Hyperbola Questions in English

(A) The given equation of the hyperbola is $\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 4$.
Dividing by $4$,we get $\frac{x^2}{4\cos^2 \alpha} - \frac{y^2}{4\sin^2 \alpha} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 4\cos^2 \alpha$ and $b^2 = 4\sin^2 \alpha$.
Thus,$a = 2|\cos \alpha|$ and $b = 2|\sin \alpha|$.
The length of the latus rectum is given by $LR = \frac{2b^2}{a}$.
Substituting the values,$LR = \frac{2(4\sin^2 \alpha)}{2|\cos \alpha|} = \frac{4\sin^2 \alpha}{|\cos \alpha|}$.
Using the identity $2\sin^2 \alpha = 1 - \cos 2\alpha$,we get $LR = \frac{2(1 - \cos 2\alpha)}{|\cos \alpha|} = 2\left| \frac{1 - \cos 2\alpha}{\cos \alpha} \right|$.

(D) The given hyperbola is $4x^2 - y^2 = 12$,which can be written as $\frac{x^2}{3} - \frac{y^2}{12} = 1$.
Here,$a^2 = 3$ and $b^2 = 12$.
The equation of a tangent with slope $m$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $y = mx \pm \sqrt{a^2m^2 - b^2}$.
Given the slope $m = 4$,we substitute the values:
$c = \pm \sqrt{3(4)^2 - 12} = \pm \sqrt{3(16) - 12} = \pm \sqrt{48 - 12} = \pm \sqrt{36} = \pm 6$.
Thus,the equations are $y = 4x + 6$ and $y = 4x - 6$,so $c_1 = 6$ and $c_2 = -6$.
Therefore,$|c_1 - c_2| = |6 - (-6)| = |12| = 12$.

(C) The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$.
Since the tangent passes through $C(0, -b)$,we have $\frac{0 \cdot x_1}{a^2} - \frac{(-b)y_1}{b^2} = 1$,which simplifies to $\frac{y_1}{b} = 1$,so $y_1 = b$.
Substituting $y_1 = b$ into the hyperbola equation: $\frac{x_1^2}{a^2} - \frac{b^2}{b^2} = 1$ $\Rightarrow \frac{x_1^2}{a^2} = 2$ $\Rightarrow x_1 = \pm a\sqrt{2}$.
Thus,the points of contact are $A(a\sqrt{2}, b)$ and $B(-a\sqrt{2}, b)$.
The triangle $\Delta ABC$ has vertices $C(0, -b)$,$A(a\sqrt{2}, b)$,and $B(-a\sqrt{2}, b)$.
The length of the base $AB$ is $2a\sqrt{2}$. The height of the triangle from $C$ to $AB$ is $h = b - (-b) = 2b$.
For $\Delta ABC$ to be a right-angled triangle,the angle at $C$ must be $90^{\circ}$ (since $AB$ is horizontal and $C$ lies on the $y$-axis,the triangle is isosceles with $CA=CB$).
If $\angle ACB = 90^{\circ}$,then the slope of $CA$ is $m_1 = \frac{b - (-b)}{a\sqrt{2} - 0} = \frac{2b}{a\sqrt{2}} = \frac{b\sqrt{2}}{a}$.
The slope of $CB$ is $m_2 = \frac{b - (-b)}{-a\sqrt{2} - 0} = -\frac{b\sqrt{2}}{a}$.
Since $m_1 \cdot m_2 = -1$,we have $-(\frac{b\sqrt{2}}{a})^2 = -1$ $\Rightarrow \frac{2b^2}{a^2} = 1$ $\Rightarrow \frac{a^2}{b^2} = 2$.

(C) The given equation is $(a - 2)x^2 + ay^2 = 4$.
Dividing by $4$,we get $\frac{x^2}{\frac{4}{a-2}} + \frac{y^2}{\frac{4}{a}} = 1$.
For a hyperbola,the coefficients of $x^2$ and $y^2$ must have opposite signs. Let the equation be in the form $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$ or $\frac{y^2}{A^2} - \frac{x^2}{B^2} = 1$.
For a rectangular hyperbola,the magnitudes of the semi-axes must be equal,i.e.,$A^2 = B^2$.
Here,the equation is $\frac{x^2}{\frac{4}{a-2}} - \frac{y^2}{-\frac{4}{a}} = 1$.
Equating the denominators: $\frac{4}{a-2} = -\frac{4}{a}$.
$a = -(a - 2)$.
$a = -a + 2$.
$2a = 2$.
$a = 1$.

(C) Any tangent to the hyperbola $x^2 - y^2 = a^2$ at point $P(a \sec \theta, a \tan \theta)$ is given by the equation:
$x \sec \theta - y \tan \theta = a$ --- $(1)$
The other two lines are:
$x - y = 0$ --- $(2)$
$x + y = 0$ --- $(3)$
Solving $(1)$ and $(2)$ simultaneously:
$x(\sec \theta - \tan \theta) = a \implies x = a(\sec \theta + \tan \theta), y = a(\sec \theta + \tan \theta)$
Solving $(1)$ and $(3)$ simultaneously:
$x(\sec \theta + \tan \theta) = a \implies x = a(\sec \theta - \tan \theta), y = -a(\sec \theta - \tan \theta)$
The intersection of $(2)$ and $(3)$ is the origin $(0, 0)$.
The area of the triangle with vertices $(0, 0)$,$(x_1, y_1)$,and $(x_2, y_2)$ is $\frac{1}{2} |x_1 y_2 - x_2 y_1|$.
Area $= \frac{1}{2} |a(\sec \theta + \tan \theta)(-a(\sec \theta - \tan \theta)) - (a(\sec \theta - \tan \theta))(a(\sec \theta + \tan \theta))|$
Area $= \frac{1}{2} |-a^2(\sec^2 \theta - \tan^2 \theta) - a^2(\sec^2 \theta - \tan^2 \theta)|$
Since $\sec^2 \theta - \tan^2 \theta = 1$,we get:
Area $= \frac{1}{2} |-a^2 - a^2| = \frac{1}{2} |-2a^2| = a^2$.

(A) The given equation of the hyperbola is $25x^2 - 16y^2 = 400$.
Dividing both sides by $400$,we get:
$\frac{25x^2}{400} - \frac{16y^2}{400} = 1$
$\frac{x^2}{16} - \frac{y^2}{25} = 1$
Comparing this with the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 16$ and $b^2 = 25$.
Thus,$a = 4$ and $b = 5$.
The length of the latus rectum is given by the formula $\frac{2b^2}{a}$.
Length $= \frac{2 \times 25}{4} = \frac{50}{4} = \frac{25}{2}$.

(C) The foci are at $(\pm ae, 0) = (\pm 2, 0)$,so $ae = 2$.
Given eccentricity $e = 2$,we have $a(2) = 2$,which implies $a = 1$.
For a hyperbola,$b^2 = a^2(e^2 - 1)$.
Substituting the values,$b^2 = 1^2(2^2 - 1) = 4 - 1 = 3$.
The standard equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Substituting $a^2 = 1$ and $b^2 = 3$,we get $\frac{x^2}{1} - \frac{y^2}{3} = 1$.
Multiplying by $3$,we get $3x^2 - y^2 = 3$.

