Hyperbola Questions in English

(C) The given equation of the hyperbola is $7x^2 - 9y^2 = 63$.
Dividing by $63$,we get $\frac{x^2}{9} - \frac{y^2}{7} = 1$.
Here,$a^2 = 9$ and $b^2 = 7$,so $a = 3$ and $b = \sqrt{7}$.
The equations of the asymptotes are $y = \pm \frac{b}{a}x$,which are $y = \frac{\sqrt{7}}{3}x$ and $y = -\frac{\sqrt{7}}{3}x$.
The slopes are $m_1 = \frac{\sqrt{7}}{3}$ and $m_2 = -\frac{\sqrt{7}}{3}$.
The angle $\theta$ between the asymptotes is given by $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.
Since the asymptotes are symmetric about the axes,the angle $2\alpha$ between them satisfies $\tan \alpha = \frac{b}{a} = \frac{\sqrt{7}}{3}$.
Then $\cos 2\alpha = \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} = \frac{1 - 7/9}{1 + 7/9} = \frac{2/9}{16/9} = \frac{2}{16} = \frac{1}{8}$.
Thus,$\cos \theta = \frac{1}{8}$.

(A) The equation of a hyperbola with asymptotes $L_1 = 0$ and $L_2 = 0$ is given by $L_1 L_2 = k$,where $k$ is a constant.
Given asymptotes are $L_1: 3x - 4y - 1 = 0$ and $L_2: 4x - 3y - 6 = 0$.
The axes of the hyperbola are the angle bisectors of the asymptotes.
The equations of the angle bisectors are given by $\frac{3x - 4y - 1}{\sqrt{3^2 + (-4)^2}} = \pm \frac{4x - 3y - 6}{\sqrt{4^2 + (-3)^2}}$.
Since the denominators are equal,we have $3x - 4y - 1 = \pm (4x - 3y - 6)$.
Case $1$: $3x - 4y - 1 = 4x - 3y - 6 \implies x + y - 5 = 0$.
Case $2$: $3x - 4y - 1 = -(4x - 3y - 6) \implies 3x - 4y - 1 = -4x + 3y + 6 \implies 7x - 7y - 7 = 0 \implies x - y - 1 = 0$.
Thus,the axes are $x + y - 5 = 0$ and $x - y - 1 = 0$.

(D) The asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are $y = \pm \frac{b}{a}x$.
Let $2\theta$ be the angle between the asymptotes. Then $\tan \theta = \frac{b}{a}$.
Given that the angle between the asymptotes is $2 \tan^{-1}\left(\frac{2}{3}\right)$,we have $\tan \theta = \frac{2}{3}$,so $\frac{b}{a} = \frac{2}{3}$,which implies $b = \frac{2}{3}a$.
Substitute $b^2 = \frac{4}{9}a^2$ into the given equation $a^2 - b^2 = 45$:
$a^2 - \frac{4}{9}a^2 = 45$
$\frac{5}{9}a^2 = 45$
$a^2 = 45 \times \frac{9}{5} = 81$,so $a = 9$.
Then $b^2 = \frac{4}{9}(81) = 36$,so $b = 6$.
Therefore,$ab = 9 \times 6 = 54$.

(D) The equation of the hyperbola is $x^2-k y^2=3$,which can be written as $\frac{x^2}{3}-\frac{y^2}{3/k}=1$.
Here $a^2=3$ and $b^2=\frac{3}{k}$.
The angle between the asymptotes is $2\tan^{-1}\left(\frac{b}{a}\right) = \frac{\pi}{3}$,so $\tan^{-1}\left(\frac{b}{a}\right) = \frac{\pi}{6}$.
Thus,$\frac{b}{a} = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$.
Squaring gives $\frac{b^2}{a^2} = \frac{1}{3}$ $\Rightarrow \frac{3/k}{3} = \frac{1}{3}$ $\Rightarrow \frac{1}{k} = \frac{1}{3}$ $\Rightarrow k=3$.
The hyperbola is $\frac{x^2}{3}-y^2=1$.
Eccentricity $e = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.
The pole of the line $lx+my+n=0$ with respect to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\left(-\frac{a^2l}{n}, \frac{b^2m}{n}\right)$.
For $x+y-1=0$,$l=1, m=1, n=-1$.
Pole is $\left(-\frac{3(1)}{-1}, \frac{1(1)}{-1}\right) = (3, -1)$.
Since $k=3$ and $e=\frac{2}{\sqrt{3}}$,we have $-1 = -\frac{\sqrt{3}}{2} \times \frac{2}{\sqrt{3}} = -\frac{\sqrt{3}e}{2}$.
Thus,the pole is $\left(k, -\frac{\sqrt{3}e}{2}\right)$.

(C) The given hyperbola is $5x^2 - 4y^2 = 20$,which can be written as $\frac{x^2}{4} - \frac{y^2}{5} = 1$.
Its asymptotes are given by $\frac{x^2}{4} - \frac{y^2}{5} = 0$,which implies $\sqrt{5}x - 2y = 0$ and $\sqrt{5}x + 2y = 0$.
Let $(x_0, y_0)$ be a point on the hyperbola,so $5x_0^2 - 4y_0^2 = 20$.
The lengths of the perpendiculars from $(x_0, y_0)$ to the asymptotes are $l_1 = \frac{|\sqrt{5}x_0 - 2y_0|}{\sqrt{(\sqrt{5})^2 + (-2)^2}} = \frac{|\sqrt{5}x_0 - 2y_0|}{3}$ and $l_2 = \frac{|\sqrt{5}x_0 + 2y_0|}{\sqrt{(\sqrt{5})^2 + 2^2}} = \frac{|\sqrt{5}x_0 + 2y_0|}{3}$.
Thus,$l_1 l_2 = \frac{|5x_0^2 - 4y_0^2|}{9} = \frac{20}{9}$.
Therefore,$\frac{l_1^2 l_2^2}{100} = \frac{(20/9)^2}{100} = \frac{400/81}{100} = \frac{4}{81}$.

