Hyperbola Questions in English

(B) Given the equations of the two lines:
$x \sqrt{3}-y=k \sqrt{3} \quad ...(1)$
$\sqrt{3} k x+k y=\sqrt{3} \quad ...(2)$
From equation $(1)$,we have $k = \frac{x \sqrt{3}-y}{\sqrt{3}}$.
From equation $(2)$,we have $k = \frac{\sqrt{3}}{\sqrt{3} x+y}$.
Equating the values of $k$ from both equations:
$\frac{x \sqrt{3}-y}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3} x+y}$
$(x \sqrt{3}-y)(\sqrt{3} x+y) = (\sqrt{3})(\sqrt{3})$
$3x^{2} - y^{2} = 3$
Dividing by $3$,we get $\frac{x^{2}}{1} - \frac{y^{2}}{3} = 1$.
This is the standard equation of a hyperbola,$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.

(B) The equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.
Since $2x+3y-6=0$ is a focal chord,it must pass through the focus $(ae, 0)$.
Substituting the coordinates of the focus into the equation of the chord:
$2(ae) + 3(0) - 6 = 0$
$2ae = 6$
$ae = 3$
Given the eccentricity $e = \frac{5}{4}$,we have:
$a \times \frac{5}{4} = 3$
$a = 3 \times \frac{4}{5} = \frac{12}{5}$
The length of the transverse axis is $2a$.
$2a = 2 \times \frac{12}{5} = \frac{24}{5}$.

(D) Given equations are $x^{2}+y^{2}=a^{2}$ and $xy=c^{2}$.
Substituting $y = \frac{c^{2}}{x}$ into the circle equation:
$x^{2} + (\frac{c^{2}}{x})^{2} = a^{2}$
$x^{2} + \frac{c^{4}}{x^{2}} = a^{2}$
$x^{4} - a^{2}x^{2} + c^{4} = 0$
This is a biquadratic equation in $x$. The roots are $x_{1}, x_{2}, x_{3}, x_{4}$.
Comparing this with the standard form $Ax^{4} + Bx^{3} + Cx^{2} + Dx + E = 0$,we have $B=0$.
By Vieta's formulas,the sum of the roots $x_{1}+x_{2}+x_{3}+x_{4} = -\frac{B}{A} = 0$.

(D) Given,the distance between the foci is $2ae = 16$.
Since the eccentricity $e = \sqrt{2}$,we have $2a(\sqrt{2}) = 16$,which implies $a = \frac{8}{\sqrt{2}} = 4\sqrt{2}$.
For a hyperbola,$b^2 = a^2(e^2 - 1)$.
Substituting the values,$b^2 = (4\sqrt{2})^2 ((\sqrt{2})^2 - 1) = 32(2 - 1) = 32$.
The standard equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Substituting $a^2 = 32$ and $b^2 = 32$,we get $\frac{x^2}{32} - \frac{y^2}{32} = 1$,which simplifies to $x^2 - y^2 = 32$.

(A) The given equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.
Distance between the foci $= 2ae$.
Distance between the directrices $= \frac{2a}{e}$.
Given the ratio is $3: 2$,so $\frac{2ae}{2a/e} = \frac{3}{2}$.
This simplifies to $e^{2} = \frac{3}{2}$.
We know that for a hyperbola,$e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$,so $e^{2} = 1 + \frac{b^{2}}{a^{2}}$.
Substituting $e^{2} = \frac{3}{2}$,we get $1 + \frac{b^{2}}{a^{2}} = \frac{3}{2}$.
$\frac{b^{2}}{a^{2}} = \frac{3}{2} - 1 = \frac{1}{2}$.
Therefore,$\frac{b}{a} = \frac{1}{\sqrt{2}}$,which implies $\frac{a}{b} = \sqrt{2}: 1$.

(D) The given equation of the hyperbola is $x^{2}-y^{2}=4$.
Dividing by $4$,we get $\frac{x^{2}}{4}-\frac{y^{2}}{4}=1$.
Here,$a^{2}=4$ and $b^{2}=4$.
For a hyperbola,$b^{2}=a^{2}(e^{2}-1)$,so $4=4(e^{2}-1)$,which implies $e^{2}-1=1$,so $e^{2}=2$ and $e=\sqrt{2}$.
The coordinates of the foci are $(\pm ae, 0) = (\pm 2\sqrt{2}, 0)$.
The equations of the directrices are $x = \pm \frac{a}{e} = \pm \frac{2}{\sqrt{2}} = \pm \sqrt{2}$.
Consider the focus $(2\sqrt{2}, 0)$ and the nearer directrix $x = \sqrt{2}$.
The distance between them is $|2\sqrt{2} - \sqrt{2}| = \sqrt{2}$.

(B) The given equation of the hyperbola is $3x^{2} - 3y^{2} = 25$,which can be written as $x^{2} - y^{2} = \frac{25}{3}$.
Comparing this with the standard form $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$,we get $a^{2} = \frac{25}{3}$ and $b^{2} = \frac{25}{3}$.
The eccentricity $e_{1}$ is given by $e_{1} = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{25/3}{25/3}} = \sqrt{1 + 1} = \sqrt{2}$.
The equation of the conjugate hyperbola is $\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1$,which is $-x^{2} + y^{2} = \frac{25}{3}$.
The eccentricity $e_{2}$ of the conjugate hyperbola is given by $e_{2} = \sqrt{1 + \frac{a^{2}}{b^{2}}} = \sqrt{1 + \frac{25/3}{25/3}} = \sqrt{1 + 1} = \sqrt{2}$.
Therefore,$e_{1}^{2} + e_{2}^{2} = (\sqrt{2})^{2} + (\sqrt{2})^{2} = 2 + 2 = 4$.

(C) Given the equation of the hyperbola: $\frac{x^{2}}{36}-\frac{y^{2}}{k^{2}}=1$.
Rearranging for $k^{2}$:
$\frac{y^{2}}{k^{2}} = \frac{x^{2}}{36}-1 = \frac{x^{2}-36}{36}$
$k^{2} = \frac{36y^{2}}{x^{2}-36}$.
Since $k^{2} > 0$,we must have $x^{2}-36 > 0$,which implies $x^{2} > 36$ or $|x| > 6$.
Checking the options:
$A$. $(-3, 1) \Rightarrow x^{2} = 9 < 36$ (False)
$B$. $(3, 1) \Rightarrow x^{2} = 9 < 36$ (False)
$C$. $(10, 4) \Rightarrow x^{2} = 100 > 36$ (True)
$D$. $(5, 2) \Rightarrow x^{2} = 25 < 36$ (False)
Therefore,the point $(10, 4)$ can lie on the hyperbola.

