Given the ellipse $C : x ^ { 2 } + 3 y ^ { 2 } = 3$, a line passing through point $D ( 1,0 )$ but not through point $E ( 2,1 )$ intersects the ellipse $C$ at points $A$ and $B$. The line $AE$ intersects the line $x = 3$ at point $M$.\n(1) Find the eccentricity of the ellipse $C$;\n(II) If $AB$ is perpendicular to the $x$-axis, find the slope of line $BM$;\n(III) Determine the positional relationship between line $BM$ and line $DE$, and explain the reason.

20. (12 points) The ellipse $\mathrm { E } : \frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > b > 0 )$ has semi-focal distance $c$. The distance from the origin O to the line passing through the points $( c , 0 )$ and $( 0 , b )$ is $\frac { 1 } { 2 } c$. (I) Find the eccentricity of ellipse E; (II) As shown in the figure, AB is a diameter of circle $\mathrm { M } : ( x + 2 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = \frac { 5 } { 2 }$. If the ellipse E passes through points A and B, find the equation of ellipse E. [Figure]

20. (This question is worth 13 points) As shown in the figure, the ellipse $\mathrm { E } : \frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > b > 0 )$ has eccentricity $\frac { \sqrt { 2 } } { 2 }$. The point $( 0,1 )$ is on the minor axis $CD$, and $\overline { P C } \overline { P D } = - 1$. (I) Find the equation of ellipse E; (II) Let O be the origin. A moving line through point P intersects the ellipse at points A and B. Does there exist a constant $\lambda$ such that $\overline { O A } \overline { O B } + \lambda \overline { P A } \overline { P B }$ is a constant? If it exists, find the value of $\lambda$; if it does not exist, explain why. [Figure]

20. As shown in the figure, the ellipse $\mathrm { E } : \frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1$ $(a > b > 0)$ has eccentricity $\frac { \sqrt { 2 } } { 2 }$. A moving line $l$ passing through point $\mathrm { P } ( 0, 1 )$ intersects the ellipse at points $\mathrm { A }$ and $\mathrm { B }$. When line $l$ is parallel to the $x$-axis, the length of the chord intercepted by line $l$ on ellipse $E$ is $2 \sqrt { 2 }$.
(1) Find the equation of ellipse $E$;
(2) In the rectangular coordinate system $xOy$, does there exist a fixed point $Q$ different from point $P$ such that $\frac { | Q A | } { | Q B | } = \frac { | P A | } { | P B | }$ always holds? If it exists, find the coordinates of point $Q$; if it does not exist, explain the reason. [Figure]