(B) For a rectangular hyperbola,the eccentricity $e = \sqrt{2}$.
The length of the latus rectum is $\frac{2b^2}{a} = 2$,which implies $b^2 = a$.
For a rectangular hyperbola,$a^2 = b^2$,so $a^2 = a$,which gives $a = 1$ (since $a > 0$).
Thus,$a = 1$ and $b = 1$.
The distance between the foci is $2ae = 2(1)(\sqrt{2}) = 2\sqrt{2}$.
Let the other focus be $(h, k)$. The given focus is $S_1 = (1, 0)$ and the point on the hyperbola is $P = (0, 0)$.
By the definition of a hyperbola,$|PS_1 - PS_2| = 2a$.
$|\sqrt{(0-1)^2 + (0-0)^2} - \sqrt{(0-h)^2 + (0-k)^2}| = 2(1)$.
$|1 - \sqrt{h^2 + k^2}| = 2$.
This implies $\sqrt{h^2 + k^2} = 3$ or $\sqrt{h^2 + k^2} = -1$ (impossible).
Therefore,$h^2 + k^2 = 9$,which represents the curve $x^2 + y^2 = 9$.

(B) For a hyperbola of the form $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$,the product of the lengths of the perpendiculars drawn from the foci to any tangent is equal to $b^{2}$.
Given the equation $x^{2} - \frac{y^{2}}{4} = 1$,we have $a^{2} = 1$ and $b^{2} = 4$.
Therefore,the product of the lengths of the perpendiculars is $b^{2} = 4$.

(C) The given equation is $\frac{(3x - 4y - z)^2}{100} - \frac{(4x + 3y - 1)^2}{225} = 1$.
Assuming the constant $z$ in the first term is a typo for $1$,the equation becomes $\frac{(3x - 4y - 1)^2}{100} - \frac{(4x + 3y - 1)^2}{225} = 1$.
Dividing the numerators by the square of the distance between the lines (which is $5^2 = 25$),we rewrite the equation as:
$\frac{(\frac{3x - 4y - 1}{5})^2}{4} - \frac{(\frac{4x + 3y - 1}{5})^2}{9} = 1$.
This is in the standard form $\frac{X^2}{a^2} - \frac{Y^2}{b^2} = 1$,where $a^2 = 4$ and $b^2 = 9$.
Thus,$a = 2$ and $b = 3$.
The length of the latus rectum is given by $\frac{2b^2}{a} = \frac{2 \times 9}{2} = 9$.

(D) The equation of the normal to the hyperbola $\frac{x^2}{p^2} - \frac{y^2}{q^2} = 1$ at the point $(p \sec \theta, q \tan \theta)$ is given by $\frac{px}{\sec \theta} + \frac{qy}{\tan \theta} = p^2 + q^2$,which simplifies to $px \cos \theta + qy \cot \theta = p^2 + q^2$.
Comparing this with the given normal equation $ax + by = 1$,we can rewrite the given equation as $(p^2 + q^2)ax + (p^2 + q^2)by = p^2 + q^2$.
Equating the coefficients,we get $p \cos \theta = a(p^2 + q^2)$ and $q \cot \theta = b(p^2 + q^2)$.
Thus,$\cos \theta = \frac{a(p^2 + q^2)}{p}$ and $\cot \theta = \frac{b(p^2 + q^2)}{q}$.
Using the identity $\csc^2 \theta - \cot^2 \theta = 1$,we have $\frac{1}{\cos^2 \theta} - \cot^2 \theta = 1$ is incorrect; rather,we use $\sec^2 \theta - \tan^2 \theta = 1$.
From the expressions,$\sec \theta = \frac{p}{a(p^2 + q^2)}$ and $\tan \theta = \frac{q}{b(p^2 + q^2)}$.
Substituting into $\sec^2 \theta - \tan^2 \theta = 1$,we get $\frac{p^2}{a^2(p^2 + q^2)^2} - \frac{q^2}{b^2(p^2 + q^2)^2} = 1$.
Therefore,$\frac{p^2}{a^2} - \frac{q^2}{b^2} = (p^2 + q^2)^2$.

(B) Given equation: $16x^2 - 9y^2 - 32x - 36y - 164 = 0$
Rearranging terms: $16(x^2 - 2x) - 9(y^2 + 4y) = 164$
Completing the square: $16(x^2 - 2x + 1) - 9(y^2 + 4y + 4) = 164 + 16 - 36$
$16(x - 1)^2 - 9(y + 2)^2 = 144$
Dividing by $144$: $\frac{(x - 1)^2}{9} - \frac{(y + 2)^2}{16} = 1$
Here,$a^2 = 9$ and $b^2 = 16$. The eccentricity $e$ of the hyperbola is $\sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$.
The conjugate hyperbola is $\frac{(y + 2)^2}{16} - \frac{(x - 1)^2}{9} = 1$.
For the conjugate hyperbola,the roles of $a$ and $b$ are swapped,so its eccentricity $e'$ is $\sqrt{1 + \frac{a^2}{b^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

(B) The given equation of the hyperbola is $4x^2 - 9y^2 - 8x - 18y = 41$.
Completing the square,we get $4(x^2 - 2x + 1) - 9(y^2 + 2y + 1) = 41 + 4 - 9$.
$4(x - 1)^2 - 9(y + 1)^2 = 36$.
Dividing by $36$,we get $\frac{(x - 1)^2}{9} - \frac{(y + 1)^2}{4} = 1$.
This is a hyperbola with center $(1, -1)$ and $a^2 = 9$.
The locus of the foot of the perpendicular from a focus to any tangent of a hyperbola is its auxiliary circle.
The auxiliary circle of the hyperbola $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$ is $(x - h)^2 + (y - k)^2 = a^2$.
Substituting $h = 1, k = -1, a^2 = 9$,we get $(x - 1)^2 + (y + 1)^2 = 9$.
Expanding this,$x^2 - 2x + 1 + y^2 + 2y + 1 = 9$,which simplifies to $x^2 + y^2 - 2x + 2y = 7$.

(D) The given curve is $\sqrt{(x + 5\sqrt{2})^2 + y^2} - \sqrt{(x - 5\sqrt{2})^2 + y^2} = 10$. This represents the right branch of a hyperbola with foci at $(\pm 5\sqrt{2}, 0)$ and $2a = 10$,so $a = 5$. The distance between foci is $2ae = 10\sqrt{2}$,so $e = \sqrt{2}$. The equation is $\frac{x^2}{25} - \frac{y^2}{25} = 1$,i.e.,$x^2 - y^2 = 25$ for $x > 0$.
For a point to have exactly one tangent drawn to a hyperbola,it must lie on the curve itself. However,the question asks for points interior to $x^2 + y^2 < 10$. Looking at the geometry,the points from which exactly one tangent can be drawn to the right branch of the hyperbola are those lying on the curve itself or in the region where the tangent is defined. Based on the provided diagram,the points are in the shaded regions where $|y| \geq |x|$ and $x^2 + y^2 < 10$.
Testing integral points $(x, y)$ such that $x^2 + y^2 < 10$ and $|y| \geq |x|$:
If $x = 0$,$y^2 < 10 \implies y \in \{-3, -2, -1, 1, 2, 3\}$ ($6$ points).
If $x = 1$,$1 + y^2 < 10 \implies y^2 < 9 \implies y \in \{-2, -1, 1, 2\}$ ($4$ points).
If $x = -1$,$1 + y^2 < 10 \implies y^2 < 9 \implies y \in \{-2, -1, 1, 2\}$ ($4$ points).
If $x = 2$,$4 + y^2 < 10 \implies y^2 < 6 \implies y \in \{-2, 2\}$ ($2$ points).
If $x = -2$,$4 + y^2 < 10 \implies y^2 < 6 \implies y \in \{-2, 2\}$ ($2$ points).
Summing these: $6 + 4 + 4 + 2 + 2 = 18$ points.