(D) The given equation of the hyperbola is $4x^2 - 9y^2 - 24x - 36y - 36 = 0$.
Rearranging the terms,we get $4(x^2 - 6x) - 9(y^2 + 4y) = 36$.
Completing the square,$4(x^2 - 6x + 9) - 9(y^2 + 4y + 4) = 36 + 36 - 36$.
This simplifies to $4(x - 3)^2 - 9(y + 2)^2 = 36$.
Dividing by $36$,we get $\frac{(x - 3)^2}{9} - \frac{(y + 2)^2}{4} = 1$.
The equation of the pair of asymptotes is obtained by setting the constant term to zero: $\frac{(x - 3)^2}{9} - \frac{(y + 2)^2}{4} = 0$.
Multiplying by $36$,we get $4(x - 3)^2 - 9(y + 2)^2 = 0$.
Expanding this,$4(x^2 - 6x + 9) - 9(y^2 + 4y + 4) = 0$.
$4x^2 - 24x + 36 - 9y^2 - 36y - 36 = 0$.
Thus,the equation is $4x^2 - 9y^2 - 24x - 36y = 0$.

(B) The equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.
It is a standard property that the area of the triangle formed by the asymptotes of a hyperbola with any tangent is constant and equal to $ab$.
Given the eccentricity $e = \sec \alpha$,we have $e^2 = 1 + \frac{b^2}{a^2} = \sec^2 \alpha$.
Since $1 + \tan^2 \alpha = \sec^2 \alpha$,we get $\frac{b^2}{a^2} = \tan^2 \alpha$,which implies $b = a \tan \alpha$.
Substituting this into the area formula,the area is $a(a \tan \alpha) = a^2 \tan \alpha$.
However,the standard result for the area is $ab$.
Given $b = a \tan \alpha$,the area is $a^2 \tan \alpha$.
Comparing with the options,if we evaluate $ab$,we get $a(a \tan \alpha) = a^2 \tan \alpha$.
Since the question asks for the area in terms of the given parameters,and $ab$ is the constant area,the correct value is $ab$.

(B) The equation of the hyperbola is $\frac{x^2}{a^2}-\frac{(y-2)^2}{4}=1$.
Given $\cos \theta = \frac{5}{13}$,we have $\tan \theta = \frac{12}{5}$ (assuming $\theta$ is acute).
The asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ are $y-k = \pm \frac{b}{a}x$.
The slopes are $m_1 = \frac{b}{a}$ and $m_2 = -\frac{b}{a}$.
The angle $\theta$ between the asymptotes is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{2b/a}{1 - b^2/a^2} \right| = \left| \frac{2ab}{a^2 - b^2} \right|$.
Here $b^2 = 4$,so $b = 2$. Thus,$\tan \theta = \left| \frac{4a}{a^2 - 4} \right|$.
Equating the two expressions for $\tan \theta$: $\frac{12}{5} = \frac{4a}{a^2 - 4}$ or $\frac{12}{5} = -\frac{4a}{a^2 - 4}$.
Case $1$: $3(a^2 - 4) = 5a$ $\Rightarrow 3a^2 - 5a - 12 = 0$ $\Rightarrow (3a + 4)(a - 3) = 0$.
Since $a^2 > 0$,$a = 3 \Rightarrow a^2 = 9$.
Case $2$: $3(a^2 - 4) = -5a$ $\Rightarrow 3a^2 + 5a - 12 = 0$ $\Rightarrow (3a - 4)(a + 3) = 0$.
Since $a^2 > 0$,$a = 4/3 \Rightarrow a^2 = 16/9$.
Thus,$a^2 = \frac{16}{9}$ or $9$.

(D) The equation of the auxiliary circle is $x^2+y^2-2x-4y+3=0$.
Rewriting it as $(x-1)^2+(y-2)^2 = 1+4-3 = 2$.
The centre of the hyperbola is the same as the centre of the auxiliary circle,which is $(1, 2)$.
The radius of the auxiliary circle is $a = \sqrt{2}$.
Since the asymptotes are at right angles,the hyperbola is a rectangular hyperbola,so $e = \sqrt{2}$.
For a rectangular hyperbola,$a = b$,so $b = \sqrt{2}$.
The equation of the hyperbola with transverse axis parallel to the $X$-axis is $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$.
Substituting the values: $\frac{(x-1)^2}{2} - \frac{(y-2)^2}{2} = 1$.
$(x-1)^2 - (y-2)^2 = 2$.
$x^2-2x+1 - (y^2-4y+4) = 2$.
$x^2-y^2-2x+4y-3-2 = 0$.
$x^2-y^2-2x+4y-5 = 0$.

(C) The given equation of the hyperbola is $x^2-2 y^2-8 x+8 y+4=0$.
Completing the square,we get:
$(x^2-8 x+16) - 2(y^2-4 y+4) + 4 - 16 + 8 = 0$
$(x-4)^2 - 2(y-2)^2 = 4$.
The equation of the pair of asymptotes is obtained by setting the constant term to zero relative to the center:
$(x-4)^2 - 2(y-2)^2 = 0$.
Expanding this,we get:
$(x^2-8 x+16) - 2(y^2-4 y+4) = 0$
$x^2-8 x+16 - 2 y^2+8 y-8 = 0$
$x^2-2 y^2-8 x+8 y+8 = 0$.

(D) The equation of the hyperbola is $\frac{(x-3)^2}{3}-\frac{(y-2)^2}{2}=1$.
Expanding this,we get $2(x^2-6x+9) - 3(y^2-4y+4) = 6$,which simplifies to $2x^2 - 3y^2 - 12x + 12y + 18 - 12 = 6$,or $2x^2 - 3y^2 - 12x + 12y = 0$.
The equation of the pair of asymptotes differs from the hyperbola equation only by a constant term.
Let the equation of the pair of asymptotes be $2x^2 - 3y^2 - 12x + 12y + \lambda = 0$.
For this to represent a pair of straight lines,the determinant $\Delta$ must be zero.
Comparing with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we have $a=2, b=-3, h=0, g=-6, f=6, c=\lambda$.
The condition $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ gives $2(-3)(\lambda) + 2(6)(-6)(0) - 2(6)^2 - (-3)(-6)^2 - \lambda(0)^2 = 0$.
This simplifies to $-6\lambda - 72 + 108 = 0$,which gives $-6\lambda + 36 = 0$,so $\lambda = 6$.
Thus,the equation of the pair of asymptotes is $2x^2 - 3y^2 - 12x + 12y + 6 = 0$.