(A) Given,$x-y=1$ $(i)$ and $\frac{x^{2}}{4}-\frac{y^{2}}{3}=1$ (ii).
Substitute $y=x-1$ from $(i)$ into (ii):
$\frac{x^{2}}{4}-\frac{(x-1)^{2}}{3}=1$
Multiply by $12$:
$3x^{2}-4(x-1)^{2}=12$
$3x^{2}-4(x^{2}-2x+1)=12$
$3x^{2}-4x^{2}+8x-4=12$
$-x^{2}+8x-16=0$
$x^{2}-8x+16=0$
$(x-4)^{2}=0$
$x=4$.
Substitute $x=4$ into $(i)$:
$4-y=1 \Rightarrow y=3$.
Therefore,the point of contact is $(4,3)$.

(D) The equation of the hyperbola is $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$.
Comparing this with $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$,we get $a^{2} = 16$ and $b^{2} = 9$.
The point $(-4, 0)$ lies on the hyperbola since $\frac{(-4)^{2}}{16} - \frac{0^{2}}{9} = 1 - 0 = 1$.
The slope of the tangent at any point $(x_{1}, y_{1})$ is given by differentiating the hyperbola equation: $\frac{2x}{16} - \frac{2y y'}{9} = 0 \Rightarrow y' = \frac{9x}{16y}$.
At $(-4, 0)$,the tangent is vertical $(y' \to \infty)$,which means the tangent line is $x = -4$.
$A$ line perpendicular to a vertical line is a horizontal line.
Since the normal passes through $(-4, 0)$ and is perpendicular to the vertical tangent $x = -4$,the normal must be the horizontal line passing through $y = 0$.
Thus,the equation of the normal is $y = 0$.

(D) The equation of the asymptotes of the hyperbola is given by $3x \pm 5y = 0$.
The equation of the hyperbola can be written as $(3x + 5y)(3x - 5y) = \lambda$,where $\lambda$ is a constant.
This simplifies to $9x^2 - 25y^2 = \lambda$.
Since the vertices of the hyperbola are $(\pm 5, 0)$,the point $(5, 0)$ must satisfy the equation of the hyperbola.
Substituting $(x, y) = (5, 0)$ into the equation $9x^2 - 25y^2 = \lambda$,we get:
$9(5)^2 - 25(0)^2 = \lambda$
$9(25) = \lambda$
$\lambda = 225$
Therefore,the equation of the hyperbola is $9x^2 - 25y^2 = 225$.

(A) The slope $m$ of the line passing through $(ct_1, c/t_1)$ and $(ct_2, c/t_2)$ is given by:
$m = \frac{\frac{c}{t_2} - \frac{c}{t_1}}{ct_2 - ct_1} = \frac{c(t_1 - t_2)}{t_1 t_2} \cdot \frac{1}{c(t_2 - t_1)} = -\frac{1}{t_1 t_2}$
Using the point-slope form $(y - y_1) = m(x - x_1)$ with point $(ct_1, c/t_1)$:
$y - \frac{c}{t_1} = -\frac{1}{t_1 t_2}(x - ct_1)$
Multiplying both sides by $t_1 t_2$:
$t_1 t_2 y - ct_2 = -(x - ct_1)$
$t_1 t_2 y - ct_2 = -x + ct_1$
$x + t_1 t_2 y = c(t_1 + t_2)$

(A) Let $P(x, y)$ be the point. Given $A(4, 0)$ and $B(-4, 0)$.
Given the condition $PA - PB = 4$.
Using the distance formula,$\sqrt{(x-4)^2 + y^2} - \sqrt{(x+4)^2 + y^2} = 4$.
$\sqrt{(x-4)^2 + y^2} = 4 + \sqrt{(x+4)^2 + y^2}$.
Squaring both sides: $(x-4)^2 + y^2 = 16 + (x+4)^2 + y^2 + 8\sqrt{(x+4)^2 + y^2}$.
$x^2 - 8x + 16 + y^2 = 16 + x^2 + 8x + 16 + y^2 + 8\sqrt{(x+4)^2 + y^2}$.
$-16x - 16 = 8\sqrt{(x+4)^2 + y^2}$.
$-2x - 2 = \sqrt{(x+4)^2 + y^2}$.
Squaring again: $4x^2 + 8x + 4 = x^2 + 8x + 16 + y^2$.
$3x^2 - y^2 = 12$.

(D) The equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{1} = 1$,where $a^2 = 9$ and $b^2 = 1$.
The equation of a tangent with slope $m$ is $y = mx \pm \sqrt{9m^2 - 1}$. Since it passes through $(3, 2)$,we have $2 = 3m \pm \sqrt{9m^2 - 1}$.
$(2 - 3m)^2 = 9m^2 - 1$ $\Rightarrow 4 - 12m + 9m^2 = 9m^2 - 1$ $\Rightarrow 12m = 5$ $\Rightarrow m = \frac{5}{12}$.
The other tangent is the vertical line $x = 3$ (since the point $(3, 2)$ lies on the line $x = 3$).
The equation of the first tangent is $y - 2 = \frac{5}{12}(x - 3)$ $\Rightarrow 12y - 24 = 5x - 15$ $\Rightarrow 5x - 12y + 9 = 0$.
The chord of contact for the point $(x_1, y_1) = (3, 2)$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$ $\Rightarrow \frac{3x}{9} - \frac{2y}{1} = 1$ $\Rightarrow x - 6y = 3$.
The vertices of the triangle are the intersection of the two tangents $(3, 2)$,the intersection of the first tangent and the chord of contact $(3, 1)$,and the intersection of the second tangent $(x=3)$ and the chord of contact $(3, 0)$. Wait,let us re-evaluate the intersection points.
Intersection of $x=3$ and $x-6y=3$ is $(3, 0)$.
Intersection of $5x-12y+9=0$ and $x-6y=3$ (i.e.,$x=6y+3$): $5(6y+3) - 12y + 9 = 0$ $\Rightarrow 30y + 15 - 12y + 9 = 0$ $\Rightarrow 18y = -24$ $\Rightarrow y = -4/3$. Then $x = 6(-4/3) + 3 = -8 + 3 = -5$.
The vertices are $(3, 2)$,$(3, 0)$,and $(-5, -4/3)$.
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = \frac{1}{2} |3(0 - (-4/3)) + 3(-4/3 - 2) + (-5)(2 - 0)| = \frac{1}{2} |3(4/3) + 3(-10/3) - 10| = \frac{1}{2} |4 - 10 - 10| = \frac{1}{2} |-16| = 8$ sq units.

