Let $P ( 3,3 )$ be a point on the hyperbola, $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$. If the normal to it at $P$ intersects the $x$-axis at $( 9,0 )$ and $e$ is its eccentricity, then the ordered pair $\left( a ^ { 2 } , e ^ { 2 } \right)$ is equal to:
(1) $\left( \frac { 9 } { 2 } , 3 \right)$
(2) $\left( \frac { 3 } { 2 } , 2 \right)$
(3) $\left( \frac { 9 } { 2 } , 2 \right)$
(4) $( 9,3 )$

If the normal at an end of latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity e of the ellipse satisfies:
(1) $\mathrm{e}^{4}+2\mathrm{e}^{2}-1=0$
(2) $\mathrm{e}^{2}+\mathrm{e}-1=0$
(3) $\mathrm{e}^{4}+\mathrm{e}^{2}-1=0$
(4) $\mathrm{e}^{2}+2\mathrm{e}-1=0$

Let $E _ { 1 } : \frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 , a > b$. Let $E _ { 2 }$ be another ellipse such that it touches the end points of major axis of $E _ { 1 }$ and the foci of $E _ { 2 }$ are the end points of minor axis of $E _ { 1 }$. If $E _ { 1 }$ and $E _ { 2 }$ have same eccentricities, then its value is:
(1) $\frac { - 1 + \sqrt { 5 } } { 2 }$
(2) $\frac { - 1 + \sqrt { 8 } } { 2 }$
(3) $\frac { - 1 + \sqrt { 3 } } { 2 }$
(4) $\frac { - 1 + \sqrt { 6 } } { 2 }$

Let the maximum area of the triangle that can be inscribed in the ellipse $\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { 4 } = 1 , a > 2$, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the $y$-axis, be $6 \sqrt { 3 }$. Then the eccentricity of the ellipse is:
(1) $\frac { \sqrt { 3 } } { 2 }$
(2) $\frac { 1 } { 2 }$
(3) $\frac { 1 } { \sqrt { 2 } }$
(4) $\frac { \sqrt { 3 } } { 4 }$

Let the hyperbola $H : \frac { x ^ { 2 } } { a ^ { 2 } } - y ^ { 2 } = 1$ and the ellipse $E : 3 x ^ { 2 } + 4 y ^ { 2 } = 12$ be such that the length of latus rectum of $H$ is equal to the length of latus rectum of $E$. If $e _ { H }$ and $e _ { E }$ are the eccentricities of $H$ and $E$ respectively, then the value of $12 \left( e _ { H } ^ { 2 } + e _ { E } ^ { 2 } \right)$ is equal to $\_\_\_\_$.

If the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ meets the line $\frac{x}{7} + \frac{y}{2\sqrt{6}} = 1$ on the $x$-axis and the line $\frac{x}{7} - \frac{y}{2\sqrt{6}} = 1$ on the $y$-axis, then the eccentricity of the ellipse is
(1) $\frac{5}{7}$
(2) $\frac{2\sqrt{6}}{7}$
(3) $\frac{3}{7}$
(4) $\frac{2\sqrt{5}}{7}$

Let the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{7} = 1$ and the hyperbola $\frac{x^2}{144} - \frac{y^2}{\alpha} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:
(1) $\frac{32}{9}$
(2) $\frac{18}{5}$
(3) $\frac{27}{4}$
(4) $\frac{27}{10}$

Let a line $L$ pass through the point of intersection of the lines $b x + 10 y - 8 = 0$ and $2 x - 3 y = 0$, $b \in R - \left\{ \frac { 4 } { 3 } \right\}$. If the line $L$ also passes through the point $( 1,1 )$ and touches the circle $17 \left( x ^ { 2 } + y ^ { 2 } \right) = 16$, then the eccentricity of the ellipse $\frac { x ^ { 2 } } { 5 } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1$ is
(1) $\frac { 2 } { \sqrt { 5 } }$
(2) $\sqrt { \frac { 3 } { 5 } }$
(3) $\frac { 1 } { \sqrt { 5 } }$
(4) $\sqrt { \frac { 2 } { 5 } }$

Let H be the hyperbola, whose foci are $(1 \pm \sqrt{2}, 0)$ and eccentricity is $\sqrt{2}$. Then the length of its latus rectum is:
(1) 3
(2) $\frac{5}{2}$
(3) 2
(4) $\frac{3}{2}$

Let the eccentricity of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is reciprocal to that of the hyperbola $2x^{2} - 2y^{2} = 1$. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is $\_\_\_\_$.