(A) The asymptotes of a hyperbola are given by the equations $x+y+1=0$ and $x-y+4=0$.
Solving these equations simultaneously gives the center of the hyperbola: $x+y=-1$ and $x-y=-4$. Adding them gives $2x = -5$,so $x = -2.5$. Subtracting gives $2y = 3$,so $y = 1.5$. Thus,the center is $(-2.5, 1.5)$.
The equation of the hyperbola is $(x+2.5+y-1.5)(x+2.5-(y-1.5)) = \lambda$,which simplifies to $(x+y+1)(x-y+4) = k$.
Since the point $(1,1)$ lies on the hyperbola,we substitute $x=1$ and $y=1$: $(1+1+1)(1-1+4) = k \Rightarrow 3 \times 4 = 12$. So $k=12$.
The equation is $(x+2.5)^2 - (y-1.5)^2 = 12$.
Comparing this to $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$,we have $a^2 = 12$ and $b^2 = 12$,so $a = \sqrt{12} = 2\sqrt{3}$ and $b = 2\sqrt{3}$.
The length of the latus rectum is $\frac{2b^2}{a} = \frac{2 \times 12}{2\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}$.

(D) The equations of the asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are $bx - ay = 0$ and $bx + ay = 0$.
Let $P(a \sec \theta, b \tan \theta)$ be any point on the hyperbola.
The perpendicular distance from $P$ to the asymptote $bx - ay = 0$ is $PQ = \frac{|b(a \sec \theta) - a(b \tan \theta)|}{\sqrt{b^2 + a^2}} = \frac{ab(\sec \theta - \tan \theta)}{\sqrt{a^2 + b^2}}$.
The perpendicular distance from $P$ to the asymptote $bx + ay = 0$ is $PR = \frac{|b(a \sec \theta) + a(b \tan \theta)|}{\sqrt{b^2 + a^2}} = \frac{ab(\sec \theta + \tan \theta)}{\sqrt{a^2 + b^2}}$.
The product of these distances is given as $6$:
$\frac{a^2 b^2(\sec^2 \theta - \tan^2 \theta)}{a^2 + b^2} = 6$
Since $\sec^2 \theta - \tan^2 \theta = 1$,we have $\frac{a^2 b^2}{a^2 + b^2} = 6$.
Given $e = \sqrt{3}$,we know $b^2 = a^2(e^2 - 1) = a^2(3 - 1) = 2a^2$.
Substituting $b^2 = 2a^2$ into the equation:
$\frac{a^2(2a^2)}{a^2 + 2a^2} = 6$ $\Rightarrow \frac{2a^4}{3a^2} = 6$ $\Rightarrow \frac{2}{3}a^2 = 6$ $\Rightarrow a^2 = 9$.
Then $b^2 = 2(9) = 18$,so $b = \sqrt{18} = 3\sqrt{2}$.
The length of the conjugate axis is $2b = 2(3\sqrt{2}) = 6\sqrt{2}$.

(A) The equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.
The asymptotes of the hyperbola are given by $\frac{x}{a} \pm \frac{y}{b} = 0$.
It is a standard property that the area of the triangle formed by any tangent to the hyperbola and its asymptotes is constant and equal to $ab$.
Given that the area is $a^2 \tan(\alpha)$,we have $ab = a^2 \tan(\alpha)$,which implies $b = a \tan(\alpha)$.
The eccentricity $e$ of the hyperbola is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$.
Substituting $b = a \tan(\alpha)$,we get $e = \sqrt{1 + \frac{a^2 \tan^2(\alpha)}{a^2}} = \sqrt{1 + \tan^2(\alpha)} = \sqrt{\sec^2(\alpha)} = \sec(\alpha)$.

(B) The equation of the asymptotes of the hyperbola $2 x^2+5 x y+2 y^2-11 x-7 y-4=0$ is of the form $2 x^2+5 x y+2 y^2-11 x-7 y+\lambda=0$.
Since this represents a pair of straight lines,the determinant of the general second-degree equation must be zero:
$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Here,$a=2, b=2, c=\lambda, h=\frac{5}{2}, g=-\frac{11}{2}, f=-\frac{7}{2}$.
Substituting these values:
$2(2)(\lambda) + 2(-\frac{7}{2})(-\frac{11}{2})(\frac{5}{2}) - 2(-\frac{7}{2})^2 - 2(-\frac{11}{2})^2 - \lambda(\frac{5}{2})^2 = 0$.
$4\lambda + \frac{385}{4} - \frac{49}{2} - \frac{121}{2} - \frac{25\lambda}{4} = 0$.
Multiplying by $4$:
$16\lambda + 385 - 98 - 242 - 25\lambda = 0$.
$-9\lambda + 45 = 0 \Rightarrow \lambda = 5$.
Thus,the equation is $2 x^2+5 x y+2 y^2-11 x-7 y+5=0$.

(B) The equation of the pair of asymptotes is given by the product of the lines: $(3x+4y-2)(2x+y+1)=0$.
Since the equation of the hyperbola differs from the equation of its asymptotes only by a constant $\lambda$,we can write: $(3x+4y-2)(2x+y+1)=\lambda$.
Given that the hyperbola passes through the point $(1,1)$,we substitute $x=1$ and $y=1$ into the equation: $(3(1)+4(1)-2)(2(1)+1+1)=\lambda$.
$(5)(4)=\lambda$,which gives $\lambda=20$.
Substituting $\lambda=20$ back into the equation: $(3x+4y-2)(2x+y+1)=20$.
Expanding the left side: $6x^2+3xy+3x+8xy+4y^2+4y-4x-2y-2=20$.
Simplifying: $6x^2+11xy+4y^2-x+2y-2=20$.
Thus,the equation is $6x^2+11xy+4y^2-x+2y-22=0$.

(B) The given hyperbola is $16x^2 - 25y^2 = 400$,which can be written as $\frac{x^2}{25} - \frac{y^2}{16} = 1$.
Here,$a^2 = 25$ and $b^2 = 16$,so $a = 5$ and $b = 4$.
The equations of the asymptotes are $y = \pm \frac{b}{a}x$,i.e.,$4x - 5y = 0$ and $4x + 5y = 0$.
Let $P(x_1, y_1)$ be any point on the hyperbola,so $16x_1^2 - 25y_1^2 = 400$.
The lengths of the perpendiculars from $P$ to the asymptotes are $d_1 = \frac{|4x_1 - 5y_1|}{\sqrt{4^2 + 5^2}} = \frac{|4x_1 - 5y_1|}{\sqrt{41}}$ and $d_2 = \frac{|4x_1 + 5y_1|}{\sqrt{4^2 + 5^2}} = \frac{|4x_1 + 5y_1|}{\sqrt{41}}$.
The product $p = d_1 d_2 = \frac{|16x_1^2 - 25y_1^2|}{41} = \frac{400}{41}$.
The angle $\theta$ between the asymptotes is given by $\tan \frac{\theta}{2} = \frac{b}{a} = \frac{4}{5}$.
Therefore,$p \tan \frac{\theta}{2} = \frac{400}{41} \times \frac{4}{5} = \frac{80 \times 4}{41} = \frac{320}{41}$.