(C) The eccentricity $e'$ of the ellipse $\frac{x^2}{4} + \frac{y^2}{2} = 1$ is given by:
$e' = \sqrt{1 - \frac{2}{4}} = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}}$
Given that the eccentricity $e$ of the hyperbola $H$ is the reciprocal of $e'$,we have:
$e = \frac{1}{e'} = \sqrt{2}$
For a hyperbola with transverse axis along the $X$-axis,$e = \sqrt{1 + \frac{b^2}{a^2}}$. Thus:
$\sqrt{1 + \frac{b^2}{a^2}} = \sqrt{2} \implies 1 + \frac{b^2}{a^2} = 2 \implies \frac{b^2}{a^2} = 1 \implies b^2 = a^2$
The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{a^2} = 1$.
Since the point $(5, 4)$ lies on $H$:
$\frac{25}{a^2} - \frac{16}{a^2} = 1 \implies \frac{9}{a^2} = 1 \implies a^2 = 9 \implies a = 3$
The length of the transverse axis is $2a = 2 \times 3 = 6$.

(B) Given equations are:
$x = \frac{a}{2} (t + \frac{1}{t}) \dots (i)$
$y = \frac{a}{2} (t - \frac{1}{t}) \dots (ii)$
Squaring both equations and subtracting $(ii)$ from $(i)$:
$x^2 - y^2 = \frac{a^2}{4} (t + \frac{1}{t})^2 - \frac{a^2}{4} (t - \frac{1}{t})^2$
$= \frac{a^2}{4} [(t^2 + \frac{1}{t^2} + 2) - (t^2 + \frac{1}{t^2} - 2)]$
$= \frac{a^2}{4} [t^2 + \frac{1}{t^2} + 2 - t^2 - \frac{1}{t^2} + 2]$
$= \frac{a^2}{4} (4) = a^2$
Therefore,the Cartesian form is $x^2 - y^2 = a^2$.

(A) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The coordinates of the endpoints of the latus rectum are $(ae, b^2/a)$ and $(ae, -b^2/a)$.
Let the angle subtended by the latus rectum at the centre $(0,0)$ be $2\theta$. Then $\tan \theta = \frac{b^2/a}{ae} = \frac{b^2}{a^2e}$.
Given $2\theta = 2 \operatorname{Tan}^{-1}\left(\frac{3}{2}\right)$,so $\tan \theta = \frac{3}{2}$.
Thus,$\frac{b^2}{a^2e} = \frac{3}{2}$.
Given $b^2 = 36$,we have $\frac{36}{a^2e} = \frac{3}{2}$,which implies $a^2e = 24$.
We know $b^2 = a^2(e^2 - 1)$,so $36 = a^2e^2 - a^2$. Since $a^2e = 24$,$a^2 = \frac{24}{e}$.
Substituting $a^2$,we get $36 = (\frac{24}{e})e^2 - \frac{24}{e} \implies 36 = 24e - \frac{24}{e}$.
Dividing by $12$,$3 = 2e - \frac{2}{e} \implies 2e^2 - 3e - 2 = 0$.
Solving the quadratic equation,$(2e+1)(e-2) = 0$. Since $e > 1$,$e = 2$.
Then $a^2 = \frac{24}{2} = 12$.
Finally,$\sqrt{a^2 + e^2} = \sqrt{12 + 2^2} = \sqrt{12 + 4} = \sqrt{16} = 4$.

(C) The foci are $F_1(8,3)$ and $F_2(0,3)$. The center of the hyperbola is the midpoint of the foci: $(\frac{8+0}{2}, \frac{3+3}{2}) = (4,3)$. Thus,$\alpha = 4$ and $\beta = 3$.
The distance between the foci is $2ae = 8 - 0 = 8$. Given $e = \frac{4}{3}$,we have $2a(\frac{4}{3}) = 8$,which implies $a = 3$.
Using the relation $b^2 = a^2(e^2 - 1)$,we get $b^2 = 3^2((\frac{4}{3})^2 - 1) = 9(\frac{16}{9} - 1) = 9(\frac{7}{9}) = 7$.
In the standard form $\frac{(x-\alpha)^2}{a^2} - \frac{(y-\beta)^2}{b^2} = 1$,we have $p = a^2 = 9$ and $q = b^2 = 7$.
Therefore,$p+q = 9+7 = 16$.
Since $\alpha = 4$ and $\beta = 3$,$\alpha^2 = 16$. Thus,$p+q = \alpha^2$.

(A) Let the hyperbola $H$ be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Distance between foci is $2ae = 26$,so $ae = 13$.
Distance between directrices is $\frac{2a}{e} = \frac{50}{13}$,so $\frac{a}{e} = \frac{25}{13}$,which means $a = \frac{25e}{13}$.
Substituting $a$ in $ae = 13$: $(\frac{25e}{13})e = 13 \implies e^2 = \frac{169}{25} \implies e = \frac{13}{5}$.
Now,$a = \frac{25}{13} \times \frac{13}{5} = 5$.
Since $b^2 = a^2(e^2 - 1)$,we have $b^2 = 25(\frac{169}{25} - 1) = 169 - 25 = 144$,so $b = 12$.
The conjugate hyperbola is $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$.
Its eccentricity $e'$ satisfies $e'^2 = 1 + \frac{a^2}{b^2} = 1 + \frac{25}{144} = \frac{169}{144}$.
Thus,$e' = \sqrt{\frac{169}{144}} = \frac{13}{12}$.

(C) The eccentricity is given by $e = \frac{2}{\sqrt{7}-\sqrt{3}}$. Rationalizing the denominator,we get $e = \frac{2(\sqrt{7}+\sqrt{3})}{7-3} = \frac{2(\sqrt{7}+\sqrt{3})}{4} = \frac{\sqrt{7}+\sqrt{3}}{2}$.
For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the coordinates of the ends of the latus rectum are $(ae, \pm \frac{b^2}{a})$.
The angle $\theta$ subtended by the latus rectum at the centre $(0,0)$ is given by $\tan(\frac{\theta}{2}) = \frac{b^2/a}{ae} = \frac{b^2}{a^2e}$.
Using $b^2 = a^2(e^2-1)$,we have $\tan(\frac{\theta}{2}) = \frac{a^2(e^2-1)}{a^2e} = \frac{e^2-1}{e}$.
Given $e = \frac{\sqrt{7}+\sqrt{3}}{2}$,then $e^2 = \frac{7+3+2\sqrt{21}}{4} = \frac{10+2\sqrt{21}}{4} = \frac{5+\sqrt{21}}{2}$.
So,$\tan(\frac{\theta}{2}) = \frac{\frac{5+\sqrt{21}}{2} - 1}{\frac{\sqrt{7}+\sqrt{3}}{2}} = \frac{3+\sqrt{21}}{\sqrt{7}+\sqrt{3}} = \frac{\sqrt{3}(\sqrt{3}+\sqrt{7})}{\sqrt{7}+\sqrt{3}} = \sqrt{3}$.
Thus,$\frac{\theta}{2} = 60^{\circ}$,which means $\theta = 120^{\circ}$.
Therefore,$\sin \theta = \sin 120^{\circ} = \frac{\sqrt{3}}{2}$.