(C) The joint equation of the asymptotes is given by $(3x - 4y + 7)(4x + 3y + 1) = 0$.
The equation of the hyperbola differs from the joint equation of its asymptotes by a constant $k$.
Thus,the equation of the hyperbola is $(3x - 4y + 7)(4x + 3y + 1) + k = 0$.
Since the hyperbola passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ into the equation:
$(3(0) - 4(0) + 7)(4(0) + 3(0) + 1) + k = 0
(7)(1) + k = 0
k = -7$.
Substituting $k = -7$ back into the equation:
$(3x - 4y + 7)(4x + 3y + 1) - 7 = 0
(12x^2 + 9xy + 3x - 16xy - 12y^2 - 4y + 28x + 21y + 7) - 7 = 0
12x^2 - 7xy - 12y^2 + 31x + 17y = 0$.

(C) Let the point on the hyperbola $xy = c^2$ be $P(ct, c/t)$.
The equation of the tangent at $P$ is $\frac{x}{ct} + \frac{y}{c/t} = 2$,which simplifies to $\frac{x}{t} + ty = 2c$.
The $x$-intercept $a_1 = 2ct$ and the $y$-intercept $b_1 = 2c/t$.
The equation of the normal at $P$ is $y - \frac{c}{t} = t^2(x - ct)$,which simplifies to $t^2x - y = ct^3 - c/t$.
The $x$-intercept $a_2 = \frac{ct^3 - c/t}{t^2} = ct - \frac{c}{t^3}$.
The $y$-intercept $b_2 = -(ct^3 - c/t) = \frac{c}{t} - ct^3$.
Now,calculate $a_1a_2 + b_1b_2$:
$a_1a_2 = (2ct)(ct - \frac{c}{t^3}) = 2c^2t^2 - \frac{2c^2}{t^2}$.
$b_1b_2 = (\frac{2c}{t})(\frac{c}{t} - ct^3) = \frac{2c^2}{t^2} - 2c^2t^2$.
Adding these gives $a_1a_2 + b_1b_2 = (2c^2t^2 - \frac{2c^2}{t^2}) + (\frac{2c^2}{t^2} - 2c^2t^2) = 0$.

(C) The equation of the hyperbola is $4x^2 - 3y^2 = 1$,which can be written as $\frac{x^2}{1/4} - \frac{y^2}{1/3} = 1$.
The condition for the line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$.
Here,$m = h$,$c = k$,$a^2 = \frac{1}{4}$,and $b^2 = \frac{1}{3}$.
Substituting these values into the condition,we get $k^2 = \frac{1}{4}h^2 - \frac{1}{3}$.
Rearranging the terms,we get $\frac{h^2}{4} - k^2 = \frac{1}{3}$,which represents a hyperbola.
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{x^2}{4/3} - \frac{y^2}{1/3} = 1$,which is a hyperbola.

(D) The equation of the ellipse is $2x^2 + 3y^2 = 6$,which can be written as $\frac{x^2}{3} + \frac{y^2}{2} = 1$.
Here $a^2 = 3$ and $b^2 = 2$.
The line $y = kx + 2h$ is tangent to the ellipse if $c^2 = a^2m^2 + b^2$,where $y = mx + c$.
Comparing $y = kx + 2h$ with $y = mx + c$,we have $m = k$ and $c = 2h$.
Substituting these into the condition: $(2h)^2 = 3(k)^2 + 2$,which simplifies to $4h^2 = 3k^2 + 2$.
The locus of $P(h, k)$ is $4x^2 - 3y^2 = 2$,or $\frac{x^2}{1/2} - \frac{y^2}{2/3} = 1$.
This is a hyperbola with $a^2 = \frac{1}{2}$ and $b^2 = \frac{2}{3}$.
The eccentricity $e$ of a hyperbola is given by $e^2 = 1 + \frac{b^2}{a^2}$.
$e^2 = 1 + \frac{2/3}{1/2} = 1 + \frac{4}{3} = \frac{7}{3}$.
Therefore,$e = \sqrt{\frac{7}{3}}$.

(C) The given hyperbola is $\frac{x^2}{4a^2} - \frac{y^2}{12} = 1$.
Here,$A^2 = 4a^2$ and $B^2 = 12$,so $A = 2a$ and $B = 2\sqrt{3}$.
The angle $\theta$ between the asymptotes is given by $2 \tan^{-1}(\frac{B}{A}) = \frac{\pi}{3}$.
Thus,$\tan^{-1}(\frac{B}{A}) = \frac{\pi}{6}$,which implies $\frac{B}{A} = \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$.
Substituting the values,$\frac{2\sqrt{3}}{2a} = \frac{1}{\sqrt{3}}$ $\Rightarrow \frac{\sqrt{3}}{a} = \frac{1}{\sqrt{3}}$ $\Rightarrow a = 3$.
Then $A^2 = 4(3)^2 = 36$.
The equation of the hyperbola is $\frac{x^2}{36} - \frac{y^2}{12} = 1$.
The conjugate hyperbola is $\frac{x^2}{36} - \frac{y^2}{12} = -1$,which is $\frac{y^2}{12} - \frac{x^2}{36} = 1$.

(B) For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the region from which two tangents can be drawn to two different branches is the region between the two asymptotes $y = \pm \frac{b}{a}x$. Here,$a^2 = 25$ and $b^2 = 16$,so the asymptotes are $y = \pm \frac{4}{5}x$.
For a point $(x_1, y_1)$ to be in this region,it must satisfy $\left| \frac{y_1}{x_1} \right| > \frac{4}{5}$.
Checking the options:
$(1, 6) \implies |6/1| = 6 > 0.8$ (True)
$(1, 3) \implies |3/1| = 3 > 0.8$ (True)
$(7, 1) \implies |1/7| \approx 0.14 < 0.8$ (False)
$(1, 0.5) \implies |0.5/1| = 0.5 < 0.8$ (False)
Now,for the circle $x^2 + y^2 = 36$,a point $(x_1, y_1)$ lies outside the circle if $x_1^2 + y_1^2 > 36$. If a point is outside,two tangents can be drawn.
For $(1, 6)$,$1^2 + 6^2 = 37 > 36$. Thus,two tangents can be drawn to the circle.
For $(1, 3)$,$1^2 + 3^2 = 10 < 36$. Thus,the point is inside the circle,and no tangents can be drawn.
Therefore,the correct point is $(1, 3)$.

(B) Let the equation of line $CP$ be $y = mx$. Substituting this into the hyperbola equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $\frac{x^2}{a^2} - \frac{m^2 x^2}{b^2} = 1$.
Solving for $x^2$,we have $x^2 = \frac{a^2 b^2}{b^2 - a^2 m^2}$.
Since $y^2 = m^2 x^2$,we have $y^2 = \frac{a^2 b^2 m^2}{b^2 - a^2 m^2}$.
Then $(CP)^2 = x^2 + y^2 = x^2(1 + m^2) = \frac{a^2 b^2(1 + m^2)}{b^2 - a^2 m^2}$.
Since $CP \perp CQ$,the slope of $CQ$ is $-\frac{1}{m}$. Replacing $m$ with $-\frac{1}{m}$ in the expression for $(CP)^2$,we get $(CQ)^2 = \frac{a^2 b^2(1 + (-1/m)^2)}{b^2 - a^2(-1/m)^2} = \frac{a^2 b^2(1 + m^2)/m^2}{(b^2 m^2 - a^2)/m^2} = \frac{a^2 b^2(1 + m^2)}{b^2 m^2 - a^2}$.
Now,$\frac{1}{(CP)^2} + \frac{1}{(CQ)^2} = \frac{b^2 - a^2 m^2}{a^2 b^2(1 + m^2)} + \frac{b^2 m^2 - a^2}{a^2 b^2(1 + m^2)} = \frac{b^2 - a^2 m^2 + b^2 m^2 - a^2}{a^2 b^2(1 + m^2)} = \frac{b^2(1 + m^2) - a^2(1 + m^2)}{a^2 b^2(1 + m^2)} = \frac{b^2 - a^2}{a^2 b^2} = \frac{1}{a^2} - \frac{1}{b^2}$.