(B) The equation of the hyperbola is $14 x^2+38 x y+20 y^2+x-7 y-91=0$.
Asymptotes of a hyperbola differ from the hyperbola equation by a constant.
Let the asymptotes be $(7 x+5 y+c_1)(2 x+4 y+c_2) = 14 x^2+38 x y+20 y^2+x-7 y+k = 0$.
Expanding the product: $14 x^2+28 x y+7 x c_2+10 x y+20 y^2+5 y c_2+2 x c_1+4 y c_1+c_1 c_2 = 14 x^2+38 x y+20 y^2+x(7 c_2+2 c_1)+y(5 c_2+4 c_1)+c_1 c_2$.
Comparing the coefficients with the given hyperbola equation:
$7 c_2+2 c_1 = 1$
$5 c_2+4 c_1 = -7$
Given one asymptote is $7 x+5 y-3=0$,so $c_1 = -3$.
Substituting $c_1 = -3$ in the first equation: $7 c_2 + 2(-3) = 1$ $\Rightarrow 7 c_2 - 6 = 1$ $\Rightarrow 7 c_2 = 7$ $\Rightarrow c_2 = 1$.
Thus,the other asymptote is $2 x+4 y+1=0$.

(C) For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the asymptotes are given by $\frac{x}{a} - \frac{y}{b} = 0$ and $\frac{x}{a} + \frac{y}{b} = 0$.
Here,$a^2 = 9$ and $b^2 = 4$,so $a = 3$ and $b = 2$.
The asymptotes are $\frac{x}{3} - \frac{y}{2} = 0$ or $2x - 3y = 0$ and $2x + 3y = 0$.
Let $P(x_0, y_0)$ be any point on the hyperbola,so $\frac{x_0^2}{9} - \frac{y_0^2}{4} = 1$.
The perpendicular distance $d_1$ from $P$ to $2x - 3y = 0$ is $d_1 = \frac{|2x_0 - 3y_0|}{\sqrt{2^2 + (-3)^2}} = \frac{|2x_0 - 3y_0|}{\sqrt{13}}$.
The perpendicular distance $d_2$ from $P$ to $2x + 3y = 0$ is $d_2 = \frac{|2x_0 + 3y_0|}{\sqrt{2^2 + 3^2}} = \frac{|2x_0 + 3y_0|}{\sqrt{13}}$.
The product $d_1 d_2 = \frac{|(2x_0 - 3y_0)(2x_0 + 3y_0)|}{13} = \frac{|4x_0^2 - 9y_0^2|}{13}$.
Since $\frac{x_0^2}{9} - \frac{y_0^2}{4} = 1$,we have $4x_0^2 - 9y_0^2 = 36$.
Thus,$d_1 d_2 = \frac{36}{13}$.

(B) The equation of the hyperbola is given by $(x+y+3)(2x-y+1) + \lambda = 0$.
Since the point $(1,-2)$ lies on the hyperbola,we substitute $x=1$ and $y=-2$:
$(1-2+3)(2(1)-(-2)+1) + \lambda = 0$
$(2)(5) + \lambda = 0 \implies \lambda = -10$.
Thus,the equation of the hyperbola is $(x+y+3)(2x-y+1) - 10 = 0$.
Expanding this: $2x^2 - xy + x + 2xy - y^2 + y + 6x - 3y + 3 - 10 = 0$,which simplifies to $2x^2 + xy - y^2 + 7x - 2y - 7 = 0$.
The equation of the conjugate hyperbola is $(x+y+3)(2x-y+1) + \lambda' = 0$.
Since the conjugate hyperbola passes through the point symmetric to $(1,-2)$ with respect to the center of the hyperbola,or by using the property that the constant term changes such that the sum of the constants of the hyperbola and its conjugate is $2 \times (\text{constant of the product of asymptotes})$,we find the constant term.
Alternatively,the equation of the conjugate hyperbola is $(x+y+3)(2x-y+1) + 10 = 0$.
Expanding this: $2x^2 + xy - y^2 + 7x - 2y + 3 + 10 = 0$,which simplifies to $2x^2 + xy - y^2 + 7x - 2y + 13 = 0$.

(B) Given,$x=35 \sec \theta$ and $y=35 \tan \theta$.
Point $P = (35 \sec \theta, 35 \tan \theta)$.
Differentiating $x$ and $y$ with respect to $\theta$:
$\frac{dx}{d\theta} = 35 \sec \theta \tan \theta$
$\frac{dy}{d\theta} = 35 \sec^2 \theta$
Slope of the tangent $m = \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{35 \sec^2 \theta}{35 \sec \theta \tan \theta} = \frac{\sec \theta}{\tan \theta} = \frac{1}{\sin \theta}$.
Equation of the tangent at $P$ is $y - y_1 = m(x - x_1)$:
$y - 35 \tan \theta = \frac{1}{\sin \theta} (x - 35 \sec \theta)$
$y \sin \theta - 35 \tan \theta \sin \theta = x - 35 \sec \theta$
$y \sin \theta - 35 \frac{\sin^2 \theta}{\cos \theta} = x - \frac{35}{\cos \theta}$
$y \sin \theta = x + 35 \frac{\sin^2 \theta}{\cos \theta} - \frac{35}{\cos \theta}$
$y \sin \theta = x + \frac{35(\sin^2 \theta - 1)}{\cos \theta}$
$y \sin \theta = x + \frac{35(-\cos^2 \theta)}{\cos \theta}$
$y \sin \theta = x - 35 \cos \theta$.
Thus,the correct option is $B$.