(A) The given hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{4} = 1$. Here,$b^2 = 4$,so $b = 2$.
Since the circle is concentric with the hyperbola and passes through its foci $(\pm c, 0)$,the radius of the circle is equal to the distance of the foci from the center $(0, 0)$.
Thus,$c = 4$.
Using the relation $c^2 = a^2 + b^2$,we have $16 = a^2 + 4$,which gives $a^2 = 12$,so $a = 2 \sqrt{3}$.
The conjugate hyperbola is $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$,which is $\frac{y^2}{4} - \frac{x^2}{12} = 1$.
For this conjugate hyperbola,the semi-transverse axis is $b = 2$ and the semi-conjugate axis is $a = 2 \sqrt{3}$.
Let $e'$ be the eccentricity of the conjugate hyperbola. The relation is $a^2 = b^2(e'^2 - 1)$ for the conjugate hyperbola,or more simply,$c^2 = a^2 + b^2$ where $c$ is the distance to the foci of the conjugate hyperbola.
For the conjugate hyperbola,the foci are on the $y$-axis at $(0, \pm c')$,where $c'^2 = a^2 + b^2 = 12 + 4 = 16$,so $c' = 4$.
The eccentricity $e'$ is given by $c' = b e'$,where $b$ is the semi-transverse axis of the conjugate hyperbola.
Thus,$4 = 2 e'$,which gives $e' = 2$.

(B) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Since it passes through $(4, -2 \sqrt{3})$,we have $\frac{16}{a^2} - \frac{12}{b^2} = 1$.
Using $b^2 = a^2(e^2 - 1)$,we get $\frac{16}{a^2} - \frac{12}{a^2(e^2 - 1)} = 1$,which simplifies to $16(e^2 - 1) - 12 = a^2(e^2 - 1) \Rightarrow 16e^2 - 28 = a^2(e^2 - 1) \quad (i)$.
The directrix is $x = \frac{a}{e}$,so $\frac{a}{e} = \frac{4}{\sqrt{5}} \Rightarrow a^2 = \frac{16e^2}{5} \quad (ii)$.
Substituting $(ii)$ into $(i)$: $16e^2 - 28 = \frac{16e^2}{5}(e^2 - 1)$.
Multiplying by $5$: $80e^2 - 140 = 16e^4 - 16e^2 \Rightarrow 16e^4 - 96e^2 + 140 = 0$.
Dividing by $4$: $4e^4 - 24e^2 + 35 = 0$.
Factoring the quadratic in $e^2$: $(2e^2 - 7)(2e^2 - 5) = 0$.
Thus,$e^2 = \frac{7}{2}$ or $e^2 = \frac{5}{2}$.
Given the options,$e^2 = \frac{7}{2}$ is the correct value.

(D) : Distance between foci is three times the distance between its directrices.
$2ae = 3 \times \frac{2a}{e}$ $\Rightarrow e^2 = 3$ $\Rightarrow e = \sqrt{3} \approx 1.732$
$B$: The transverse axis is twice the conjugate axis.
$2a = 2(2b)$ $\Rightarrow a = 2b$ $\Rightarrow b = \frac{a}{2}$.
We know that $a^2e^2 = a^2 + b^2 = a^2 + \frac{a^2}{4} = \frac{5a^2}{4}$.
$e^2 = \frac{5}{4} \Rightarrow e = \frac{\sqrt{5}}{2} \approx 1.118$
$C$: Slopes of asymptotes are $m_1 = -1$ and $m_2 = 1$.
Since $m_1 \cdot m_2 = -1$,it is a rectangular hyperbola.
$e = \sqrt{2} \approx 1.414$
Comparing the values: $1.732 > 1.414 > 1.118$.
Thus,the descending order is $A, C, B$.

(A) For the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,the eccentricity is $e_1 = \sqrt{1 + \frac{b^2}{a^2}} = \frac{\sqrt{a^2+b^2}}{a}$.
For its conjugate hyperbola $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$,the eccentricity is $e_2 = \sqrt{1 + \frac{a^2}{b^2}} = \frac{\sqrt{a^2+b^2}}{b}$.
The given line is $\frac{x}{2 e_1} + \frac{y}{2 e_2} = 1$.
Substituting the values of $e_1$ and $e_2$:
$\frac{x}{2(\frac{\sqrt{a^2+b^2}}{a})} + \frac{y}{2(\frac{\sqrt{a^2+b^2}}{b})} = 1$
$\frac{ax}{2\sqrt{a^2+b^2}} + \frac{by}{2\sqrt{a^2+b^2}} = 1$
$ax + by = 2\sqrt{a^2+b^2}$.
Since this line touches the circle $x^2 + y^2 = r^2$ with center at the origin,the perpendicular distance from the origin $(0,0)$ to the line must equal the radius $r$.
$r = \frac{|0 + 0 - 2\sqrt{a^2+b^2}|}{\sqrt{a^2+b^2}} = \frac{2\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}} = 2$.
Thus,the radius is $2$.

(A) The standard form of a conic section is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
For this equation to represent a hyperbola,the coefficients of $x^2$ and $y^2$ must have opposite signs.
Let $A = \frac{1}{7 - k}$ and $B = \frac{1}{5 - k}$.
For a hyperbola,$A \times B < 0$.
$\left( \frac{1}{7 - k} \right) \left( \frac{1}{5 - k} \right) < 0$
Multiply by $-1$ to simplify the denominators:
$\left( \frac{1}{k - 7} \right) \left( \frac{1}{k - 5} \right) < 0$
Using the sign scheme method,the product is negative when $k$ lies between the roots $5$ and $7$.
Therefore,$5 < k < 7$.

