Let $A$ be a point on the parabola $C : y ^ { 2 } = 2 p x$ ($p > 0$). The distance from point $A$ to the focus of $C$ is 12, and the distance to the $y$-axis is 9. Then $p =$
A. 2
B. 3
C. 6
D. 9

Let $O$ be the origin of coordinates. The line $x = 2$ intersects the parabola $C : y ^ { 2 } = 2 p x ( p > 0 )$ at points $D$ and $E$ . If $O D \perp O E$ , then the focus coordinates of $C$ are
A. $\left( \frac { 1 } { 4 } , 0 \right)$
B. $\left( \frac { 1 } { 2 } , 0 \right)$
C. $( 1,0 )$
D. $( 2,0 )$

In the plane, $A , B$ are two fixed points and $C$ is a moving point. If $\overrightarrow { A C } \cdot \overrightarrow { B C } = 1$, then the locus of point $C$ is
A. a circle
B. an ellipse
C. a parabola
D. a line

Let $O$ be the origin of coordinates. The line $x = 2$ intersects the parabola $C : y ^ { 2 } = 2 p x ( p > 0 )$ at points $D , E$. If $O D \perp O E$, then the focus coordinates of $C$ are
A. $\left( \frac { 1 } { 4 } , 0 \right)$
B. $\left( \frac { 1 } { 2 } , 0 \right)$
C. $( 1,0 )$
D. $( 2,0 )$

Let $O$ be the origin of coordinates. The line $x = a$ intersects the two asymptotes of the hyperbola $C : \frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > 0 , b > 0 )$ at points $D$ and $E$ respectively. If the area of $\triangle O D E$ is 8, then the minimum value of the focal distance of $C$ is
A． 4
B． 8
C． 16
D． 32

For the ellipse $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 3 } = 1$, a line $l$ passes through the right focus $F$ and intersects the ellipse at points $P$ and $Q$, with $P$ in the second quadrant. Given $Q \left( x _ { Q } , y _ { Q } \right)$ and $Q ^ { \prime } \left( x _ { Q } ^ { \prime } , y _ { Q } ^ { \prime } \right)$ both on the ellipse, with $y _ { Q } + y _ { Q } ^ { \prime } = 0$ and $F Q ^ { \prime } \perp P Q$, find the equation of line $l$ as $\_\_\_\_$

Let $F _ { 1 } , F _ { 2 }$ be the two foci of the hyperbola $C : x ^ { 2 } - \frac { y ^ { 2 } } { 3 } = 1$ , $O$ be the origin, and point $P$ on $C$ with $| O P | = 2$ . The area of $\triangle P F _ { 1 } F _ { 2 }$ is
A. $\frac { 7 } { 2 }$
B. 3
C. $\frac { 5 } { 2 }$
D. 2

For the hyperbola $C : \frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > 0 , b > 0 )$ with left and right foci $F _ { 1 } , F _ { 2 }$ respectively, the eccentricity is $\sqrt { 5 }$ . $P$ is a point on $C$ such that $F _ { 1 } P \perp F _ { 2 } P$ . If the area of $\triangle P F _ { 1 } F _ { 2 }$ is 4 , then $a =$
A. $1$
B. $2$
C. $4$
D. $8$

For the hyperbola $C : \frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > 0 , b > 0 )$, one asymptote is $y = \sqrt { 2 } x$. Then the eccentricity of $C$ is $\_\_\_\_$ .

Let $A , B$ be the left and right vertices of the ellipse $E : \frac { x ^ { 2 } } { a ^ { 2 } } + y ^ { 2 } = 1 ( a > 1 )$ respectively, $G$ be the upper vertex of $E$ , and $\overrightarrow { A G } \cdot \overrightarrow { G B } = 8$ . $P$ is a moving point on the line $x = 6$ , the other intersection point of $P A$ with $E$ is $C$ , and the other intersection point of $P B$ with $E$ is $D$ .
(1) Find the equation of $E$ ;
(2) Prove that the line $C D$ passes through a fixed point.

Given the ellipse $C : \frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { m ^ { 2 } } = 1 ( 0 < m < 5 )$ with eccentricity $\frac { \sqrt { 15 } } { 4 }$, where $A , B$ are the left and right vertices of $C$ respectively.
(1) Find the equation of $C$;
(2) If point $P$ is on $C$, point $Q$ is on the line $x = 6$, and $| B P | = | B Q | , B P \perp B Q$, find the area of $\triangle A P Q$ .

3. For the parabola $y ^ { 2 } = 2 p x ( p > 0 )$, the distance from its focus to the line $y = x + 1$ is $\sqrt { 2 }$. Then $p =$
A. 1
B. 2
C. $2 \sqrt { 2 }$
D. 4
【Answer】B 【Solution】 【Analysis】First determine the coordinates of the focus of the parabola, then use the point-to-line distance formula to find the value of $p$.
【Detailed Solution】The focus of the parabola has coordinates $\left( \frac { p } { 2 } , 0 \right)$. The distance from this point to the line $x - y + 1 = 0$ is: $\quad d = \frac { \left| \frac { p } { 2 } - 0 + 1 \right| } { \sqrt { 1 + 1 } } = \sqrt { 2 }$, Solving: $p = 2$ (we discard $p = -6$). Therefore, the answer is: B.

