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

8. 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. $8$

The right focus of the hyperbola $C : \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 2 } = 1$ is $F$. Point $P$ is on one of the asymptotes of $C$, and $O$ is the origin. If $| PO | = | PF |$, then the area of $\triangle PFO$ is
A. $\frac { 3 \sqrt { 2 } } { 4 }$
B. $\frac { 3 \sqrt { 2 } } { 2 }$
C. $2 \sqrt { 2 }$
D. $3 \sqrt { 2 }$

10. Given that the foci of ellipse $C$ are $F _ { 1 } ( - 1,0 ) , F _ { 2 } ( 1,0 )$, and a line through $F _ { 2 }$ intersects $C$ at points $A , B$. If $\left| A F _ { 2 } \right| = 2 \left| F _ { 2 } B \right| , | A B | = \left| B F _ { 1 } \right|$, then the equation of $C$ is
A. $\frac { x ^ { 2 } } { 2 } + y ^ { 2 } = 1$
B. $\frac { x ^ { 2 } } { 3 } + \frac { y ^ { 2 } } { 2 } = 1$
C. $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 3 } = 1$
D. $\frac { x ^ { 2 } } { 5 } + \frac { y ^ { 2 } } { 4 } = 1$

10. Let $F$ be a focus of the hyperbola $C : \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 5 } = 1$ . Point $P$ is on $C$ , $O$ is the origin. If $| O P | = | O F |$ , then the area of $\triangle O P F$ is
A. $\frac { 3 } { 2 }$
B. $\frac { 5 } { 2 }$
C. $\frac { 7 } { 2 }$
D. $\frac { 9 } { 2 }$

10. For the hyperbola $C : \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 2 } = 1$ with right focus $F$ , if point $P$ is on one of the asymptotes of $C$ , $O$ is the origin, and $| P O | = | P F |$ , then the area of $\triangle P F O$ is
A. $\frac { 3 \sqrt { 2 } } { 4 }$
B. $\frac { 3 \sqrt { 2 } } { 2 }$
C. $2 \sqrt { 2 }$
D. $3 \sqrt { 2 }$

11. Let $F$ be the right focus of the hyperbola $C : \frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > 0 , b > 0 )$, and $O$ be the origin. The circle with $O F$ as diameter intersects the circle $x ^ { 2 } + y ^ { 2 } = a ^ { 2 }$ at points $P , Q$. If $| P Q | = | O F |$, then the eccentricity of $C$ is
A. $\sqrt { 2 }$
B. $\sqrt { 3 }$
C. $2$
D. $\sqrt { 5 }$

16. Given 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. A line through $F _ { 1 }$ intersects the two asymptotes of $C$ at points $A , B$ respectively. If $\overrightarrow { F _ { 1 } A } = \overrightarrow { A B } , \overrightarrow { F _ { 1 } B } \cdot \overrightarrow { F _ { 2 } B } = 0$, then the eccentricity of $C$ is $\_\_\_\_$.
III. Solution Questions: Total 70 points. Solutions should include explanations, proofs, or calculation steps. Questions 17-21 are required for all students. Questions 22 and 23 are optional; students should choose one to answer. If more than one is answered, only the first one will be graded.
(I) Required Questions: Total 60 points.

16. Given 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. A line through $F _ { 1 }$ intersects the two asymptotes of $C$ at points $A , B$ respectively. If $\overrightarrow { F _ { 1 } A } = \overrightarrow { A B }$ and $\overrightarrow { F _ { 1 } B } \cdot \overrightarrow { F _ { 2 } B } = 0$ , then the eccentricity of $C$ is $\_\_\_\_$ .
Section III: Solution Questions: Total 70 points. Show all work, proofs, and calculations. Questions 17-21 are required for all students. Questions 22 and 23 are optional; choose one to answer.