If the maximum distance of normal to the ellipse $\frac{x^2}{4} + \frac{y^2}{b^2} = 1$, $b < 2$, from the origin is 1, then the eccentricity of the ellipse is:
(1) $\frac{1}{\sqrt{2}}$
(2) $\frac{\sqrt{3}}{2}$
(3) $\frac{1}{2}$
(4) $\frac{\sqrt{3}}{4}$

If the length of the minor axis of ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is :
(1) $\frac { \sqrt { 5 } } { 3 }$
(2) $\frac { \sqrt { 3 } } { 2 }$
(3) $\frac { 1 } { \sqrt { 3 } }$
(4) $\frac { 2 } { \sqrt { 5 } }$

Let the foci of a hyperbola be $( 1,14 )$ and $( 1 , - 12 )$. If it passes through the point $( 1,6 )$, then the length of its latus-rectum is:
(1) $\frac { 24 } { 5 }$
(2) $\frac { 25 } { 6 }$
(3) $\frac { 144 } { 5 }$
(4) $\frac { 288 } { 5 }$

Let the product of the focal distances of the point $\left(\sqrt{3}, \frac{1}{2}\right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $(a > b)$, be $\frac{7}{4}$. Then the absolute difference of the eccentricities of two such ellipses is
(1) $\frac{1 - \sqrt{3}}{\sqrt{2}}$
(2) $\frac{3 - 2\sqrt{2}}{2\sqrt{3}}$
(3) $\frac{3 - 2\sqrt{2}}{3\sqrt{2}}$
(4) $\frac{1 - 2\sqrt{2}}{\sqrt{3}}$

Let $\mathrm{H}_{1} : \frac{x^{2}}{\mathrm{a}^{2}} - \frac{y^{2}}{\mathrm{b}^{2}} = 1$ and $\mathrm{H}_{2} : -\frac{x^{2}}{\mathrm{A}^{2}} + \frac{y^{2}}{\mathrm{B}^{2}} = 1$ be two hyperbolas having length of latus rectums $15\sqrt{2}$ and $12\sqrt{5}$ respectively. Let their eccentricities be $e_{1} = \sqrt{\frac{5}{2}}$ and $e_{2}$ respectively. If the product of the lengths of their transverse axes is $100\sqrt{10}$, then $25\mathrm{e}_{2}^{2}$ is equal to $\_\_\_\_$.

On the coordinate plane, let $\Gamma$ be an ellipse with center at the origin and major axis on the $y$-axis. It is known that a linear transformation of counterclockwise rotation by angle $\theta$ about the origin (where $0 < \theta < \pi$) transforms $\Gamma$ to a new ellipse $\Gamma ^ { \prime } : 40 x ^ { 2 } + 4 \sqrt { 5 } x y + 41 y ^ { 2 } = 180$. The point $\left( - \frac { 5 } { 3 } , \frac { 2 \sqrt { 5 } } { 3 } \right)$ is one of the two points on $\Gamma ^ { \prime }$ farthest from the origin.
The length of the major axis of ellipse $\Gamma ^ { \prime }$ is (15-1) $\sqrt{\underline{(15-2)}}$. (Express as a simplified radical)

On the coordinate plane, let $\Gamma$ be an ellipse with center at the origin and major axis on the $y$-axis. It is known that a linear transformation of counterclockwise rotation by angle $\theta$ about the origin (where $0 < \theta < \pi$) transforms $\Gamma$ to a new ellipse $\Gamma ^ { \prime } : 40 x ^ { 2 } + 4 \sqrt { 5 } x y + 41 y ^ { 2 } = 180$. The point $\left( - \frac { 5 } { 3 } , \frac { 2 \sqrt { 5 } } { 3 } \right)$ is one of the two points on $\Gamma ^ { \prime }$ farthest from the origin.
Find the equation of the line containing the minor axis of $\Gamma ^ { \prime }$ and the length of the minor axis.