21. (14 marks) (I) Let point $D ( t , 0 ) ( | t | \leq 2 ) , N \left( x _ { 0 } , y _ { 0 } \right) , M ( x , y )$. According to the problem,
$$\overrightarrow { M D } = 2 \overrightarrow { D N } \text {, and } | \overrightarrow { D N } | = | \overrightarrow { O N } | = 1,$$
[Figure]
so $( t - x , - y ) = 2 \left( x _ { 0 } - t , y _ { 0 } \right)$, and $\left\{ \begin{array} { l } \left( x _ { 0 } - t \right) ^ { 2 } + y _ { 0 } ^ { 2 } = 1 , \\ x _ { 0 } ^ { 2 } + y _ { 0 } ^ { 2 } = 1 . \end{array} \right.$ That is, $\left\{ \begin{array} { l } t - x = 2 x _ { 0 } - 2 t , \\ y = - 2 y _ { 0 } . \end{array} \right.$ and $t \left( t - 2 x _ { 0 } \right) = 0$. Since when point $D$ is fixed, point $N$ is also fixed, $t$ is not identically zero, thus $t = 2 x _ { 0 }$, so $x _ { 0 } = \frac { x } { 4 } , y _ { 0 } = - \frac { y } { 2 }$. Substituting into $x _ { 0 } ^ { 2 } + y _ { 0 } ^ { 2 } = 1$, we obtain $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 1$, that is, the equation of the required curve $C$ is $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 1$. (II) (1) When the slope of line $l$ does not exist, line $l$ is $x = 4$ or $x = - 4$, and we have $S _ { \triangle O P Q } = \frac { 1 } { 2 } \times 4 \times 4 = 8$.
(2) When the slope of line $l$ exists, let line $l : y = k x + m \left( k \neq \pm \frac { 1 } { 2 } \right)$. From $\left\{ \begin{array} { l } y = k x + m , \\ x ^ { 2 } + 4 y ^ { 2 } = 16 , \end{array} \right.$ eliminating $y$, we obtain $\left( 1 + 4 k ^ { 2 } \right) x ^ { 2 } + 8 k m x + 4 m ^ { 2 } - 16 = 0$. Since line $l$ always has exactly one common point with ellipse $C$, we have $\Delta = 64 k ^ { 2 } m ^ { 2 } - 4 \left( 1 + 4 k ^ { 2 } \right) \left( 4 m ^ { 2 } - 16 \right) = 0$, that is, $m ^ { 2 } = 16 k ^ { 2 } + 4$.
From $\left\{ \begin{array} { l } y = k x + m , \\ x - 2 y = 0 , \end{array} \right.$ we obtain $P \left( \frac { 2 m } { 1 - 2 k } , \frac { m } { 1 - 2 k } \right)$; similarly, we obtain $Q \left( \frac { - 2 m } { 1 + 2 k } , \frac { m } { 1 + 2 k } \right)$. From the distance from origin $O$ to line $P Q$ being $d = \frac { | m | } { \sqrt { 1 + k ^ { 2 } } }$ and $| P Q | = \sqrt { 1 + k ^ { 2 } } \left| x _ { P } - x _ { Q } \right|$, we obtain $S _ { \triangle O P Q } = \frac { 1 } { 2 } | P Q | \cdot d = \frac { 1 } { 2 } | m | \left| x _ { P } - x _ { Q } \right| = \frac { 1 } { 2 } \cdot | m | \left| \frac { 2 m } { 1 - 2 k } + \frac { 2 m } { 1 + 2 k } \right| = \left| \frac { 2 m ^ { 2 } } { 1 - 4 k ^ { 2 } } \right|$.
Substituting (1) into (2), we obtain $S _ { \triangle O P Q } = \left| \frac { 2 m ^ { 2 } } { 1 - 4 k ^ { 2 } } \right| = 8 \frac { \left| 4 k ^ { 2 } + 1 \right| } { \left| 4 k ^ { 2 } - 1 \right| }$. When $k ^ { 2 } > \frac { 1 } { 4 }$, $S _ { \triangle O P Q } = 8 \left( \frac { 4 k ^ { 2 } + 1 } { 4 k ^ { 2 } - 1 } \right) = 8 \left( 1 + \frac { 2 } { 4 k ^ { 2 } - 1 } \right) > 8$; When $0 \leq k ^ { 2 } < \frac { 1 } { 4 }$, $S _ { \triangle O P Q } = 8 \left( \frac { 4 k ^ { 2 } + 1 } { 1 - 4 k ^ { 2 } } \right) = 8 \left( - 1 + \frac

A circle with the vertex of parabola $C$ as its center intersects $C$ at points $A , B$, and intersects the directrix of $C$ at points $D , E$. If $|DE| = 2 \sqrt { 5 }$, then the distance from the focus of $C$ to the directrix is
(A) 2
(B) 4
(C) 6
(D) 8

The line $x + y + 2 = 0$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively. Point $P$ is on the circle $( x - 2 ) ^ { 2 } + y ^ { 2 } = 2$. The range of the area of $\triangle ABP$ is
A. $[ 2,6 ]$
B. $[ 4,8 ]$
C. $[ \sqrt { 2 } , 3 \sqrt { 2 } ]$
D. $[ 2 \sqrt { 2 } , 3 \sqrt { 2 } ]$

