let $P_1$ be the length of the perpendicular from $(a \sec \theta, b \tan \theta)$ to $\frac{x}{a} + \frac{y}{b} = 0$.
$P_1 = \frac{|\sec \theta + \tan \theta|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$.
Let $P_2$ be the length of the perpendicular from $(a \sec \theta, b \tan \theta)$ to $\frac{x}{a} - \frac{y}{b} = 0$.
$P_2 = \frac{|\sec \theta - \tan \theta|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$.
The product $P_1 P_2 = \frac{|\sec^2 \theta - \tan^2 \theta|}{\frac{1}{a^2} + \frac{1}{b^2}} = \frac{1}{\frac{a^2 + b^2}{a^2 b^2}} = \frac{a^2 b^2}{a^2 + b^2}$.

(C) The equation of a hyperbola with given asymptotes $L_1=0$ and $L_2=0$ is of the form $L_1 \times L_2 + k = 0$.
Given asymptotes are $(4x+3y-7)=0$ and $(x-2y-1)=0$.
So,the equation of the hyperbola is $(4x+3y-7)(x-2y-1)+k=0$ ...$(i)$
Since the hyperbola passes through the point $(2,3)$,we substitute $x=2$ and $y=3$ into Eq. $(i)$:
$(4(2)+3(3)-7)(2-2(3)-1)+k=0$
$(8+9-7)(2-6-1)+k=0$
$(10)(-5)+k=0$
$-50+k=0 \Rightarrow k=50$
Substituting $k=50$ back into Eq. $(i)$:
$(4x+3y-7)(x-2y-1)+50=0$
$4x^2-8xy-4x+3xy-6y^2-3y-7x+14y+7+50=0$
$4x^2-5xy-6y^2-11x+11y+57=0$