5. Let $F_1, F_2$ be the two foci of hyperbola $C$. Let $P$ be a point on $C$, and $\angle F_1 P F_2 = 60°$, $|PF_1| = 3|PF_2|$. Then the eccentricity of $C$ is
A. $\frac{\sqrt{7}}{2}$
B. $\frac{\sqrt{13}}{2}$
C. $\sqrt{7}$
D. $\sqrt{13}$

The parabola $y ^ { 2 } = 2 p x ( p > 0 )$ has its focus at a distance of $\sqrt { 2 }$ from the line $y = x + 1$. Then $p =$
A. 1
B. 2
C. $2 \sqrt { 2 }$
D. 4

Let $F$ be the focus of the parabola $C: y^2 = 4x$, point $A$ is on $C$, point $B(3,0)$. If $|AF| = |BF|$, then $|AB| =$
A. $2$
B. $2\sqrt{2}$
C. $3$
D. $3\sqrt{2}$

Let $F$ be the focus of the parabola $C : y ^ { 2 } = 4 x$ , point $A$ is on $C$ , point $B ( 3,0 )$ , if $| A F | = | B F |$ , then $| A B | =$
A. 2
B. $2 \sqrt { 2 }$
C. 3
D. $3 \sqrt { 2 }$

For the ellipse $C : \frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > b > 0 )$, let $A$ be the left vertex. Points $P$ and $Q$ are both on $C$ and symmetric about the $y$-axis. If the product of the slopes of $AP$ and $AQ$ is $\frac { 1 } { 4 }$, then the eccentricity of $C$ is:
A. $\frac { \sqrt { 3 } } { 2 }$
B. $\frac { \sqrt { 2 } } { 2 }$
C. $\frac { 1 } { 2 }$
D. $\frac { 1 } { 3 }$

The ellipse $C : \frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > b > 0 )$ has eccentricity $\frac { 1 } { 3 }$. Let $A _ { 1 } , A _ { 2 }$ be the left and right vertices of $C$ respectively, and $B$ be the upper vertex. If $\overrightarrow { B A _ { 1 } } \cdot \overrightarrow { B A _ { 2 } } = - 1$ , then the equation of $C$ is
A. $\frac { x ^ { 2 } } { 18 } + \frac { y ^ { 2 } } { 16 } = 1$
B. $\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 8 } = 1$
C. $\frac { x ^ { 2 } } { 3 } + \frac { y ^ { 2 } } { 2 } = 1$
D. $\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 32 } = 1$

For an ellipse $C$, $F_1, F_2$ are its two foci, $M, N$ are two points on the ellipse. If $\cos \angle F_1NF_2 = \frac{3}{5}$, then the eccentricity of $C$ is
A. $\frac{1}{2}$
B. $\frac{3}{2}$
C. $\frac{\sqrt{13}}{2}$
D. $\frac{\sqrt{17}}{2}$

11. Let $O$ be the origin. Point $A ( 1,1 )$ lies on the parabola $C : x ^ { 2 } = 2 p y$ ( $p > 0$ ). A line through point $B ( 0 , - 1 )$ intersects $C$ at points $P$ and $Q$. Then
A. The directrix of $C$ is $y = - 1$
B. Line $A B$ is tangent to $C$
C. $| O P | \cdot | O Q | > | O A | ^ { 2 }$
D. $| B P | \cdot | B Q | > | B A | ^ { 2 }$

If the asymptotes of the hyperbola $y ^ { 2 } - \frac { x ^ { 2 } } { m ^ { 2 } } = 1 ( m > 0 )$ are tangent to the circle $x ^ { 2 } + y ^ { 2 } - 4 y + 3 = 0$, then $m =$ $\_\_\_\_$

For the hyperbola $C : \frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > 0 , b > 0 )$ with eccentricity $e$, write out a value of $e$ that satisfies the condition ``the line $y = kx$ intersects the hyperbola at four distinct points'' and give one such value $\_\_\_\_$ .

(1) Find the equation of $C$;
(2) Let the lines $MD$ and $ND$ intersect $C$ at another point $A$ and $B$ respectively. Denote the inclination angles of lines $MN$ and $AB$ as $\alpha$ and $\beta$ respectively. When $\alpha - \beta$ attains its maximum value, find the equation of line $AB$.

An ellipse $E$ has its center at the origin, with axes of symmetry along the $x$-axis and $y$-axis, and passes through points $A ( 0 , - 2 ) , B \left( \frac { 3 } { 2 } , 1 \right)$.
(The remainder of this question was cut off in the source document.)

Given the ellipse $\frac{x^2}{3}+y^2=1$ with left and right foci $F_1, F_2$ respectively, the line $y=x+m$ intersects $C$ at points $A$ and $B$. If the area of $\triangle F_1AB$ is 2 times the area of $\triangle F_2AB$, then $m=$
A. $\frac{2}{3}$
B. $\frac{\sqrt{2}}{3}$
C. $-\frac{\sqrt{2}}{3}$
D. $-\frac{2}{3}$