(I) Required Questions: Total 60 points

20. (1) Solution: $\because e = \frac{c}{a} = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{m}{m+2}} = \sqrt{\frac{2}{m+2}}$
Since $m > 1$, $\therefore 0 < e < \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}$.
\begin{tabular}{|l|l|} \hline $\therefore e \in \left(0, \frac{\sqrt{6}}{3}\right)$ & (2) Proof: Since the major axis length of the ellipse is $2\sqrt{m+2} = 4$, $\therefore m = 2$. \hline The equation of line $BM$ is $y = -\frac{y_0}{4}(x-2)$, i.e., $y = -\frac{y_0}{4}x + \frac{1}{2}y_0$. & Substituting into the ellipse equation $x^2 + 2y^2 = 4$, \hline By Vieta's formulas, $2x_1 = \frac{4(y_0^2 - 8)}{y_0^2 + 8}$, & 9 marks \hline 10 marks & \hline $\therefore x_1 = \frac{2(y_0^2 - 8)}{y_0^2 + 8}$, $\therefore y_1 = \frac{8y_0}{y_0^2 + 8}$, $\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots$ $\therefore \overrightarrow{OP} \cdot \overrightarrow{OM} = -2x_1 + y_0y_1 = -\frac{4(y_0^2 - 8)}{y_0^2 + 8} + \frac{8y_0^2}{y_0^2 + 8} = \frac{4y_0^2 + 32}{y_0^2 + 8} = 4 = \mathbf{2m}$. & 12 marks \hline

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 $D , E$. 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

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 $\_\_\_\_$ .

5. The distance from the point $( 3,0 )$ to an asymptote of the hyperbola $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 9 } = 1$ is
A. $\frac { 9 } { 5 }$
B. $\frac { 8 } { 5 }$
C. $\frac { 6 } { 5 }$
D. $\frac { 4 } { 5 }$

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. $\sqrt{13}$

14. $x = - \frac { 3 } { 2 }$
Solution: The focus has coordinates $F \left( \frac { p } { 2 } , 0 \right)$, and point $P$ has coordinates $P \left( \frac { p } { 2 } , p \right)$. Thus $| P F | = p , | O F | = \frac { p } { 2 }$. By the focal chord property $| P F | ^ { 2 } = | O F | \cdot | F Q |$, we get $p ^ { 2 } = 3 p$, so $p = 3$. The directrix equation is $x = - \frac { 3 } { 2 }$.
15. 1
Solution: When $x \geq \frac { 1 } { 2 }$, $f ( x ) = 2 x - 2 \ln ( x ) - 1$, and $f ^ { \prime } ( x ) = \frac { 2 ( x - 1 ) } { x } > 0$, so $f ( x ) \geq f ( 1 ) = 1$. When $x < \frac { 1 } { 2 }$, $f ( x ) = 1 - 2 \ln ( x ) - 2 x$, and $f ^ { \prime } ( x ) = \frac { - 2 ( x + 1 ) } { x } < 0$, so $f ( x ) > f \left( \frac { 1 } { 2 } \right) = 2 \ln 2 > 1$. Therefore, the minimum value of function $f ( x )$ is 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. $\frac{\sqrt{17}}{2}$

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 $\_\_\_\_$ .

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. $-\frac{2}{3}$

The eccentricity of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \ (a > 0 , b > 0)$ is $\sqrt{5}$ . One of its asymptotes intersects the circle $(x - 2)^{2} + (y - 3)^{2} = 1$ at points $A , B$ , then $|AB| =$
A. $\frac{1}{5}$
B. $\frac{\sqrt{5}}{5}$
C. $\frac{2\sqrt{5}}{5}$
D. $\frac{4\sqrt{5}}{5}$

Let $O$ be the origin of coordinates. The line $y=-\sqrt{3}(x-1)$ passes through the focus of the parabola $C: y^2=2px$ $(p>0)$ and intersects $C$ at points $M$ and $N$. Let $l$ be the directrix of $C$. Then
A. $p=2$
B. $|MN|=\frac{8}{3}$
C. the circle with $MN$ as diameter is tangent to $l$
D. $\triangle OMN$ is an isosceles triangle

The ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{6} = 1$ has foci $F_{1} , F_{2}$ and center $O$ . Let $P$ be a point on the ellipse. If $\cos \angle F_{1}PF_{2} = \frac{3}{5}$ , then $|OP| =$
A. $\frac{2}{5}$
B. $\frac{\sqrt{30}}{2}$
C. $\frac{3}{5}$
D. $\frac{\sqrt{35}}{2}$