The line $y = x + 1$ intersects the circle $x ^ { 2 } + y ^ { 2 } + 2 y - 3 = 0$ at points $A$ and $B$. Then $| A B | = $ \_\_\_\_

Given that the eccentricity of the ellipse $\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > b > 0 )$ is $\frac { 1 } { 2 }$, then (A) $a ^ { 2 } = 2 b ^ { 2 }$ (B) $3 a ^ { 2 } = 4 b ^ { 2 }$ (C) $a = 2 b$ (D) $3 a = 4 b$

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. $8$

9. Let $S _ { n }$ denote the sum of the first $n$ terms of an arithmetic sequence $\left\{ a _ { n } \right\}$. Given $S _ { 4 } = 0 , a _ { 5 } = 5$, then
A. $a _ { n } = 2 n - 5$
B. $a _ { n } = 3 n - 10$
C. $S _ { n } = 2 n ^ { 2 } - 8 n$
D. $S _ { n } = \frac { 1 } { 2 } n ^ { 2 } - 2 n$

Mathematics (Science) Test Paper Page 2 (Total 5 Pages)

9. 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

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 $\left| A F _ { 2 } \right| = 2 \left| F _ { 2 } B \right|$ and $| A B | = \left| B F _ { 1 } \right|$ , 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. $\frac { x ^ { 2 } } { 5 } + \frac { y ^ { 2 } } { 4 } = 1$

12. In $\triangle A B C$, the sides opposite to angles $A , B , C$ are $a , b , c$ respectively. Given $10 \sin A - 5 \sin C = 2 \sqrt { 6 }$ and $\cos B = \frac { 1 } { 5 }$, then $\frac { c } { a } =$
$$\text { A. } \frac { 6 } { 7 } \quad \text{B.} \frac { 7 } { 6 } \quad \text{C.} \frac { 5 } { 6 } \quad \text{D.} \frac { 6 } { 5 }$$

Section II

II. Fill-in-the-Blank Questions: This section contains 4 questions, each worth 5 points, totaling 20 points. Write your answers on the answer sheet.

15. A quadrangular pyramid $P - A B C D$ has all vertices on the surface of sphere $O$. $PA$ is perpendicular to the plane containing rectangle $A B C D$. $AB = 3$, $AD = \sqrt { 3 }$. The surface area of sphere $O$ is $13 \pi$. The length of segment $PA$ is \_\_\_\_.

15. Let $F _ { 1 } , F _ { 2 }$ be the two foci of the ellipse $C : \frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 20 } = 1$ . Let $M$ be a point on $C$ in the first quadrant. If $\triangle M F _ { 1 } F _ { 2 }$ is an isosceles triangle, then the coordinates of $M$ are $\_\_\_\_$

15. Let $F _ { 1 } , F _ { 2 }$ be the two foci of the ellipse $C : \frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 20 } = 1$ , and let $M$ be a point on $C$ in the first quadrant. If $\triangle M F _ { 1 } F _ { 2 }$ is an isosceles triangle, then the coordinates of $M$ are $\_\_\_\_$ .

The parabola $C : x ^ { 2 } = - 2 p y$ passes through the point $( 2 , - 1 )$. (I) Find the equation of parabola $C$ and the equation of its directrix; (II) Let $O$ be the origin. A line $l$ with non-zero slope passes through the focus of parabola $C$ and intersects parabola $C$ at two points $M , N$. The line $y = - 1$ intersects lines $O M$ and $O N$ at points $A$ and $B$ respectively. Prove that the circle with $A B$ as diameter passes through two fixed points on the $y$-axis.

20. (12 points)
Let $F _ { 1 } , F _ { 2 }$ be the two foci of the ellipse $C : \frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > b > 0 )$, $P$ be a point on $C$, and $O$ be the origin.
(1) If $\triangle P O F _ { 2 }$ is an equilateral triangle, find the eccentricity of $C$;
(2) If there exists a point $P$ such that $P F _ { 1 } \perp P F _ { 2 }$ and the area of $\triangle F _ { 1 } P F _ { 2 }$ equals 16, find the value of $b$ and the range of values for $a$.