(D) The equation of the normal to the rectangular hyperbola $xy = c^2$ at the point $t_1$ is given by $x t_1^2 - y = c(t_1^4 - 1)/t_1$.
Since this normal meets the curve again at the point $t_2$,the point $(c t_2, c/t_2)$ must satisfy the equation of the normal.
Substituting $x = c t_2$ and $y = c/t_2$ into the normal equation:
$(c t_2) t_1^2 - c/t_2 = c(t_1^4 - 1)/t_1$
Dividing by $c$:
$t_2 t_1^2 - 1/t_2 = (t_1^4 - 1)/t_1$
$(t_2^2 t_1^2 - 1)/t_2 = (t_1^4 - 1)/t_1$
Since $t_1 \neq t_2$,we use the property that for a normal at $t_1$ meeting at $t_2$,the relation is $t_1^3 t_2 = -1$.

(D) Let the coordinates of $P$ be $(\alpha, \beta)$. Since $PQ$ is a double ordinate,$Q$ is $(\alpha, -\beta)$.
$PQ = 2|\beta|$ and $OP = \sqrt{\alpha^2 + \beta^2}$.
Since $\triangle OPQ$ is equilateral,$OP = PQ$,so $OP^2 = PQ^2$.
$\alpha^2 + \beta^2 = (2\beta)^2 = 4\beta^2 \Rightarrow \alpha^2 = 3\beta^2$.
Since $P(\alpha, \beta)$ lies on the hyperbola,$\frac{\alpha^2}{a^2} - \frac{\beta^2}{b^2} = 1$.
Substituting $\alpha^2 = 3\beta^2$,we get $\frac{3\beta^2}{a^2} - \frac{\beta^2}{b^2} = 1$.
$\beta^2 \left( \frac{3}{a^2} - \frac{1}{b^2} \right) = 1$.
Since $\beta^2 > 0$,we must have $\frac{3}{a^2} - \frac{1}{b^2} > 0$,which implies $\frac{3}{a^2} > \frac{1}{b^2} \Rightarrow \frac{b^2}{a^2} > \frac{1}{3}$.
Using $b^2 = a^2(e^2 - 1)$,we get $e^2 - 1 > \frac{1}{3}$.
$e^2 > \frac{4}{3} \Rightarrow e > \frac{2}{\sqrt{3}}$.

(B) The given hyperbola is $x^2 - 2y^2 = 2$,which can be written as $\frac{x^2}{2} - y^2 = 1$.
The asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are given by $\frac{x}{a} \pm \frac{y}{b} = 0$.
Here,$a^2 = 2$ and $b^2 = 1$,so $a = \sqrt{2}$ and $b = 1$.
The asymptotes are $\frac{x}{\sqrt{2}} - y = 0$ and $\frac{x}{\sqrt{2}} + y = 0$,or $x - \sqrt{2}y = 0$ and $x + \sqrt{2}y = 0$.
Let $P(x_1, y_1)$ be any point on the hyperbola,so $x_1^2 - 2y_1^2 = 2$.
The product of the perpendiculars from $P(x_1, y_1)$ to the lines $x - \sqrt{2}y = 0$ and $x + \sqrt{2}y = 0$ is:
$p_1 p_2 = \left| \frac{x_1 - \sqrt{2}y_1}{\sqrt{1^2 + (-\sqrt{2})^2}} \right| \times \left| \frac{x_1 + \sqrt{2}y_1}{\sqrt{1^2 + (\sqrt{2})^2}} \right|$
$= \frac{|x_1^2 - 2y_1^2|}{\sqrt{3} \times \sqrt{3}} = \frac{|x_1^2 - 2y_1^2|}{3}$
Since $x_1^2 - 2y_1^2 = 2$,the product is $\frac{2}{3}$.

(B) Any tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $y = mx \pm \sqrt{a^2m^2 - b^2}$.
For this line to be a tangent to the hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$,we substitute $y = mx + c$ into the second equation:
$\frac{(mx + c)^2}{a^2} - \frac{x^2}{b^2} = 1$
$b^2(m^2x^2 + 2mcx + c^2) - a^2x^2 = a^2b^2$
$x^2(b^2m^2 - a^2) + 2b^2mcx + (b^2c^2 - a^2b^2) = 0$.
Since the line is a tangent,the discriminant of this quadratic equation must be zero:
$D = (2b^2mc)^2 - 4(b^2m^2 - a^2)(b^2c^2 - a^2b^2) = 0$
$4b^4m^2c^2 - 4b^2(b^2m^2 - a^2)(c^2 - a^2) = 0$
$b^2m^2c^2 - (b^2m^2c^2 - a^2b^2m^2 - a^2c^2 + a^4) = 0$
$a^2b^2m^2 + a^2c^2 - a^4 = 0$
$b^2m^2 + c^2 - a^2 = 0 \Rightarrow c^2 = a^2 - b^2m^2$.
Equating the two expressions for $c^2$:
$a^2m^2 - b^2 = a^2 - b^2m^2$
$m^2(a^2 + b^2) = a^2 + b^2$ $\Rightarrow m^2 = 1$ $\Rightarrow m = \pm 1$.
Substituting $m^2 = 1$ into $c^2 = a^2m^2 - b^2$:
$c^2 = a^2 - b^2 \Rightarrow c = \pm \sqrt{a^2 - b^2}$.
Thus,the common tangents are $y = \pm x \pm \sqrt{a^2 - b^2}$.

(A) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Differentiating with respect to $x$,we get $\frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = \frac{b^2 x}{a^2 y}$.
At $P(6, 3)$,the slope of the tangent is $m_t = \frac{b^2(6)}{a^2(3)} = \frac{2b^2}{a^2}$.
The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{a^2}{2b^2}$.
The equation of the normal at $P(6, 3)$ is $y - 3 = -\frac{a^2}{2b^2}(x - 6)$.
Since the normal passes through $(10, 0)$,we have $0 - 3 = -\frac{a^2}{2b^2}(10 - 6)$.
$-3 = -\frac{a^2}{2b^2}(4)$ $\Rightarrow 3 = \frac{2a^2}{b^2}$ $\Rightarrow \frac{a^2}{b^2} = \frac{3}{2}$.
The eccentricity $e$ is given by $e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{2}{3} = \frac{5}{3}$.
Therefore,$e = \sqrt{\frac{5}{3}}$.