(A) Let the third vertex be $P(x, y)$. The other two vertices are $A(-5, 0)$ and $B(6, 0)$.
The perimeter is the sum of the lengths of the sides: $PA + PB + AB = 20$.
The distance $AB = \sqrt{(6 - (-5))^2 + (0 - 0)^2} = \sqrt{11^2} = 11$.
So,$PA + PB = 20 - 11 = 9$.
Substituting the coordinates,$\sqrt{(x + 5)^2 + y^2} + \sqrt{(x - 6)^2 + y^2} = 9$.
Rearranging: $\sqrt{(x + 5)^2 + y^2} = 9 - \sqrt{(x - 6)^2 + y^2}$.
Squaring both sides: $(x + 5)^2 + y^2 = 81 + (x - 6)^2 + y^2 - 18\sqrt{(x - 6)^2 + y^2}$.
$x^2 + 10x + 25 + y^2 = 81 + x^2 - 12x + 36 + y^2 - 18\sqrt{(x - 6)^2 + y^2}$.
$22x - 92 = -18\sqrt{(x - 6)^2 + y^2}$.
Dividing by $-2$: $46 - 11x = 9\sqrt{(x - 6)^2 + y^2}$.
Squaring again: $(46 - 11x)^2 = 81((x - 6)^2 + y^2)$.
$2116 - 1012x + 121x^2 = 81(x^2 - 12x + 36 + y^2)$.
$2116 - 1012x + 121x^2 = 81x^2 - 972x + 2916 + 81y^2$.
$40x^2 - 81y^2 - 40x - 800 = 0$.

(C) The given equation is $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$.
Multiplying by $8r$,we get:
$8 = r + 3r \cos \theta$
Since $x = r \cos \theta$,we have $r = 8 - 3x$.
Squaring both sides:
$r^2 = (8 - 3x)^2$
$x^2 + y^2 = 64 + 9x^2 - 48x$
$8x^2 - y^2 - 48x + 64 = 0$.
Comparing this with the general equation of a conic section $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$,here $A = 8$ and $C = -1$.
Since $A$ and $C$ have opposite signs $(AC < 0)$,the equation represents a hyperbola.

(A) For $f(x) = -3x^2+4x+1$,the maximum value $l$ occurs at $x = -b/(2a) = -4/(2 \times -3) = 2/3$.
$l = f(2/3) = -3(4/9) + 4(2/3) + 1 = -4/3 + 8/3 + 1 = 7/3$.
For $g(x) = 3x^2+4x+1$,the minimum value $m$ occurs at $x = -4/(2 \times 3) = -2/3$.
$m = g(-2/3) = 3(4/9) + 4(-2/3) + 1 = 4/3 - 8/3 + 1 = -1/3$.
The foci are $(l, 0) = (7/3, 0)$ and $(7m, 0) = (7 \times -1/3, 0) = (-7/3, 0)$.
The center of the hyperbola is $(0, 0)$. The distance between foci is $2ae = 7/3 - (-7/3) = 14/3$.
Given $e = 2$,we have $2a(2) = 14/3 \implies 4a = 14/3 \implies a = 7/6$.
Also,$b^2 = a^2(e^2-1) = (49/36)(4-1) = (49/36)(3) = 49/12$.
The equation of the hyperbola is $x^2/a^2 - y^2/b^2 = 1$.
$x^2/(49/36) - y^2/(49/12) = 1 \implies 36x^2/49 - 12y^2/49 = 1 \implies 36x^2 - 12y^2 = 49$.

(B) For the first hyperbola,the transverse axis is $2a$ and the conjugate axis is $2b$. Given $2a = 2(2b)$,so $a = 2b$. The eccentricity $x$ is given by $x^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{b^2}{(2b)^2} = 1 + \frac{1}{4} = \frac{5}{4}$.
For the second hyperbola,the distance between the foci is $2ae$ and the distance between the directrices is $\frac{2a}{e}$. Given $2ae = 3 \times \frac{2a}{e}$,which simplifies to $e^2 = 3$. Thus,$y^2 = 3$.
Therefore,$y^2 - x^2 = 3 - \frac{5}{4} = \frac{12-5}{4} = \frac{7}{4}$.

(D) For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 9$,so $a = 3$. The foci are $(\pm ae, 0)$ and vertices are $(\pm a, 0)$.
Let the focus be $S(ae, 0)$ and the farther vertex be $A'(-a, 0)$.
The latus rectum is the line $x = ae$. The endpoints of the latus rectum are $L(ae, \frac{b^2}{a})$ and $L'(ae, -\frac{b^2}{a})$.
The vector $\vec{A'L} = (ae - (-a), \frac{b^2}{a} - 0) = (a(e+1), \frac{b^2}{a})$.
The vector $\vec{A'L'} = (ae - (-a), -\frac{b^2}{a} - 0) = (a(e+1), -\frac{b^2}{a})$.
Since the angle $\angle L A' L'$ is $90^\circ$,the dot product $\vec{A'L} \cdot \vec{A'L'} = 0$.
$a^2(e+1)^2 - \frac{b^4}{a^2} = 0 \implies a^4(e+1)^2 = b^4$.
Since $b^2 = a^2(e^2-1) = a^2(e-1)(e+1)$,we have $b^4 = a^4(e-1)^2(e+1)^2$.
Equating the two expressions for $b^4$: $a^4(e+1)^2 = a^4(e-1)^2(e+1)^2$.
Dividing by $a^4(e+1)^2$,we get $1 = (e-1)^2$,so $e-1 = 1$,which means $e = 2$.
Then $b^2 = a^2(e^2-1) = 9(2^2-1) = 9(3) = 27$.

(C) The given equation is $\frac{x^2}{k-\frac{5}{2}} + \frac{y^2}{\frac{7}{3}-k} = 1$.
For this to represent a hyperbola,the denominators must have opposite signs,i.e.,their product must be negative:
$(k - \frac{5}{2})(\frac{7}{3} - k) < 0$.
Multiplying by $-1$,we get $(k - \frac{5}{2})(k - \frac{7}{3}) > 0$.
Since $\frac{7}{3} \approx 2.33$ and $\frac{5}{2} = 2.5$,we have $\frac{7}{3} < \frac{5}{2}$.
The inequality $(k - \frac{7}{3})(k - \frac{5}{2}) > 0$ holds when $k < \frac{7}{3}$ or $k > \frac{5}{2}$.
Thus,the set of all values of $k$ is $\left(-\infty, \frac{7}{3}\right) \cup \left(\frac{5}{2}, \infty\right)$.

(C) The given hyperbola is $\frac{x^2}{25} - \frac{y^2}{9} = 1$.
Any point on this hyperbola is given by $(5 \sec \theta, 3 \tan \theta)$.
Given $P(\theta) = (x_1, \frac{3 \sqrt{5}}{2})$,we equate the $y$-coordinates: $3 \tan \theta = \frac{3 \sqrt{5}}{2} \implies \tan \theta = \frac{\sqrt{5}}{2}$.
Since $1 + \tan^2 \theta = \sec^2 \theta$,we have $\sec^2 \theta = 1 + \frac{5}{4} = \frac{9}{4}$,so $\sec \theta = \frac{3}{2}$ (as $0 < \theta < \frac{\pi}{2}$).
Thus,$x_1 = 5 \sec \theta = 5 \times \frac{3}{2} = \frac{15}{2}$.
Also,$\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{1}{\sec^2 \theta} = 1 - \frac{4}{9} = \frac{5}{9}$.
Finally,$2 x_1 + 9 \sin^2 \theta = 2(\frac{15}{2}) + 9(\frac{5}{9}) = 15 + 5 = 20$.

