If the focus of the parabola $y ^ { 2 } = 2 p x \ ( p > 0 )$ is a focus of the ellipse $\frac { x ^ { 2 } } { 3 p } + \frac { y ^ { 2 } } { p } = 1$, then $p =$
A． 2
B． 3
C． 4
D． 8

A hyperbola, having the transverse axis of length $2\sin\theta$, is confocal with the ellipse $3x^2 + 4y^2 = 12$. Then its equation is
(A) $x^2\csc^2\theta - y^2\sec^2\theta = 1$
(B) $x^2\sec^2\theta - y^2\csc^2\theta = 1$
(C) $x^2\sin^2\theta - y^2\cos^2\theta = 1$
(D) $x^2\cos^2\theta - y^2\sin^2\theta = 1$

Let $P \left( x _ { 1 } , y _ { 1 } \right)$ and $Q \left( x _ { 2 } , y _ { 2 } \right) , y _ { 1 } < 0 , y _ { 2 } < 0$, be the end points of the latus rectum of the ellipse $x ^ { 2 } + 4 y ^ { 2 } = 4$. The equations of parabolas with latus rectum $P Q$ are
(A) $x ^ { 2 } + 2 \sqrt { 3 } \quad y = 3 + \sqrt { 3 }$
(B) $x ^ { 2 } - 2 \sqrt { 3 } \quad y = 3 + \sqrt { 3 }$
(C) $x ^ { 2 } + 2 \sqrt { 3 } \quad y = 3 - \sqrt { 3 }$
(D) $x ^ { 2 } - 2 \sqrt { 3 } \quad y = 3 - \sqrt { 3 }$

Let the hyperbola $H : \frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$ pass through the point $( 2 \sqrt { 2 } , - 2 \sqrt { 2 } )$. A parabola is drawn whose focus is same as the focus of $H$ with positive abscissa and the directrix of the parabola passes through the other focus of $H$. If the length of the latus rectum of the parabola is $e$ times the length of the latus rectum of $H$, where $e$ is the eccentricity of $H$, then which of the following points lies on the parabola?
(1) $( 2 \sqrt { 3 } , 3 \sqrt { 2 } )$
(2) $( 3 \sqrt { 3 } , - 6 \sqrt { 2 } )$
(3) $( \sqrt { 3 } , - \sqrt { 6 } )$
(4) $( 3 \sqrt { 6 } , 6 \sqrt { 2 } )$

Let $\mathrm { E } : \frac { x ^ { 2 } } { \mathrm { a } ^ { 2 } } + \frac { y ^ { 2 } } { \mathrm {~b} ^ { 2 } } = 1 , \mathrm { a } > \mathrm { b }$ and $\mathrm { H } : \frac { x ^ { 2 } } { \mathrm {~A} ^ { 2 } } - \frac { y ^ { 2 } } { \mathrm {~B} ^ { 2 } } = 1$. Let the distance between the foci of E and the foci of $H$ be $2 \sqrt { 3 }$. If $a - A = 2$, and the ratio of the eccentricities of $E$ and $H$ is $\frac { 1 } { 3 }$, then the sum of the lengths of their latus rectums is equal to:
(1) 10
(2) 9
(3) 8
(4) 7

Answer the following questions regarding curves on the $x y$-plane.
(1) Show that the foci of an ellipse:
$$\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 \quad ( a > b > 0 )$$
and those of a hyperbola:
$$\frac { x ^ { 2 } } { c ^ { 2 } } - \frac { y ^ { 2 } } { d ^ { 2 } } = 1 \quad ( c > d > 0 )$$
are $\left( \pm \sqrt { a ^ { 2 } - b ^ { 2 } } , 0 \right)$ and $\left( \pm \sqrt { c ^ { 2 } + d ^ { 2 } } , 0 \right)$, respectively. Note that an ellipse (hyperbola) is a curve such that the sum (difference) of the distances from the foci to any point on the curve is constant.
(2) As for the ellipse equation, consider the set $E _ { u }$ of ellipses such that $a ^ { 2 } - b ^ { 2 } = u ^ { 2 }$ ($u$ is a positive constant). By writing the simultaneous equations that consist of the ellipse equation and the differential equation obtained by taking the derivative of the ellipse equation with respect to $x$, show that any ellipse in $E _ { u }$ satisfies
$$x y y ^ { \prime 2 } + \left( x ^ { 2 } - y ^ { 2 } - u ^ { 2 } \right) y ^ { \prime } - x y = 0 , \quad ( * * * )$$
where $y ^ { \prime } = \frac { \mathrm { d} y } { \mathrm {~d} x }$.
(3) As for the hyperbola equation, consider the set $H _ { u }$ of hyperbolae such that $c ^ { 2 } + d ^ { 2 } = u ^ { 2 }$. Show that any hyperbola in $H _ { u }$ satisfies Eq. $(***) $.
(4) Let $C _ { u }$ be the set of curves perpendicular to any ellipse in $E _ { u }$. Let $D _ { u }$ be the set of curves obtained by removing from $C _ { u }$ the line $x = 0$ as well as all the curves including a point such that $y ^ { \prime } = 0$. Find a differential equation that any curve in $D _ { u }$ satisfies.
(5) Solve the differential equation that you found in Question (4). If necessary, rewrite the differential equation into a differential equation with respect to $p$ with replacement such that $\alpha = x ^ { 2 } , \beta = y ^ { 2 }$, and $p = \frac { \mathrm { d} \beta } { \mathrm { d} \alpha }$.