(D) Let the midpoint of the chord be $(h, k)$. The equation of the chord of the circle $x^2 + y^2 = a^2$ with midpoint $(h, k)$ is given by $T = S_1$,which is $xh + yk = h^2 + k^2$.
This can be rewritten as $y = -\frac{h}{k}x + \frac{h^2 + k^2}{k}$.
Comparing this with the line $y = mx + c$,we have $m = -\frac{h}{k}$ and $c = \frac{h^2 + k^2}{k}$.
Since this chord touches the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the condition for tangency $c^2 = a^2m^2 - b^2$ must be satisfied.
Substituting the values of $m$ and $c$:
$\left(\frac{h^2 + k^2}{k}\right)^2 = a^2\left(-\frac{h}{k}\right)^2 - b^2$
$(h^2 + k^2)^2 = a^2h^2 - b^2k^2$.
Replacing $(h, k)$ with $(x, y)$,the locus is $(x^2 + y^2)^2 = a^2x^2 - b^2y^2$.

(A) Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Given the ratio of the length of the conjugate axis $(2b)$ to the length of the transverse axis $(2a)$ is $3:2$,we have $\frac{2b}{2a} = \frac{3}{2}$,which implies $\frac{b}{a} = \frac{3}{2}$.
The distance between the foci is $2ae$.
The distance between the two directrices is $\frac{2a}{e}$.
The ratio of the distance between the foci to the distance between the directrices is $\frac{2ae}{2a/e} = e^2$.
For a hyperbola,$e^2 = 1 + \frac{b^2}{a^2}$.
Substituting $\frac{b}{a} = \frac{3}{2}$,we get $e^2 = 1 + (\frac{3}{2})^2 = 1 + \frac{9}{4} = \frac{13}{4}$.
Thus,the required ratio is $13:4$.

(B) Let $P(x, y)$ be a point on the hyperbola $x^2 - y^2 = 4$,so $x^2 = y^2 + 4$.
Let $Q$ be the point $(0, -1)$. The distance $PQ$ is given by $PQ^2 = (x - 0)^2 + (y - (-1))^2 = x^2 + (y + 1)^2$.
Substituting $x^2 = y^2 + 4$,we get $PQ^2 = y^2 + 4 + y^2 + 2y + 1 = 2y^2 + 2y + 5$.
To minimize $PQ^2$,we differentiate with respect to $y$: $f(y) = 2y^2 + 2y + 5$.
$f'(y) = 4y + 2 = 0 \implies y = -\frac{1}{2}$.
The distance of $P$ from the $x$-axis is $|y| = |-\frac{1}{2}| = \frac{1}{2}$.

(A) The given hyperbola is $4x^2 - 9y^2 = 36$,which can be written as $\frac{x^2}{9} - \frac{y^2}{4} = 1$. Here $a^2 = 9$ and $b^2 = 4$.
The slope of the line $5x + 2y - 10 = 0$ is $m_1 = -\frac{5}{2}$.
The slope of the tangent perpendicular to this line is $m = -\frac{1}{m_1} = \frac{2}{5}$.
The condition for a line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$.
Substituting the values: $c^2 = 9\left(\frac{2}{5}\right)^2 - 4 = 9\left(\frac{4}{25}\right) - 4 = \frac{36}{25} - 4 = \frac{36 - 100}{25} = -\frac{64}{25}$.
Since $c^2 < 0$,there is no real value of $c$ for which the line is a tangent.
Thus,the number of possible tangents is $0$.

(D) For a rectangular hyperbola,the asymptotes are mutually perpendicular. Given one asymptote is $3x - 4y - 6 = 0$,the other must be of the form $4x + 3y + \lambda = 0$.
The centre of the hyperbola is the intersection point of the two asymptotes.
Solving $3x - 4y - 6 = 0$ and $4x + 3y + \lambda = 0$:
Multiply the first by $3$ and the second by $4$: $9x - 12y - 18 = 0$ and $16x + 12y + 4\lambda = 0$.
Adding these gives $25x + 4\lambda - 18 = 0$,so $x = \frac{18 - 4\lambda}{25}$.
Substituting $x$ into $3x - 4y - 6 = 0$: $4y = 3(\frac{18 - 4\lambda}{25}) - 6 = \frac{54 - 12\lambda - 150}{25} = \frac{-96 - 12\lambda}{25}$,so $y = \frac{-24 - 3\lambda}{25}$.
Since the centre lies on $x - y - 1 = 0$,we have $\frac{18 - 4\lambda}{25} - (\frac{-24 - 3\lambda}{25}) - 1 = 0$.
$18 - 4\lambda + 24 + 3\lambda - 25 = 0$ $\Rightarrow 17 - \lambda = 0$ $\Rightarrow \lambda = 17$.
Thus,the equation of the other asymptote is $4x + 3y + 17 = 0$.

(A) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at the point $(a \sec \theta, b \tan \theta)$ is given by $ax \cos \theta + by \cot \theta = a^2 + b^2$.
Here,$a^2 = 25$ and $b^2 = 16$,so $a = 5$ and $b = 4$.
The equation of the normal is $5x \cos \theta + 4y \cot \theta = 25 + 16 = 41$ .......$(i)$
This line is perpendicular to the line $2x + y = 1$,which has a slope of $-2$.
The slope of the normal line $(i)$ is $m = -\frac{5 \cos \theta}{4 \cot \theta} = -\frac{5 \sin \theta}{4}$.
Since the lines are perpendicular,the product of their slopes is $-1$:
$(-\frac{5 \sin \theta}{4})(-2) = -1 \implies \frac{5 \sin \theta}{2} = -1 \implies \sin \theta = -\frac{2}{5}$.
Then,$\cos \theta = \pm \sqrt{1 - \sin^2 \theta} = \pm \sqrt{1 - \frac{4}{25}} = \pm \frac{\sqrt{21}}{5}$.
Also,$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\pm \sqrt{21}/5}{-2/5} = \mp \frac{\sqrt{21}}{2}$.
Substituting these into equation $(i)$:
$5x(\pm \frac{\sqrt{21}}{5}) + 4y(\mp \frac{\sqrt{21}}{2}) = 41$
$\pm \sqrt{21}x \mp 2\sqrt{21}y = 41$
$\pm \sqrt{21}(x - 2y) = 41$.
Taking the positive root,we get $\sqrt{21}(x - 2y) = 41$.

(D) The equation of the hyperbola is $4y^2 - x^2 = 1$.
Let the point of tangency be $P(x_1, y_1)$. The equation of the tangent at $P$ is $4yy_1 - xx_1 = 1$.
This tangent intersects the $x$-axis at $A$ where $y=0$,so $-xx_1 = 1 \Rightarrow x = -1/x_1$. Thus,$A = (-1/x_1, 0)$.
This tangent intersects the $y$-axis at $B$ where $x=0$,so $4yy_1 = 1 \Rightarrow y = 1/(4y_1)$. Thus,$B = (0, 1/(4y_1))$.
Let the midpoint of $AB$ be $(h, k)$.
Then $h = -1/(2x_1) \Rightarrow x_1 = -1/(2h)$ and $k = 1/(8y_1) \Rightarrow y_1 = 1/(8k)$.
Since $(x_1, y_1)$ lies on the hyperbola $4y_1^2 - x_1^2 = 1$,we substitute the values:
$4(1/(8k))^2 - (-1/(2h))^2 = 1$
$4/(64k^2) - 1/(4h^2) = 1$
$1/(16k^2) - 1/(4h^2) = 1$
Multiplying by $16h^2k^2$:
$h^2 - 4k^2 = 16h^2k^2$
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 - 4y^2 - 16x^2y^2 = 0$.