(C) Given the hyperbola equation $\frac{x^2}{16}-\frac{y^2}{9}=1$,we have $a^2=16$ and $b^2=9$.
Since point $P(5, y_1)$ lies on the hyperbola,we substitute $x=5$:
$\frac{25}{16}-\frac{y_1^2}{9}=1$
$\Rightarrow \frac{y_1^2}{9}=\frac{25}{16}-1 = \frac{9}{16}$
$\Rightarrow y_1^2 = \frac{81}{16}$ $\Rightarrow y_1 = \pm \frac{9}{4}$.
Thus,$P = (5, \pm \frac{9}{4})$.
The eccentricity $e$ is given by $e = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{9}{16}} = \frac{5}{4}$.
The focus $S$ on the positive $X$-axis is $(ae, 0) = (4 \times \frac{5}{4}, 0) = (5, 0)$.
Now,the distance $SP$ is:
$SP = \sqrt{(5-5)^2 + (0 - (\pm \frac{9}{4}))^2} = \sqrt{0 + \frac{81}{16}} = \frac{9}{4}$.

(C) The given equations are in parametric form $x = a \sec \theta, y = b \tan \theta$,which represents the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
The eccentricity $e$ is given by $e^2 = 1 + \frac{b^2}{a^2} = \frac{a^2 + b^2}{a^2}$.
For the first hyperbola,$a = 1$ and $b = \sqrt{2}$,so $e_1^2 = 1 + \frac{(\sqrt{2})^2}{1^2} = 1 + 2 = 3$.
For the second hyperbola,$a = \sqrt{2}$ and $b = 1$,so $e_2^2 = 1 + \frac{1^2}{(\sqrt{2})^2} = 1 + \frac{1}{2} = \frac{3}{2}$.
Therefore,$\frac{e_2^2}{e_1^2} = \frac{3/2}{3} = \frac{1}{2}$.

(C) The equation of the 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 the length of the transverse axis $2a = 8$,we have $a = 4$,so $a^2 = 16$.
The hyperbola passes through $(5, 2)$,so $\frac{25}{16} - \frac{4}{b^2} = 1$.
$\frac{4}{b^2} = \frac{25}{16} - 1 = \frac{9}{16}$,which gives $b^2 = \frac{64}{9}$.
The conjugate hyperbola is $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$.
The eccentricity $e'$ of the conjugate hyperbola is given by $e' = \sqrt{1 + \frac{a^2}{b^2}}$.
$e' = \sqrt{1 + \frac{16}{64/9}} = \sqrt{1 + \frac{16 \times 9}{64}} = \sqrt{1 + \frac{9}{4}} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2}$.

(B) For the first hyperbola,the distance between foci is $2ae_1$ and the distance between directrices is $\frac{2a}{e_1}$.
Given $2ae_1 = 2 \times \frac{2a}{e_1}$,which simplifies to $e_1^2 = 2$. Since $e_1 > 1$,we have $e_1 = \sqrt{2}$.
For the second hyperbola,the length of the transverse axis is $2a_2$ and the length of the conjugate axis is $2b_2$.
Given $2a_2 = 2(2b_2)$,so $a_2 = 2b_2$ or $b_2 = \frac{a_2}{2}$.
The eccentricity $e_2$ is given by $e_2 = \sqrt{1 + \frac{b_2^2}{a_2^2}} = \sqrt{1 + \frac{(a_2/2)^2}{a_2^2}} = \sqrt{1 + \frac{1}{4}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$.
Therefore,$e_1 e_2 = \sqrt{2} \times \frac{\sqrt{5}}{2} = \frac{\sqrt{10}}{2}$.
Thus,option $B$ is correct.

(C) For a point $P$ on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the focal distances are $SP = |ex - a|$ and $S^{\prime}P = |ex + a|$.
Given the condition $SP + S^{\prime}P - 2|SP - S^{\prime}P| = 0$,we know that $|SP - S^{\prime}P| = 2a$.
Substituting this into the equation: $SP + S^{\prime}P = 2(2a) = 4a$.
For a hyperbola,the sum of focal distances is $SP + S^{\prime}P = 2ex$ (assuming $x > 0$).
Thus,$2ex = 4a$,which implies $ex = 2a$,or $x = \frac{2a}{e}$.
The point $P$ is given as $P(\frac{\pi}{6})$,which in parametric form is $(a \sec \theta, b \tan \theta)$.
Here,$x = a \sec(\frac{\pi}{6}) = a \cdot \frac{2}{\sqrt{3}}$.
Equating the two expressions for $x$: $\frac{2a}{e} = \frac{2a}{\sqrt{3}}$.
Therefore,$e = \sqrt{3}$.

(C) The equation of the hyperbola is $x^2 - 2y^2 = 1$. Comparing with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $a^2 = 1$ and $b^2 = \frac{1}{2}$.
The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1/2}{1}} = \sqrt{\frac{3}{2}}$.
The focus $S$ is $(ae, 0) = (\sqrt{\frac{3}{2}}, 0)$.
The line passing through $P(-1, 1)$ and $S(\sqrt{\frac{3}{2}}, 0)$ has slope $m = \frac{0 - 1}{\sqrt{\frac{3}{2}} - (-1)} = \frac{-1}{\sqrt{\frac{3}{2}} + 1} = \frac{-\sqrt{2}}{\sqrt{3} + \sqrt{2}}$.
The equation of line $PS$ is $y - 0 = m(x - \sqrt{\frac{3}{2}})$,which is $y = \frac{-\sqrt{2}}{\sqrt{3} + \sqrt{2}}(x - \sqrt{\frac{3}{2}})$.
To find the $Y$-intercept $(B)$,set $x = 0$: $y = \frac{-\sqrt{2}}{\sqrt{3} + \sqrt{2}}(-\sqrt{\frac{3}{2}}) = \frac{\sqrt{3}}{\sqrt{3} + \sqrt{2}}$.
The area of the triangle formed by the line with coordinate axes is $\frac{1}{2} \times |OS| \times |OB| = \frac{1}{2} \times \sqrt{\frac{3}{2}} \times \frac{\sqrt{3}}{\sqrt{3} + \sqrt{2}} = \frac{1}{2} \times \frac{3}{\sqrt{2}(\sqrt{3} + \sqrt{2})} = \frac{3}{2(\sqrt{6} + 2)}$.
Thus,the correct option is $C$.