(D) For the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$,we have $a^2=16$ and $b^2=9$. The eccentricity $e_1 = \sqrt{1+\frac{9}{16}} = \frac{5}{4}$. The foci are $(\pm ae_1, 0) = (\pm 5, 0)$.
For the conjugate hyperbola $\frac{y^2}{9}-\frac{x^2}{16}=1$ (or $\frac{x^2}{-16} - \frac{y^2}{-9} = 1$),the equation is $\frac{y^2}{9}-\frac{x^2}{16}=1$. Here $a^2=9$ and $b^2=16$. The eccentricity $e_2 = \sqrt{1+\frac{16}{9}} = \frac{5}{3}$. The foci are $(0, \pm be_2) = (0, \pm 5)$.
The four foci are $(\pm 5, 0)$ and $(0, \pm 5)$.
These points form a rhombus with diagonals of length $10$ and $10$.
Area of the quadrilateral $= \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 10 \times 10 = 50$ square units.

(C) The given equation of the hyperbola is $5x^2 - 6y^2 = 30$,which can be written as $\frac{x^2}{6} - \frac{y^2}{5} = 1$.
Here,$a^2 = 6$ and $b^2 = 5$.
Let the equation of the tangent be $y = mx + c$. Since it passes through $(2, 3)$,we have $3 = 2m + c$,so $c = 3 - 2m$.
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 $c^2 = a^2m^2 - b^2$.
Substituting the values,we get $(3 - 2m)^2 = 6m^2 - 5$.
$9 - 12m + 4m^2 = 6m^2 - 5$.
$2m^2 + 12m - 14 = 0$,which simplifies to $m^2 + 6m - 7 = 0$.
Factoring gives $(m + 7)(m - 1) = 0$,so the slopes are $m_1 = 1$ and $m_2 = -7$.
The angle $\theta$ between the tangents is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right|$.
$\tan \theta = \left| \frac{1 - (-7)}{1 + (1)(-7)} \right| = \left| \frac{8}{1 - 7} \right| = \left| \frac{8}{-6} \right| = \frac{4}{3}$.

(D) The difference between the focal distances of any point on a hyperbola is equal to the length of the transverse axis,which is $2a$.
Given $2a = 6$,we find $a = 3$.
The coordinates of the endpoints of the latus rectum are $(\pm ae, \pm \frac{b^2}{a})$.
Since the point $(\sqrt{13}, k)$ is an endpoint of the latus rectum,we have $ae = \sqrt{13}$.
Using the relation $c^2 = a^2 + b^2$,where $c = ae = \sqrt{13}$,we get $13 = a^2 + b^2$.
Substituting $a = 3$,we have $13 = 3^2 + b^2$,which gives $13 = 9 + b^2$,so $b^2 = 4$.
The $y$-coordinate of the endpoint of the latus rectum is $k = \pm \frac{b^2}{a}$.
Substituting the values,$k = \pm \frac{4}{3}$.

(A) The given equation of the hyperbola is $4 x^2-y^2-8 x-2 y-13=0$.
Rewriting it by completing the square:
$4(x^2-2x+1) - (y^2+2y+1) = 13 + 4 - 1$
$4(x-1)^2 - (y+1)^2 = 16$
$\frac{(x-1)^2}{4} - \frac{(y+1)^2}{16} = 1$.
This is a hyperbola of the form $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ where $h=1, k=-1, a=2, b=4$.
The parametric coordinates are $x = 1 + 2 \sec \theta$ and $y = -1 + 4 \tan \theta$.
For point $P(\frac{\pi}{4})$:
$x_P = 1 + 2 \sec(\frac{\pi}{4}) = 1 + 2\sqrt{2}$
$y_P = -1 + 4 \tan(\frac{\pi}{4}) = -1 + 4 = 3$.
So,$P = (1+2\sqrt{2}, 3)$.
For point $Q(\frac{3\pi}{4})$:
$x_Q = 1 + 2 \sec(\frac{3\pi}{4}) = 1 + 2(-\sqrt{2}) = 1 - 2\sqrt{2}$
$y_Q = -1 + 4 \tan(\frac{3\pi}{4}) = -1 + 4(-1) = -5$.
So,$Q = (1-2\sqrt{2}, -5)$.
The distance $PQ = \sqrt{(x_P - x_Q)^2 + (y_P - y_Q)^2}$
$PQ = \sqrt{(1+2\sqrt{2} - (1-2\sqrt{2}))^2 + (3 - (-5))^2}$
$PQ = \sqrt{(4\sqrt{2})^2 + (8)^2} = \sqrt{32 + 64} = \sqrt{96} = 4\sqrt{6}$.

(A) Let the equation of the hyperbola be $S(x, y) = \frac{x^2}{a^2} - y^2 - 1 = 0$.
Since the origin $(0,0)$ and the point $(1,1)$ lie in the same region,the value of $S(0,0)$ and $S(1,1)$ must have the same sign.
$S(0,0) = \frac{0^2}{a^2} - 0^2 - 1 = -1$.
Since $S(0,0) < 0$,we must have $S(1,1) < 0$.
$S(1,1) = \frac{1^2}{a^2} - 1^2 - 1 = \frac{1}{a^2} - 2$.
Setting $\frac{1}{a^2} - 2 < 0$,we get $\frac{1}{a^2} < 2$,which implies $a^2 > \frac{1}{2}$.
Since $a > 0$,we have $a > \frac{1}{\sqrt{2}}$.
Thus,the range of $a$ is $\left(\frac{1}{\sqrt{2}}, \infty\right)$.

(A) Given the hyperbola equation $\frac{x^2}{9}-\frac{y^2}{16}=1$,we have $a=3$ and $b=4$.
The parametric coordinates are given by $x=a \sec \theta$ and $y=b \tan \theta$,so $x=3 \sec \theta$ and $y=4 \tan \theta$.
Calculating the coordinates for each point:
$P_1\left(\frac{\pi}{4}\right) = (3 \sec \frac{\pi}{4}, 4 \tan \frac{\pi}{4}) = (3\sqrt{2}, 4)$
$P_2\left(\frac{3\pi}{4}\right) = (3 \sec \frac{3\pi}{4}, 4 \tan \frac{3\pi}{4}) = (-3\sqrt{2}, -4)$
$P_3\left(\frac{5\pi}{4}\right) = (3 \sec \frac{5\pi}{4}, 4 \tan \frac{5\pi}{4}) = (-3\sqrt{2}, 4)$
$P_4\left(\frac{7\pi}{4}\right) = (3 \sec \frac{7\pi}{4}, 4 \tan \frac{7\pi}{4}) = (3\sqrt{2}, -4)$
Plotting these points,we see they form a rectangle with vertices at $(3\sqrt{2}, 4), (-3\sqrt{2}, 4), (-3\sqrt{2}, -4),$ and $(3\sqrt{2}, -4)$.
The sides are parallel to the coordinate axes,and the lengths are $6\sqrt{2}$ and $8$. Thus,it is a rectangle.