(C) The equation of the hyperbola is $4x^2 - 9y^2 = 36$,which can be written as $\frac{x^2}{9} - \frac{y^2}{4} = 1$.
Let the point on the hyperbola be $(x_0, y_0)$. The equation of the normal at $(x_0, y_0)$ is $\frac{a^2x}{x_0} + \frac{b^2y}{y_0} = a^2 + b^2$.
Here $a^2 = 9$ and $b^2 = 4$,so the equation is $\frac{9x}{x_0} + \frac{4y}{y_0} = 9 + 4 = 13$.
This normal meets the $x$-axis at $A(\frac{13x_0}{9}, 0)$ and the $y$-axis at $B(0, \frac{13y_0}{4})$.
Since $OABP$ is a parallelogram with $O(0,0)$,the coordinates of $P(x, y)$ are given by $P = A + B = (\frac{13x_0}{9}, \frac{13y_0}{4})$.
Thus,$x = \frac{13x_0}{9} \Rightarrow x_0 = \frac{9x}{13}$ and $y = \frac{13y_0}{4} \Rightarrow y_0 = \frac{4y}{13}$.
Since $(x_0, y_0)$ lies on the hyperbola $4x_0^2 - 9y_0^2 = 36$,we substitute the values:
$4(\frac{9x}{13})^2 - 9(\frac{4y}{13})^2 = 36$
$4 \cdot \frac{81x^2}{169} - 9 \cdot \frac{16y^2}{169} = 36$
$\frac{324x^2 - 144y^2}{169} = 36$
$324x^2 - 144y^2 = 36 \cdot 169$
Dividing by $36$,we get $9x^2 - 4y^2 = 169$.

(A) Given the equation for eccentricity: $9e^2 - 18e + 5 = 0$.
Solving for $e$: $(3e - 1)(3e - 5) = 0$,so $e = 1/3$ or $e = 5/3$. Since for a hyperbola $e > 1$,we have $e = 5/3$.
For a hyperbola with focus $S(ae, 0)$ and directrix $x = a/e$,we have $ae = 5$ and $a/e = 9/5$.
Multiplying these: $(ae)(a/e) = 5 \times (9/5) \Rightarrow a^2 = 9$.
Using $ae = 5$ and $e = 5/3$,we get $a(5/3) = 5 \Rightarrow a = 3$.
For a hyperbola,$b^2 = a^2(e^2 - 1)$.
$b^2 = 9((5/3)^2 - 1) = 9(25/9 - 1) = 9(16/9) = 16$.
Thus,$a^2 - b^2 = 9 - 16 = -7$.

(C) The given conic is $\frac{x^2}{3} + \frac{y^2}{4} = 4$,which can be rewritten as $\frac{x^2}{12} + \frac{y^2}{16} = 1$.
Here,$a^2 = 12$ and $b^2 = 16$. Since $b^2 > a^2$,the major axis is along the $y$-axis.
The eccentricity of the ellipse is $e = \sqrt{1 - \frac{12}{16}} = \sqrt{\frac{4}{16}} = \frac{1}{2}$.
The foci of the ellipse are $(0, \pm be) = (0, \pm 4 \times \frac{1}{2}) = (0, \pm 2)$.
The hyperbola has its vertices at $(0, \pm 2)$,so its transverse axis is along the $y$-axis and $a = 2$.
The equation of the hyperbola is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$,where $a = 2$.
Given the eccentricity $e_h = \frac{3}{2}$,we use $b^2 = a^2(e_h^2 - 1) = 4(\frac{9}{4} - 1) = 4(\frac{5}{4}) = 5$.
The equation of the hyperbola is $\frac{y^2}{4} - \frac{x^2}{5} = 1$.
Checking the points:
For $(0, 2)$: $\frac{4}{4} - 0 = 1$ (Lies on it).
For $(\sqrt{5}, 2\sqrt{2})$: $\frac{8}{4} - \frac{5}{5} = 2 - 1 = 1$ (Lies on it).
For $(\sqrt{10}, 2\sqrt{3})$: $\frac{12}{4} - \frac{10}{5} = 3 - 2 = 1$ (Lies on it).
For $(5, 2\sqrt{3})$: $\frac{12}{4} - \frac{25}{5} = 3 - 5 = -2 \neq 1$ (Does not lie on it).

(D) 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 given by $ax \cos \theta + by \cot \theta = a^2 + b^2$.
For the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$,we have $a=3$ and $b=2$. Thus,the normal at $P(3 \sec \theta, 2 \tan \theta)$ is $3x \cos \theta + 2y \cot \theta = 13 \dots (1)$.
The normal at $Q(3 \sec \phi, 2 \tan \phi)$ is $3x \cos \phi + 2y \cot \phi = 13 \dots (2)$.
Given $\phi = \frac{\pi}{2} - \theta$,we have $\cos \phi = \sin \theta$ and $\cot \phi = \tan \theta$.
Substituting these into $(2)$,we get $3x \sin \theta + 2y \tan \theta = 13 \dots (3)$.
Let the intersection point be $(h, k)$. From $(1)$ and $(3)$:
$3h \cos \theta + 2k \cot \theta = 13$ and $3h \sin \theta + 2k \tan \theta = 13$.
Equating the two expressions for $13$:
$3h \cos \theta + 2k \cot \theta = 3h \sin \theta + 2k \tan \theta$
$3h(\cos \theta - \sin \theta) = 2k(\tan \theta - \cot \theta) = 2k \left( \frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta \cos \theta} \right) = -2k \left( \frac{\cos \theta - \sin \theta}{\sin \theta \cos \theta} \right)(\cos \theta + \sin \theta)$.
Assuming $\cos \theta \neq \sin \theta$,we get $3h = -2k \frac{(\cos \theta + \sin \theta)}{\sin \theta \cos \theta} = -2k(\sec \theta + \csc \theta)$.
Substituting $3h$ back into $(1)$: $(-2k(\sec \theta + \csc \theta)) \cos \theta + 2k \cot \theta = 13$ $\Rightarrow -2k(1 + \cot \theta) + 2k \cot \theta = 13$ $\Rightarrow -2k = 13$ $\Rightarrow k = -\frac{13}{2}$.

(A) Given the hyperbola $\frac{x^2}{4} - \frac{y^2}{5} = 1$.
Here,$a^2 = 4$ and $b^2 = 5$.
The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{5}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.
The extremity of the latus rectum in the first quadrant is $L = (ae, \frac{b^2}{a}) = (2 \times \frac{3}{2}, \frac{5}{2}) = (3, \frac{5}{2})$.
The equation of the tangent at $(x_1, y_1)$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$.
Substituting $(x_1, y_1) = (3, \frac{5}{2})$,we get $\frac{3x}{4} - \frac{y(5/2)}{5} = 1$,which simplifies to $\frac{3x}{4} - \frac{y}{2} = 1$.
This can be written as $\frac{x}{4/3} + \frac{y}{-2} = 1$.
The $x$-intercept is $OA = \frac{4}{3}$ and the $y$-intercept is $OB = -2$.
Therefore,$(OA)^2 - (OB)^2 = (\frac{4}{3})^2 - (-2)^2 = \frac{16}{9} - 4 = \frac{16 - 36}{9} = -\frac{20}{9}$.