(B) The equation of the hyperbola is $S = 4x^2 - 3y^2 - 1 = 0$.
For a point $(\alpha, \beta)$ to be in the interior of the hyperbola,the condition is $S_1 < 0$ if the point is between the branches,but for the standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the interior region containing the foci is defined by $S_1 < 0$.
However,checking the standard definition for $4x^2 - 3y^2 - 1 = 0$ at $(0,0)$,we get $S_1 = -1 < 0$,which is the interior region.
Substituting $(\alpha, -1)$ into $S < 0$:
$4\alpha^2 - 3(-1)^2 - 1 < 0$
$4\alpha^2 - 3 - 1 < 0$
$4\alpha^2 - 4 < 0$
$\alpha^2 < 1$
$-1 < \alpha < 1$
Thus,$\alpha \in (-1, 1)$.

(D) For the equation $\frac{x^2}{\alpha+3} + \frac{y^2}{2-\alpha} = 1$ to represent a hyperbola,the coefficients of $x^2$ and $y^2$ must have opposite signs.
This implies that their product must be negative:
$(\alpha+3)(2-\alpha) < 0$
Multiplying both sides by $-1$ reverses the inequality sign:
$(\alpha+3)(\alpha-2) > 0$
Solving this inequality,we find the roots are $\alpha = -3$ and $\alpha = 2$.
The expression is positive outside the interval between the roots.
Therefore,$\alpha \in (-\infty, -3) \cup (2, \infty)$.

(D) The foci of the hyperbola are $S(ae, 0)$ and $S'(-ae, 0)$.
The endpoints of the latus rectum passing through $S$ are $A(ae, \frac{b^2}{a})$ and $B(ae, -\frac{b^2}{a})$.
The angle subtended by $AB$ at $S'$ is $60^{\circ}$.
Since the triangle $\triangle AS'B$ is isosceles,the line $S'S$ bisects the angle $\angle AS'B$,so $\angle AS'S = 30^{\circ}$.
The slope of $S'A$ is $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$.
Thus,$\frac{b^2/a}{ae - (-ae)} = \frac{b^2}{2a^2e} = \frac{1}{\sqrt{3}}$.
This gives $\frac{b^2}{a^2} = \frac{2e}{\sqrt{3}}$.
Using the relation $b^2 = a^2(e^2 - 1)$,we have $\frac{b^2}{a^2} = e^2 - 1$.
Equating the two expressions: $e^2 - 1 = \frac{2e}{\sqrt{3}}$,which simplifies to $\sqrt{3}e^2 - 2e - \sqrt{3} = 0$.
Solving the quadratic equation: $(\sqrt{3}e + 1)(e - \sqrt{3}) = 0$.
Since the eccentricity $e > 1$,we get $e = \sqrt{3}$.

(D) Given equation: $5x^2 - 6y^2 - 10x - 24y - 34 = 0$
Rearranging terms: $5(x^2 - 2x) - 6(y^2 + 4y) = 34$
Completing the square: $5(x^2 - 2x + 1) - 6(y^2 + 4y + 4) = 34 + 5 - 24$
$5(x - 1)^2 - 6(y + 2)^2 = 15$
Dividing by $15$: $\frac{(x - 1)^2}{3} - \frac{(y + 2)^2}{2.5} = 1$
Here,$a^2 = 3$ and $b^2 = 2.5 = \frac{5}{2}$.
Eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{5/2}{3}} = \sqrt{1 + \frac{5}{6}} = \sqrt{\frac{11}{6}}$.
Foci are given by $(h \pm ae, k)$,where $(h, k) = (1, -2)$.
$ae = \sqrt{3} \times \sqrt{\frac{11}{6}} = \sqrt{\frac{33}{6}} = \sqrt{\frac{11}{2}}$.
Thus,the foci are $\left(1 \pm \sqrt{\frac{11}{2}}, -2\right)$.

(C) The eccentricity $e$ of the hyperbola is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$.
Given $e = \frac{5}{4}$,we have $1 + \frac{b^2}{a^2} = \left(\frac{5}{4}\right)^2 = \frac{25}{16}$.
Thus,$\frac{b^2}{a^2} = \frac{25}{16} - 1 = \frac{9}{16}$,which implies $b^2 = \frac{9}{16} a^2$.
The length of the latus rectum is $\frac{2b^2}{a} = 9$.
Substituting $b^2 = \frac{9}{16} a^2$ into the latus rectum equation:
$\frac{2}{a} \left(\frac{9}{16} a^2\right) = 9$
$\frac{9}{8} a = 9 \Rightarrow a = 8$.
Now,$b^2 = \frac{9}{16} (8)^2 = \frac{9}{16} \times 64 = 36$,so $b = 6$.
Therefore,$ab = 8 \times 6 = 48$.

(D) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Given that the eccentricity $e = \sqrt{2}$ and the distance between the foci is $2ae = 16$.
Substituting $e = \sqrt{2}$ into the equation $2ae = 16$,we get $2a(\sqrt{2}) = 16$.
Thus,$a = \frac{16}{2\sqrt{2}} = 4\sqrt{2}$,so $a^2 = (4\sqrt{2})^2 = 32$.
For a hyperbola,$b^2 = a^2(e^2 - 1)$.
Substituting the values,$b^2 = 32((\sqrt{2})^2 - 1) = 32(2 - 1) = 32$.
Therefore,the equation of the hyperbola is $\frac{x^2}{32} - \frac{y^2}{32} = 1$,which simplifies to $x^2 - y^2 = 32$.

(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Given that the length of the transverse axis is $2a = 12$,so $a = 6$.
Since the point $(8, 2)$ lies on the hyperbola,it must satisfy the equation:
$\frac{8^2}{a^2} - \frac{2^2}{b^2} = 1$
Substituting $a = 6$:
$\frac{64}{36} - \frac{4}{b^2} = 1$
$\frac{16}{9} - 1 = \frac{4}{b^2}$
$\frac{7}{9} = \frac{4}{b^2} \implies b^2 = \frac{36}{7}$.
The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$.
$e = \sqrt{1 + \frac{36/7}{36}} = \sqrt{1 + \frac{1}{7}} = \sqrt{\frac{8}{7}} = \frac{2 \sqrt{2}}{\sqrt{7}}$.

