23. (Total Score: 18 points) Subproblem 1: 4 points, Subproblem 2: 6 points, Subproblem 3: 8 points.

Given that the equation of ellipse $\Gamma$ is $\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > b > 0 )$ , with $A ( 0 , b ) , B ( 0 , - b )$ and $Q ( a , 0 )$ being three vertices of $\Gamma$.\\
(1) If point $M$ satisfies $\overrightarrow { A M } = \frac { 1 } { 2 } ( \overrightarrow { A Q } + \overrightarrow { A B } )$ , find the coordinates of point $M$;\\
(2) Let line $l _ { 1 } : y = k _ { 1 } x + p$ intersect ellipse $\Gamma$ at points $C , D$ and intersect line $l _ { 2 } : y = k _ { 2 } x$ at point $E$ . If $k _ { 1 } \cdot k _ { 2 } = - \frac { b ^ { 2 } } { a ^ { 2 } }$,\\
prove that $E$ is the midpoint of $C D$;\\
(3) Let point $P$ be inside ellipse $\Gamma$ and not on the $x$-axis. How should one construct a line $l$ passing through the midpoint $F$ of $P Q$ such that the two intersection points $P _ { 1 } , P _ { 2 }$ of $l$ with ellipse $\Gamma$ satisfy $\overrightarrow { P P _ { 1 } } + \overrightarrow { P P _ { 2 } } = \overrightarrow { P Q }$ ? Let $a = 10 , b = 5$ , and the coordinates of point $P$ are $( - 8 , - 1 )$ . If points $P _ { 1 } , P _ { 2 }$ on ellipse $\Gamma$ satisfy $\overrightarrow { P P _ { 1 } } + \overrightarrow { P P _ { 2 } } = \overrightarrow { P Q }$ , find the coordinates of points $P _ { 1 } , P _ { 2 }$ .

\section*{2010 National College Entrance Examination Mathematics (Science) Shanghai Test}
2010-6-7\\
Class $\_\_\_\_$ , Student ID $\_\_\_\_$ , Name $\_\_\_\_$\\
I. Fill in the Blanks (Total Score: 56 points, 4 points each)\\
1. The solution set of the inequality $\frac { 2 - x } { x + 4 } > 0$ is $\_\_\_\_$.

2. If the complex number $z = 1 - 2 i$ ($i$ is the imaginary unit), then $z \cdot \bar { z } + z =$ $\_\_\_\_$.

3. A moving
From the system of equations $\left\{ \begin{array} { l } y = k _ { 1 } x + p \\ y = k _ { 2 } x \end{array} \right.$ , eliminating $y$ gives the equation $\left( k _ { 2 } - k _ { 1 } \right) x = p$ ,\\
Since $k _ { 2 } = - \frac { b ^ { 2 } } { a ^ { 2 } k _ { 1 } }$ , we have $\left\{ \begin{array} { l } x = \frac { p } { k _ { 2 } - k _ { 1 } } = - \frac { a ^ { 2 } k _ { 1 } p } { a ^ { 2 } k _ { 1 } ^ { 2 } + b ^ { 2 } } = x _ { 0 } \\ y = k _ { 2 } x = \frac { b ^ { 2 } p } { a ^ { 2 } k _ { 1 } ^ { 2 } + b ^ { 2 } } = y _ { 0 } \end{array} \right.$ ,\\
Therefore $E$ is the midpoint of $C D$ ;\\
(3) Since point $P$ is inside the ellipse $\Gamma$ and not on the $x$-axis, point $F$ is inside the ellipse $\Gamma$ . We can find the slope $k _ { 2 }$ of line $O F$ . From $\overrightarrow { P P _ { 1 } } + \overrightarrow { P P _ { 2 } } = \overrightarrow { P Q }$ we know that $F$ is the midpoint of $P _ { 1 } P _ { 2 }$ . According to (2), we can obtain the slope of line $l$ as $k _ { 1 } = - \frac { b ^ { 2 } } { a ^ { 2 } k _ { 2 } }$ , and thus obtain the equation of line $l$ .\\
$F \left( 1 , - \frac { 1 } { 2 } \right)$ , the slope of line $O F$ is $k _ { 2 } = - \frac { 1 } { 2 }$ , the slope of line $l$ is $k _ { 1 } = - \frac { b ^ { 2 } } { a ^ { 2 } k _ { 2 } } = \frac { 1 } { 2 }$ ,\\
Solving the system of equations $\left\{ \begin{array} { l } y = \frac { 1 } { 2 } x - 1 \\ \frac { x ^ { 2 } } { 100 } + \frac { y ^ { 2 } } { 25 } = 1 \end{array} \right.$ , eliminating $y$ : $x ^ { 2 } - 2 x - 48 = 0$ , we obtain $P _ { 1 } ( - 6 , - 4 ) , P _ { 2 } ( 8,3 )$ .

\section*{Reference Answers for Science}
\section*{I. Fill in the Blanks}
$1 . ( - 4,2 )$ ;\\
2. $6 - 2i$ ;\\
3. $y ^ { 2 } = 8 x$ ;\\
4. $0$ ;\\
5. $3$ ;\\
6. $8.2$ ;

\section*{7. $S \leftarrow S + a$ ;}
$8 . ( 0 , - 2 )$ ;\\
9. $\frac { 7 } { 26 }$ ;\\
10. $45$ ;\\
11. $1$ ;\\
12. $\frac { 8 \sqrt { 2 } } { 3 }$ ;\\
13. $4 a b = 1$ ;\\
14. $36$ .

\section*{II. Multiple Choice}
15. A;\\