(D) The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at the point $(a \sec \theta, b \tan \theta)$ is $\frac{x \sec \theta}{a} - \frac{y \tan \theta}{b} = 1$.
Setting $y=0$,we get $x = a \cos \theta$,so $P = (a \cos \theta, 0)$.
Setting $x=0$,we get $y = -b \cot \theta$,so $Q = (0, -b \cot \theta)$.
Since $OPRQ$ is a rectangle with $O(0,0)$,the coordinates of $R$ are $(h, k) = (a \cos \theta, -b \cot \theta)$.
From this,$\cos \theta = \frac{h}{a}$ and $\cot \theta = -\frac{k}{b}$.
Using the identity $\csc^2 \theta - \cot^2 \theta = 1$,we have $\frac{1}{\sin^2 \theta} - \cot^2 \theta = 1$,which is $\frac{1}{1 - \cos^2 \theta} - \cot^2 \theta = 1$.
Alternatively,$\frac{1}{\cos^2 \theta} = 1 + \tan^2 \theta$ is not helpful here. Let's use $\csc^2 \theta = 1 + \cot^2 \theta$.
$\frac{1}{\sin^2 \theta} = 1 + \frac{k^2}{b^2} = \frac{b^2 + k^2}{b^2}$.
Since $\cos^2 \theta = \frac{h^2}{a^2}$,then $\sin^2 \theta = 1 - \frac{h^2}{a^2} = \frac{a^2 - h^2}{a^2}$.
Substituting these into $\frac{1}{\sin^2 \theta} = 1 + \frac{k^2}{b^2}$ gives $\frac{a^2}{a^2 - h^2} = \frac{b^2 + k^2}{b^2}$.
This simplifies to $a^2 b^2 = (a^2 - h^2)(b^2 + k^2) = a^2 b^2 + a^2 k^2 - b^2 h^2 - h^2 k^2$.
$a^2 k^2 - b^2 h^2 - h^2 k^2 = 0 \Rightarrow \frac{a^2}{h^2} - \frac{b^2}{k^2} = 1$.
Given $a^2 = 4$ and $b^2 = 2$,the locus of $R(x, y)$ is $\frac{4}{x^2} - \frac{2}{y^2} = 1$.

(A) The given equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{b^2} = 1.$
Since the hyperbola passes through $(K, 2),$ we have $\frac{K^2}{9} - \frac{4}{b^2} = 1$ $(1).$
The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \frac{\sqrt{13}}{3}.$
Here,$a^2 = 9,$ so $\sqrt{1 + \frac{b^2}{9}} = \frac{\sqrt{13}}{3}.$
Squaring both sides,$1 + \frac{b^2}{9} = \frac{13}{9}.$
$\frac{b^2}{9} = \frac{13}{9} - 1 = \frac{4}{9},$ which implies $b^2 = 4.$
Substituting $b^2 = 4$ into equation $(1),$ we get $\frac{K^2}{9} - \frac{4}{4} = 1.$
$\frac{K^2}{9} - 1 = 1 \Rightarrow \frac{K^2}{9} = 2.$
Therefore,$K^2 = 18.$

(A) Given the hyperbola equation: $\frac{x^2}{\cos^2 \theta} - \frac{y^2}{\sin^2 \theta} = 1$.
Here,$a^2 = \cos^2 \theta$ and $b^2 = \sin^2 \theta$.
The eccentricity $e$ is given by $e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{\sin^2 \theta}{\cos^2 \theta} = 1 + \tan^2 \theta = \sec^2 \theta$.
Given $e > 2$,so $e^2 > 4$,which implies $\sec^2 \theta > 4$,or $\tan^2 \theta > 3$.
Since $0 < \theta < \frac{\pi}{2}$,we have $\tan \theta > \sqrt{3}$,which means $\theta \in (\frac{\pi}{3}, \frac{\pi}{2})$.
The length of the latus rectum $L$ is given by $L = \frac{2b^2}{a} = \frac{2 \sin^2 \theta}{\cos \theta} = 2 \tan \theta \sin \theta$.
As $\theta$ increases from $\frac{\pi}{3}$ to $\frac{\pi}{2}$,$\tan \theta$ increases from $\sqrt{3}$ to $\infty$ and $\sin \theta$ increases from $\frac{\sqrt{3}}{2}$ to $1$.
Thus,$L = 2 \tan \theta \sin \theta > 2(\sqrt{3})(\frac{\sqrt{3}}{2}) = 3$.
As $\theta \to \frac{\pi}{2}$,$L \to \infty$.
Therefore,the length of the latus rectum lies in the interval $(3, \infty)$.

(A) The standard equation of a hyperbola with centre at the origin and transverse axis along the $x$-axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Given that the length of the transverse axis is $2a = 4$,we have $a = 2$,so $a^2 = 4$.
The equation becomes $\frac{x^2}{4} - \frac{y^2}{b^2} = 1$.
Since the hyperbola passes through the point $(4, 2)$,we substitute these values into the equation:
$\frac{4^2}{4} - \frac{2^2}{b^2} = 1$
$\frac{16}{4} - \frac{4}{b^2} = 1$
$4 - \frac{4}{b^2} = 1$
$3 = \frac{4}{b^2}$
$b^2 = \frac{4}{3}$.
The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$.
Substituting the values of $a^2$ and $b^2$:
$e = \sqrt{1 + \frac{4/3}{4}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.

(A) The given hyperbola is $4x^2 - 5y^2 = 20$,which can be written as $\frac{x^2}{5} - \frac{y^2}{4} = 1$.
Comparing this with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $a^2 = 5$ and $b^2 = 4$.
The given line is $x - y = 2$,which can be written as $y = x - 2$.
The slope of this line is $m = 1$.
The equation of a tangent to the hyperbola with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$.
Substituting $m = 1$,$a^2 = 5$,and $b^2 = 4$,we get $y = 1(x) \pm \sqrt{5(1)^2 - 4} = x \pm \sqrt{5 - 4} = x \pm 1$.
Thus,the tangents are $y = x + 1$ or $y = x - 1$,which can be written as $x - y + 1 = 0$ or $x - y - 1 = 0$.
Comparing with the given options,$x - y + 1 = 0$ is the correct choice.

(B) The given equation is $\frac{y^2}{1+r} - \frac{x^2}{1-r} = 1$.
Case $1$: If $r > 1$,then $1-r < 0$. Let $1-r = -k$ where $k = r-1 > 0$. The equation becomes $\frac{y^2}{1+r} + \frac{x^2}{r-1} = 1$,which is an ellipse.
The eccentricity $e$ of an ellipse $\frac{y^2}{a^2} + \frac{x^2}{b^2} = 1$ (where $a^2 > b^2$) is $e = \sqrt{1 - \frac{b^2}{a^2}}$.
Here $a^2 = 1+r$ and $b^2 = r-1$. Thus,$e = \sqrt{1 - \frac{r-1}{r+1}} = \sqrt{\frac{r+1-r+1}{r+1}} = \sqrt{\frac{2}{r+1}}$.
Case $2$: If $0 < r < 1$,then $1-r > 0$. The equation is of the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$,which is a hyperbola.
The eccentricity $e$ of a hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ is $e = \sqrt{1 + \frac{b^2}{a^2}}$.
Here $a^2 = 1+r$ and $b^2 = 1-r$. Thus,$e = \sqrt{1 + \frac{1-r}{1+r}} = \sqrt{\frac{1+r+1-r}{1+r}} = \sqrt{\frac{2}{1+r}}$.
Comparing with the options,option $B$ is correct.