(B) The given equation of the hyperbola is $x^2-y^2-4x+2y+c=0$.
Completing the square,we get $(x-2)^2 - (y-1)^2 = 3-c$.
Let $a^2 = 3-c$. The equation becomes $\frac{(x-2)^2}{a^2} - \frac{(y-1)^2}{a^2} = 1$.
For this rectangular hyperbola,the eccentricity $e = \sqrt{2}$.
The focus is given by $x = 2 \pm ae = 2 \pm \sqrt{a^2} \cdot \sqrt{2} = 2 \pm \sqrt{2a^2}$.
Given the focus $S(2+2\sqrt{2}, k)$,we have $\sqrt{2a^2} = 2\sqrt{2} \implies 2a^2 = 8 \implies a^2 = 4$.
Since $a^2 = 3-c$,we have $4 = 3-c$,which gives $c = -1$.
The directrix is $x = 2 \pm \frac{a}{e} = 2 \pm \frac{2}{\sqrt{2}} = 2 \pm \sqrt{2}$.
The directrix adjacent to the focus $x = 2+2\sqrt{2}$ is $x = 2+\sqrt{2}$,which matches the given condition.
Thus,$c = -1$.

(B) For the ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$,we have $a^2=16$ and $b^2=25$. Since $b^2 > a^2$,the eccentricity $e_1$ is given by $e_1 = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
Given $e_1 e_2 = 1$,we have $\frac{3}{5} e_2 = 1$,which implies $e_2 = \frac{5}{3}$.
The foci of the ellipse are $(0, \pm be_1) = (0, \pm 5 \times \frac{3}{5}) = (0, \pm 3)$.
The hyperbola passes through $(0, \pm 3)$,so its transverse axis is along the $y$-axis. The equation is of the form $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$.
Since it passes through $(0, 3)$,we have $b^2 = 3^2 = 9$.
For a hyperbola,$e_2^2 = 1 + \frac{a^2}{b^2}$. Substituting $e_2 = \frac{5}{3}$ and $b^2 = 9$:
$(\frac{5}{3})^2 = 1 + \frac{a^2}{9}$ $\Rightarrow \frac{25}{9} = 1 + \frac{a^2}{9}$ $\Rightarrow \frac{16}{9} = \frac{a^2}{9}$ $\Rightarrow a^2 = 16$.
Thus,the equation of the hyperbola is $\frac{y^2}{9} - \frac{x^2}{16} = 1$.

(B) The line $2x + y = 1$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
The equation of the directrix is $x = \frac{a}{e}$.
The line passes through the point $(\frac{a}{e}, 0)$,so $2(\frac{a}{e}) + 0 = 1$,which gives $2a = e$.
The condition for the line $y = mx + c$ to be a tangent to the hyperbola is $c^2 = a^2m^2 - b^2$.
Here,$y = -2x + 1$,so $m = -2$ and $c = 1$.
Thus,$1^2 = a^2(-2)^2 - b^2$,which simplifies to $4a^2 - b^2 = 1$.
We know that for a hyperbola,$b^2 = a^2(e^2 - 1)$.
Substituting $b^2$ into the tangent condition: $4a^2 - a^2(e^2 - 1) = 1$.
Since $e = 2a$,we have $a = \frac{e}{2}$,so $a^2 = \frac{e^2}{4}$.
Substituting $a^2$: $4(\frac{e^2}{4}) - \frac{e^2}{4}(e^2 - 1) = 1$.
$e^2 - \frac{e^4}{4} + \frac{e^2}{4} = 1$.
Multiply by $4$: $4e^2 - e^4 + e^2 = 4$.
$e^4 - 5e^2 + 4 = 0$.
$(e^2 - 4)(e^2 - 1) = 0$.
Since $e > 1$ for a hyperbola,$e^2 = 4$,so $e = 2$.

(D) Let $f(e) = 2e^3 + 10e - 13$.
Since $f(1) = 2(1)^3 + 10(1) - 13 = 2 + 10 - 13 = -1 < 0$ and $f(2) = 2(8) + 10(2) - 13 = 16 + 20 - 13 = 23 > 0$.
By the Intermediate Value Theorem,there exists a root $e$ in the interval $(1, 2)$.
Since the eccentricity $e$ of a conic section satisfies $e > 1$,the conic is a hyperbola.

(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
The given equation of the ellipse is $\frac{x^2}{13^2} + \frac{y^2}{5^2} = 1$.
For the ellipse,$a_e = 13$ and $b_e = 5$.
The eccentricity of the ellipse $e$ is given by $e = \sqrt{1 - \frac{b_e^2}{a_e^2}} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$.
The foci of the ellipse are $(\pm a_e e, 0) = (\pm 13 \times \frac{12}{13}, 0) = (\pm 12, 0)$.
Since the hyperbola passes through $(\pm 12, 0)$,we have $\frac{12^2}{a^2} - \frac{0}{b^2} = 1$,which gives $a^2 = 144$.
The eccentricity of the hyperbola $e'$ is given by $e' = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{b^2}{144}}$.
Given that the product of eccentricities is $1$,we have $e \times e' = 1$.
$\frac{12}{13} \times \sqrt{1 + \frac{b^2}{144}} = 1$.
$\sqrt{1 + \frac{b^2}{144}} = \frac{13}{12}$.
Squaring both sides,$1 + \frac{b^2}{144} = \frac{169}{144}$.
$\frac{b^2}{144} = \frac{169}{144} - 1 = \frac{25}{144}$.
Thus,$b^2 = 25$.
The equation of the hyperbola is $\frac{x^2}{144} - \frac{y^2}{25} = 1$.

(C) Given the equation of the hyperbola is $x^2 - 3y^2 - 4x - 6y - 11 = 0$.
Rearranging the terms: $(x^2 - 4x) - 3(y^2 + 2y) = 11$.
Completing the square: $(x^2 - 4x + 4) - 3(y^2 + 2y + 1) = 11 + 4 - 3$.
$(x - 2)^2 - 3(y + 1)^2 = 12$.
Dividing by $12$: $\frac{(x - 2)^2}{12} - \frac{(y + 1)^2}{4} = 1$.
Here,$a^2 = 12$ and $b^2 = 4$.
The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{12}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.
The distance between the foci is $2ae = 2 \times \sqrt{12} \times \frac{2}{\sqrt{3}} = 2 \times 2\sqrt{3} \times \frac{2}{\sqrt{3}} = 8$.