If a circle passing through point $(2,1)$ is tangent to both coordinate axes, then the distance from the center of the circle to the line $2 x - y - 3 = 0$ is
A．$\frac { \sqrt { 5 } } { 5 }$
B．$\frac { 2 \sqrt { 5 } } { 5 }$
C．$\frac { 3 \sqrt { 5 } } { 5 }$
D．$\frac { 4 \sqrt { 5 } } { 5 }$

Given the circle $x ^ { 2 } + y ^ { 2 } - 6 x = 0$ , the minimum length of the chord cut by this circle from a line passing through the point $( 1,2 )$ is
A. 1
B. 2
C. 3
D. 4

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. C

Solution: By the property of ellipses, $\left| M F _ { 1 } \right| + \left| M F _ { 2 } \right| = 2 a = 6$. By the AM-GM inequality:
$$\left| M F _ { 1 } \right| \cdot \left| M F _ { 2 } \right| \leq \frac { 1 } { 4 } \left( \left| M F _ { 1 } \right| + \left| M F _ { 2 } \right| \right) ^ { 2 } = 9 .$$
Equality holds when $\left| M F _ { 1 } \right| = \left| M F _ { 2 } \right| = 3$, and $M$ is the upper or lower vertex of the ellipse. So the answer is $C$.

11. ACD

Solution: The radius of the circle is $r = 4$. The equation of line $AB$ is $y = - \frac { 1 } { 2 } x + 2$. Drawing a line through $P$ parallel to $AB$, the distance between the two lines is $d = \frac { \left| \frac { 15 } { 2 } - 2 \right| } { \sqrt { 1 + \left( - \frac { 1 } { 2 } \right) ^ { 2 } } } = \frac { 11 } { \sqrt { 5 } }$. Since $d < 6$, the maximum distance from $P$ to line $AB$ is $d + r$, so A is correct; the minimum distance from $P$ to line $AB$ is $d - r < 2$, so B is incorrect; the extremum of $\angle P A B$ is attained when $PB$ is tangent to the circle. We have $O B = \sqrt { 5 ^ { 2 } + 3 ^ { 2 } } = \sqrt { 34 }$. Since $PB$ is tangent to $OB$, we have $PB \perp O B$. By the Pythagorean theorem, $| P B | = \sqrt { 34 - 4 ^ { 2 } } = \sqrt { 18 }$. The two tangent lines from a point to a circle have equal length, so C and D are correct. The answer is $ACD$.
When $A _ { 1 } P \perp B P$, there are two points $P$ satisfying the condition, so C is incorrect. For option D, let $E$ be the midpoint of $C C _ { 1 }$, and let $G$ be the center of rectangle $A A _ { 1 } B _ { 1 } B$. When $\mu = \frac { 1 } { 2 }$, $P$ is a point on $EF$. Since $E G \perp$ plane $A A _ { 1 } B _ { 1 }$ and $A _ { 1 } B \perp A B _ { 1 }$, we have $A _ { 1 } B \perp$ plane $E A B _ { 1 }$. When $P$ coincides with $E$, the condition is satisfied. There is a unique plane perpendicular to line $A _ { 1 } B$ passing through such points.

III. Fill in the Blank Questions

13. 1
Solution: Setting $f ( x ) = f ( - x )$ gives $x ^ { 3 } \left( 2 ^ { x } + 2 ^ { - x } \right) ( a - 1 ) = 0$ for all $x$, so $a = 1$.

15. Let $F_1, F_2$ be the two foci of the ellipse $C: \frac{x^2}{16} + \frac{y^2}{4} = 1$. Let $P, Q$ be two points on $C$ that are symmetric about the origin, and $|PQ| = |F_1F_2|$. Then the area of quadrilateral $PF_1QF_2$ is $\_\_\_\_$. [Figure]