(A) The length of the conjugate axis is $2b = 5$,so $b = \frac{5}{2}$.
The distance between the foci is $2ae = 13$,so $ae = \frac{13}{2}$.
For a hyperbola,the relationship between $a$,$b$,and $e$ is given by $a^2e^2 = a^2 + b^2$.
Substituting the values,we get $(ae)^2 = a^2 + b^2$.
$\left(\frac{13}{2}\right)^2 = a^2 + \left(\frac{5}{2}\right)^2$.
$\frac{169}{4} = a^2 + \frac{25}{4}$.
$a^2 = \frac{169 - 25}{4} = \frac{144}{4} = 36$.
Thus,$a = 6$.
The eccentricity $e$ is given by $e = \frac{ae}{a} = \frac{13/2}{6} = \frac{13}{12}$.

(D) The standard equation of a hyperbola with vertices at $(\pm a, 0)$ is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Given vertices are at $(\pm 2, 0)$,so $a = 2$,which implies $a^2 = 4$.
The focus is at $(\pm ae, 0)$. Given one focus is at $(-3, 0)$,we have $ae = 3$.
For a hyperbola,the relationship between $a, b,$ and $e$ is $b^2 = a^2(e^2 - 1) = a^2e^2 - a^2$.
Substituting the values,$b^2 = 3^2 - 2^2 = 9 - 4 = 5$.
Thus,the equation of the hyperbola is $\frac{x^2}{4} - \frac{y^2}{5} = 1$.
Now,we check which point does not satisfy this equation:
For option $A$: $\frac{(-6)^2}{4} - \frac{(2\sqrt{10})^2}{5} = \frac{36}{4} - \frac{40}{5} = 9 - 8 = 1$ (Lies on hyperbola).
For option $B$: $\frac{(2\sqrt{6})^2}{4} - \frac{5^2}{5} = \frac{24}{4} - \frac{25}{5} = 6 - 5 = 1$ (Lies on hyperbola).
For option $C$: $\frac{4^2}{4} - \frac{(\sqrt{15})^2}{5} = \frac{16}{4} - \frac{15}{5} = 4 - 3 = 1$ (Lies on hyperbola).
For option $D$: $\frac{6^2}{4} - \frac{(5\sqrt{2})^2}{5} = \frac{36}{4} - \frac{50}{5} = 9 - 10 = -1 \neq 1$ (Does not lie on hyperbola).
Therefore,the point $(6, 5\sqrt{2})$ does not lie on the hyperbola.

(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Since it passes through $(4, 6)$,we have $\frac{16}{a^2} - \frac{36}{b^2} = 1 \dots (i)$.
Given eccentricity $e = 2$,we have $e^2 = 1 + \frac{b^2}{a^2}$ $\Rightarrow 4 = 1 + \frac{b^2}{a^2}$ $\Rightarrow b^2 = 3a^2 \dots (ii)$.
Substituting $(ii)$ into $(i)$,we get $\frac{16}{a^2} - \frac{36}{3a^2} = 1$ $\Rightarrow \frac{16 - 12}{a^2} = 1$ $\Rightarrow a^2 = 4$.
Then $b^2 = 3(4) = 12$.
The equation of the hyperbola is $\frac{x^2}{4} - \frac{y^2}{12} = 1$.
The equation of the tangent at $(x_1, y_1) = (4, 6)$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$.
$\frac{4x}{4} - \frac{6y}{12} = 1$ $\Rightarrow x - \frac{y}{2} = 1$ $\Rightarrow 2x - y = 2$ $\Rightarrow 2x - y - 2 = 0$.

(A) The equation of the hyperbola is $\frac{x^2}{24} - \frac{y^2}{18} = 1$,where $a^2 = 24$ and $b^2 = 18$.
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 $c^2 = \frac{m^2(a^2 + b^2)^2}{a^2m^2 - b^2}$.
Here,$c = 7\sqrt{3}$,so $c^2 = 49 \times 3 = 147$.
Substituting the values: $147 = \frac{m^2(24 + 18)^2}{24m^2 - 18} = \frac{m^2(42)^2}{24m^2 - 18} = \frac{1764m^2}{24m^2 - 18}$.
Dividing by $147$: $1 = \frac{12m^2}{24m^2 - 18}$.
$24m^2 - 18 = 12m^2$ $\Rightarrow 12m^2 = 18$ $\Rightarrow m^2 = \frac{18}{12} = \frac{3}{2}$.
Wait,re-evaluating the normal condition: The normal at $(x_0, y_0)$ is $\frac{a^2x}{x_0} + \frac{b^2y}{y_0} = a^2 + b^2$. Comparing with $y = mx + c$,we get $m = -\frac{a^2y_0}{b^2x_0}$.
Using the slope form $m = \pm \frac{c}{\sqrt{a^2 - b^2m^2}}$ is not applicable here. The correct condition for $y = mx + c$ to be normal is $c^2 = \frac{m^2(a^2 + b^2)^2}{a^2m^2 - b^2}$ is incorrect. The correct condition is $c^2 = \frac{m^2(a^2 + b^2)^2}{a^2m^2 - b^2}$ is for ellipse. For hyperbola,it is $c^2 = \frac{m^2(a^2 + b^2)^2}{a^2m^2 - b^2}$ is wrong. The correct condition is $c = \pm \frac{m(a^2 + b^2)}{\sqrt{a^2 - b^2m^2}}$.
$c^2(a^2 - b^2m^2) = m^2(a^2 + b^2)^2$ $\Rightarrow 147(24 - 18m^2) = m^2(42)^2$ $\Rightarrow 147(24 - 18m^2) = 1764m^2$.
Divide by $147$: $24 - 18m^2 = 12m^2$ $\Rightarrow 30m^2 = 24$ $\Rightarrow m^2 = \frac{24}{30} = \frac{4}{5}$.
Thus,$m = \pm \frac{2}{\sqrt{5}}$.

(B) Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Since it passes through $(4, -2\sqrt{3})$,we have $\frac{16}{a^2} - \frac{12}{b^2} = 1 \dots (i)$.
Given the directrix $x = \frac{a}{e}$,we have $\frac{a}{e} = \frac{4\sqrt{5}}{5} = \frac{4}{\sqrt{5}}$,so $a^2 = \frac{16e^2}{5} \dots (ii)$.
Using $b^2 = a^2(e^2 - 1)$,substitute $a^2$ and $b^2$ into $(i)$:
$\frac{16}{a^2} - \frac{12}{a^2(e^2 - 1)} = 1$.
Substitute $a^2 = \frac{16e^2}{5}$:
$\frac{16}{16e^2/5} - \frac{12}{(16e^2/5)(e^2 - 1)} = 1 \Rightarrow \frac{5}{e^2} - \frac{60}{16e^2(e^2 - 1)} = 1$.
Multiply by $4e^2(e^2 - 1)$:
$20(e^2 - 1) - 15 = 4e^2(e^2 - 1) \Rightarrow 20e^2 - 35 = 4e^4 - 4e^2$.
Rearranging gives $4e^4 - 24e^2 + 35 = 0$.

(C) The given equation of the hyperbola is $16x^2 - 9y^2 = 144.$ Dividing by $144,$ we get $\frac{x^2}{9} - \frac{y^2}{16} = 1.$
Here,$a^2 = 9$ and $b^2 = 16,$ so $a = 3$ and $b = 4.$
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 $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $x = \pm \frac{a}{e}.$
Given directrix is $5x + 9 = 0,$ which is $x = -\frac{9}{5}.$
Since $\frac{a}{e} = \frac{3}{5/3} = \frac{9}{5},$ the directrix $x = -\frac{9}{5}$ corresponds to the focus at $(-ae, 0).$
Thus,the focus is $(-3 \times \frac{5}{3}, 0) = (-5, 0).$