(D) Given the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$.
Since $b^2 > a^2$ $(9 > 4)$,the eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.
Let $e_H$ be the eccentricity of the hyperbola,where $e_H = \frac{1}{e} = \frac{3}{\sqrt{5}}$.
For a hyperbola with eccentricity $e_H$ and its conjugate hyperbola with eccentricity $e_C$,the relation is $\frac{1}{e_H^2} + \frac{1}{e_C^2} = 1$.
Substituting $e_H = \frac{3}{\sqrt{5}}$,we get $\frac{1}{(3/\sqrt{5})^2} + \frac{1}{e_C^2} = 1$.
$\frac{5}{9} + \frac{1}{e_C^2} = 1 \Rightarrow \frac{1}{e_C^2} = 1 - \frac{5}{9} = \frac{4}{9}$.
Therefore,$e_C^2 = \frac{9}{4}$,which gives $e_C = \frac{3}{2}$.

(B) Given,the centre is $(0, 0)$,the foci are $(\pm 3, 0)$,and the eccentricity $e = \frac{3}{2}$.
Since the foci are $(\pm ae, 0) = (\pm 3, 0)$,we have $ae = 3$.
Substituting $e = \frac{3}{2}$,we get $a(\frac{3}{2}) = 3$,which implies $a = 2$.
Using the relation $b^2 = a^2(e^2 - 1)$,we have $b^2 = 2^2((\frac{3}{2})^2 - 1) = 4(\frac{9}{4} - 1) = 4(\frac{5}{4}) = 5$.
The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{4} - \frac{y^2}{5} = 1$,or $5x^2 - 4y^2 = 20$.
To check the intersection with the line $y = 2x - 1$,substitute $y$ into the hyperbola equation:
$5x^2 - 4(2x - 1)^2 = 20$
$5x^2 - 4(4x^2 - 4x + 1) = 20$
$5x^2 - 16x^2 + 16x - 4 = 20$
$-11x^2 + 16x - 24 = 0$
$11x^2 - 16x + 24 = 0$.
The discriminant $D = b^2 - 4ac = (-16)^2 - 4(11)(24) = 256 - 1056 = -800$.
Since $D < 0$,the line does not intersect the hyperbola.

(D) Given that the vertices are $(\pm a, 0) = (\pm 3, 0)$,so $a = 3$.
Given that the foci are $(\pm ae, 0) = (\pm 4, 0)$,so $ae = 4$.
Substituting $a=3$,we get $3e = 4$,which implies $e = \frac{4}{3}$.
We know the relation $b^2 = a^2(e^2 - 1)$.
Substituting the values,$b^2 = 3^2 \left( (\frac{4}{3})^2 - 1 \right) = 9 \left( \frac{16}{9} - 1 \right) = 16 - 9 = 7$.
Thus,$b = \sqrt{7}$.
The standard equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{9} - \frac{y^2}{7} = 1$.
The parametric equations for a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are $x = a \sec \theta$ and $y = b \tan \theta$.
Substituting $a=3$ and $b=\sqrt{7}$,we get $x = 3 \sec \theta$ and $y = \sqrt{7} \tan \theta$.

(B) Given,the hyperbola is $16 x^2 - 9 y^2 = 1$.
This can be written as $\frac{x^2}{(1/4)^2} - \frac{y^2}{(1/3)^2} = 1$.
Here,$a^2 = \frac{1}{16}$ and $b^2 = \frac{1}{9}$.
The eccentricity $e_1$ of the hyperbola is given by $e_1^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{1/9}{1/16} = 1 + \frac{16}{9} = \frac{25}{9}$.
Thus,$e_1 = \frac{5}{3}$.
The eccentricity $e_2$ of the conjugate hyperbola is given by $e_2^2 = 1 + \frac{a^2}{b^2} = 1 + \frac{1/16}{1/9} = 1 + \frac{9}{16} = \frac{25}{16}$.
Thus,$e_2 = \frac{5}{4}$.
Now,$3 e_1 = 3 \times \frac{5}{3} = 5$.
Also,$4 e_2 = 4 \times \frac{5}{4} = 5$.
Therefore,$3 e_1 = 4 e_2$.

(A) The given equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} + 1 = 0$,which can be rewritten as $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$.
This is a conjugate hyperbola.
The equation of the directrix for this hyperbola is $y = \frac{b}{e}$.
Comparing this with the given directrix $\sqrt{5} y - \sqrt{8} = 0$,we get $y = \frac{\sqrt{8}}{\sqrt{5}}$.
Thus,$\frac{b}{e} = \frac{\sqrt{8}}{\sqrt{5}}$.
Given $e = \frac{\sqrt{5}}{2}$,we have $b = \frac{\sqrt{8}}{\sqrt{5}} \times \frac{\sqrt{5}}{2} = \frac{\sqrt{8}}{2} = \sqrt{2}$.
For a conjugate hyperbola,$e^2 = 1 + \frac{a^2}{b^2}$.
Substituting the values,$(\frac{\sqrt{5}}{2})^2 = 1 + \frac{a^2}{(\sqrt{2})^2} \implies \frac{5}{4} = 1 + \frac{a^2}{2}$.
$\frac{a^2}{2} = \frac{5}{4} - 1 = \frac{1}{4} \implies a^2 = \frac{1}{2} \implies a = \frac{1}{\sqrt{2}}$.
Therefore,$\frac{1}{a} = \sqrt{2}$.

(B) The coordinates of the focus $S$ are $(ae, 0)$ and the end of the latus rectum $L$ in the first quadrant is $(ae, b^2/a)$.
Given $S(8, y_1)$,we have $ae = 8$. Since $y_1$ is the $y$-coordinate of the focus,$y_1 = 0$.
Given $L(x_1, 4)$,we have $b^2/a = 4$,which implies $b^2 = 4a$.
For a hyperbola,$c^2 = a^2 + b^2$,where $c = ae = 8$.
Substituting $c = 8$ and $b^2 = 4a$ into the equation $c^2 = a^2 + b^2$:
$64 = a^2 + 4a$
$a^2 + 4a - 64 = 0$
Using the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$a = \frac{-4 \pm \sqrt{16 - 4(1)(-64)}}{2} = \frac{-4 \pm \sqrt{16 + 256}}{2} = \frac{-4 \pm \sqrt{272}}{2} = \frac{-4 \pm 4\sqrt{17}}{2} = -2 \pm 2\sqrt{17}$.
Since $a > 0$,we take $a = 2\sqrt{17} - 2 = 2(\sqrt{17} - 1)$.
The length of the transverse axis is $2a = 2 \times 2(\sqrt{17} - 1) = 4(\sqrt{17} - 1)$.

(A) Given,the length of the transverse axis is $2a = 8$,which implies $a = 4$.
The distance between the foci is $2ae = 2\sqrt{41}$,which implies $ae = \sqrt{41}$.
Using the relation $a^2e^2 = a^2 + b^2$,we substitute the values:
$(\sqrt{41})^2 = 4^2 + b^2$
$41 = 16 + b^2$
$b^2 = 41 - 16 = 25$.
The length of the latus rectum is given by $\frac{2b^2}{a}$.
Substituting the values,we get $\frac{2 \times 25}{4} = \frac{50}{4} = \frac{25}{2}$.

(B) The definition of a conic section is $SP^2 = e^2 PM^2$,where $S$ is the focus,$P(x,y)$ is a point on the curve,$e$ is the eccentricity,and $PM$ is the perpendicular distance from $P$ to the directrix.
Given focus $S = (1,2)$,directrix $x+2y=0$,and $e = \sqrt{2}$.
$(x-1)^2 + (y-2)^2 = (\sqrt{2})^2 \frac{(x+2y)^2}{1^2+2^2}$
$(x^2-2x+1) + (y^2-4y+4) = 2 \frac{(x+2y)^2}{5}$
$5(x^2+y^2-2x-4y+5) = 2(x^2+4y^2+4xy)$
$5x^2+5y^2-10x-20y+25 = 2x^2+8y^2+8xy$
$3x^2-8xy-3y^2-10x-20y+25 = 0$

(B) The given quadratic equation is $x^2 - 5x - 14 = 0$.
Factoring the equation: $(x - 7)(x + 2) = 0$,so the roots are $x_1 = 7$ and $x_2 = -2$.
Since the length of an axis must be positive,we take the absolute value or consider the valid root. Given the context,let the semi-conjugate axis $b = 7$.
The square of the other root is the semi-transverse axis $a = (-2)^2 = 4$.
For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the distance of the focus from the center is $c = \sqrt{a^2 + b^2}$.
Substituting the values: $c = \sqrt{4^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}$.
Thus,the focus on the positive $x$-axis is $(\sqrt{65}, 0)$.

(A) Given equation: $9x^2-16y^2-72x+96y-144=0$
Rearranging terms: $9(x^2-8x)-16(y^2-6y)=144$
Completing the square: $9(x^2-8x+16)-16(y^2-6y+9)=144+144-144$
$9(x-4)^2-16(y-3)^2=144$
Dividing by $144$: $\frac{(x-4)^2}{16}-\frac{(y-3)^2}{9}=1$
Comparing with $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$,we get $a^2=16$ and $b^2=9$.
The eccentricity $e$ is given by $\sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.
Thus,Statement $I$ is true and Statement $II$ is the correct formula used to derive it.

(B) The distance between the focus and the directrix is given by $d = \frac{a}{e} - ae = a(\frac{1}{e} - e)$ (for hyperbola,distance is $a/e - ae$ or $ae - a/e$ depending on orientation). Given $e = 5/4$,the distance is $a(5/4 - 4/5) = a(9/20)$.
The perpendicular distance from the focus $(3,0)$ to the directrix $4x - 3y - 3 = 0$ is $\frac{|4(3) - 3(0) - 3|}{\sqrt{4^2 + (-3)^2}} = \frac{9}{5}$.
Equating the two: $\frac{9a}{20} = \frac{9}{5} \implies a = 4$.
The axis of the hyperbola is perpendicular to the directrix. Since the directrix slope is $4/3$,the axis slope is $-3/4$.
The unit vector along the axis towards the vertex is $(-\frac{4}{5}, \frac{3}{5})$.
The distance from the focus to the vertex is $ae - a = 4(5/4) - 4 = 1$.
Thus,the vertex is $(3, 0) + 1 \times (-\frac{4}{5}, \frac{3}{5}) = (3 - \frac{4}{5}, 0 + \frac{3}{5}) = (\frac{11}{5}, \frac{3}{5})$.

(C) Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The focus is at $(ae, 0)$.
Since the focal chord is perpendicular to the transverse axis,its equation is $x = ae$.
Substituting $x = ae$ in the hyperbola equation: $\frac{a^2e^2}{a^2} - \frac{y^2}{b^2} = 1$ $\Rightarrow e^2 - 1 = \frac{y^2}{b^2}$ $\Rightarrow y^2 = b^2(e^2 - 1)$.
Since $e^2 = 1 + \frac{b^2}{a^2}$,we have $e^2 - 1 = \frac{b^2}{a^2}$,so $y^2 = b^2(\frac{b^2}{a^2}) = \frac{b^4}{a^2} \Rightarrow y = \pm \frac{b^2}{a}$.
The endpoints of the focal chord are $A(ae, \frac{b^2}{a})$ and $B(ae, -\frac{b^2}{a})$.
The center is $O(0, 0)$. The chord subtends a right angle at the center,so the slope of $OA$ multiplied by the slope of $OB$ is $-1$.
Slope of $OA = \frac{b^2/a}{ae} = \frac{b^2}{a^2e}$. Slope of $OB = \frac{-b^2/a}{ae} = -\frac{b^2}{a^2e}$.
$(\frac{b^2}{a^2e}) \times (-\frac{b^2}{a^2e}) = -1$ $\Rightarrow \frac{b^4}{a^4e^2} = 1$ $\Rightarrow b^4 = a^4e^2$.
Since $b^2 = a^2(e^2 - 1)$,we have $(a^2(e^2 - 1))^2 = a^4e^2 \Rightarrow (e^2 - 1)^2 = e^2$.
$e^4 - 2e^2 + 1 = e^2 \Rightarrow e^4 - 3e^2 + 1 = 0$.
Let $u = e^2$. Then $u^2 - 3u + 1 = 0$. Using the quadratic formula: $u = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}$.
Since $e > 1$,$e^2 > 1$. $\frac{3 - \sqrt{5}}{2} \approx 0.38 < 1$,so we take $e^2 = \frac{3 + \sqrt{5}}{2}$.
$e^2 = \frac{6 + 2\sqrt{5}}{4} = \frac{(\sqrt{5} + 1)^2}{4} \Rightarrow e = \frac{\sqrt{5} + 1}{2}$.

(B) Given that the transverse axis and conjugate axis of the hyperbola are equal.
Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
The length of the transverse axis is $2a$ and the length of the conjugate axis is $2b$.
According to the problem,$2a = 2b$,which implies $a = b$.
We know that for a hyperbola,$b^2 = a^2(e^2 - 1)$,where $e$ is the eccentricity.
Substituting $b = a$ into the relation,we get $a^2 = a^2(e^2 - 1)$.
Dividing both sides by $a^2$ (since $a \neq 0$),we get $1 = e^2 - 1$.
Therefore,$e^2 = 2$,which gives $e = \sqrt{2}$ (since eccentricity $e > 1$ for a hyperbola).

(A) Given,in a hyperbola,the length of the transverse axis is twice that of the conjugate axis. The equation of the standard hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Since the length of the transverse axis $= 2 \times$ length of the conjugate axis,we have $2a = 2(2b)$,which implies $a = 2b$.
We know that $b^2 = a^2(e^2 - 1)$.
Substituting $b^2 = \frac{a^2}{4}$,we get $\frac{a^2}{4} = a^2(e^2 - 1)$.
Dividing by $a^2$,we get $\frac{1}{4} = e^2 - 1$,so $e^2 = 1 + \frac{1}{4} = \frac{5}{4}$.
Thus,$e = \frac{\sqrt{5}}{2}$.
The distance between the directrices is given by $\frac{2a}{e}$.
Substituting $a = 2b$ and $e = \frac{\sqrt{5}}{2}$,the distance is $\frac{2(2b)}{\sqrt{5}/2} = \frac{4b \times 2}{\sqrt{5}} = \frac{8b}{\sqrt{5}}$ units.

(D) Given,eccentricity $e = \frac{5}{3}$ and distance between foci $2ae = 10$.
$2a(\frac{5}{3}) = 10 \Rightarrow a = 3$.
For a hyperbola,$c^2 = a^2 + b^2$,where $c = ae$.
$(ae)^2 = a^2 + b^2 \Rightarrow 5^2 = 3^2 + b^2$.
$25 = 9 + b^2 \Rightarrow b^2 = 16$.
The standard equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Substituting the values,we get $\frac{x^2}{9} - \frac{y^2}{16} = 1$.
Multiplying by $144$,we get $16x^2 - 9y^2 = 144$.

(B) Given,the distance between the foci is $2ae = 10$ and the length of the transverse axis is $2a = 8$.
From these,we get $ae = 5$ and $a = 4$.
In a hyperbola,the relationship between $a, b,$ and $e$ is given by $(ae)^2 = a^2 + b^2$.
Substituting the values,we get $25 = 16 + b^2$,which implies $b^2 = 9$.
The length of the latus rectum is given by the formula $\frac{2b^2}{a}$.
Substituting the values,we get $\frac{2 \times 9}{4} = \frac{18}{4} = \frac{9}{2}$.

(C) Given,the latus rectum of the hyperbola subtends $90^{\circ}$ at its centre.
Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
So,the eccentricity is $e = \sqrt{1 + \frac{b^2}{a^2}}$,which implies $b^2 = a^2(e^2 - 1)$.
The end points of the latus rectum are $L = (ae, \frac{b^2}{a})$ and $L' = (ae, -\frac{b^2}{a})$.
The centre is $C = (0, 0)$.
Since $\angle LCL' = 90^{\circ}$,the slopes of $CL$ and $CL'$ are $m_1 = \frac{b^2/a}{ae} = \frac{b^2}{a^2e}$ and $m_2 = \frac{-b^2/a}{ae} = -\frac{b^2}{a^2e}$.
Since $CL \perp CL'$,we have $m_1 \times m_2 = -1$.
$\left(\frac{b^2}{a^2e}\right) \times \left(-\frac{b^2}{a^2e}\right) = -1$ $\Rightarrow \frac{b^4}{a^4e^2} = 1$ $\Rightarrow b^4 = a^4e^2$.
Substituting $b^2 = a^2(e^2 - 1)$:
$[a^2(e^2 - 1)]^2 = a^4e^2 \Rightarrow a^4(e^2 - 1)^2 = a^4e^2$.
$(e^2 - 1)^2 = e^2 \Rightarrow e^2 - 1 = \pm e$.
Case $1$: $e^2 - e - 1 = 0 \Rightarrow e = \frac{1 + \sqrt{5}}{2}$ (since $e > 1$).
Case $2$: $e^2 + e - 1 = 0 \Rightarrow e = \frac{-1 + \sqrt{5}}{2}$ (rejected as $e > 1$).
Thus,the eccentricity is $\frac{\sqrt{5} + 1}{2}$.

(C) Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The foci are $F_1(-ae, 0)$ and $F_2(ae, 0)$. The latus rectum through $F_1$ has endpoints $A(-ae, b^2/a)$ and $A'(-ae, -b^2/a)$.
The angle subtended by the latus rectum at the other focus $F_2$ is $60^{\circ}$.
Considering the right-angled triangle formed by the focus $F_2$,the midpoint of the latus rectum $B(-ae, 0)$,and one endpoint $A(-ae, b^2/a)$,the angle at $F_2$ is $30^{\circ}$.
In $\triangle ABF_2$,we have $AB = \frac{b^2}{a}$ and $BF_2 = ae - (-ae) = 2ae$.
Thus,$\tan 30^{\circ} = \frac{AB}{BF_2} = \frac{b^2/a}{2ae} = \frac{b^2}{2a^2e}$.
Since $\frac{1}{\sqrt{3}} = \frac{b^2}{2a^2e}$ and $b^2 = a^2(e^2 - 1)$,we substitute $b^2$:
$\frac{1}{\sqrt{3}} = \frac{a^2(e^2 - 1)}{2a^2e} = \frac{e^2 - 1}{2e}$.
This gives $2e = \sqrt{3}(e^2 - 1)$,or $\sqrt{3}e^2 - 2e - \sqrt{3} = 0$.
Solving the quadratic equation: $e = \frac{2 \pm \sqrt{4 - 4(\sqrt{3})(-\sqrt{3})}}{2\sqrt{3}} = \frac{2 \pm \sqrt{4 + 12}}{2\sqrt{3}} = \frac{2 \pm 4}{2\sqrt{3}}$.
Since $e > 1$,we take $e = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.