As shown in the figure, there is a point $\mathrm { A } ( 0 , a )$ on the $y$-axis and a point P moving on the ellipse $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 9 } = 1$ with foci $\mathrm { F } , \mathrm { F } ^ { \prime }$. When the minimum value of $\overline { \mathrm { AP } } - \overline { \mathrm { FP } }$ is 1, find the value of $a ^ { 2 }$. [4 points]

10. In the rectangular coordinate system $x O y$, among all circles with center at point $( 1,0 )$ and tangent to the line $m x - y - 2 m - 1 = 0 ( m \in R )$, the standard equation of the circle with the largest radius is $\_\_\_\_$.

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]

19. Given the ellipse $\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > b > 0 )$ with upper vertex $B$, left focus $F$, and eccentricity $\frac { \sqrt { 5 } } { 5 }$.
(1) Find the slope of line $BF$;
(2) Let line $BF$ intersect the ellipse at point $P$ (where $P$ is different from $B$). A line passing through $B$ and perpendicular to $BF$ intersects the ellipse at point $Q$ (where $Q$ is different from $B$). Line $PQ$ intersects the $x$-axis at point $M$, and $|PM| = l|MQ|$.
1) Find the value of $l$;
2) If $|PM| \sin \angle BQP = \frac { 7 \sqrt { 5 } } { 9 }$, find the equation of the ellipse.

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. $[ 2 \sqrt { 2 } , 3 \sqrt { 2 } ]$

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 $C$.

Circle $\odot O$ has radius 1. Line $PA$ is tangent to $\odot O$ at point $A$. Line $PB$ intersects $\odot O$ at points $B$ and $C$. $D$ is the midpoint of $BC$. If $| P O | = \sqrt { 2 }$, then the maximum value of $\overrightarrow { P A } \cdot \overrightarrow { P D }$ is
A. $\frac { 1 + \sqrt { 2 } } { 2 }$
B. $\frac { 1 + 2 \sqrt { 2 } } { 2 }$
C. $1 + \sqrt { 2 }$
D. $2 + \sqrt { 2 }$

If two real numbers $x$ and $y$ satisfy $( x + 5 ) ^ { 2 } + ( y - 10 ) ^ { 2 } = 196$, then the minimum possible value of $x ^ { 2 } + 2 x + y ^ { 2 } - 4 y$ is
(A) $271 - 112 \sqrt { 5 }$.
(B) $14 - 4 \sqrt { 5 }$.
(C) $276 - 112 \sqrt { 5 }$.
(D) $9 - 4 \sqrt { 5 }$.

Let $P$ be the point on the parabola $y ^ { 2 } = 4 x$ which is at the shortest distance from the center $S$ of the circle $x ^ { 2 } + y ^ { 2 } - 4 x - 16 y + 64 = 0$. Let $Q$ be the point on the circle dividing the line segment $S P$ internally. Then
(A) $S P = 2 \sqrt { 5 }$
(B) $S Q : Q P = ( \sqrt { 5 } + 1 ) : 2$
(C) the $x$-intercept of the normal to the parabola at $P$ is 6
(D) the slope of the tangent to the circle at $Q$ is $\frac { 1 } { 2 }$

Consider the region $R = \left\{ ( x , y ) \in \mathbb { R } \times \mathbb { R } : x \geq 0 \right.$ and $\left. y ^ { 2 } \leq 4 - x \right\}$. Let $\mathcal { F }$ be the family of all circles that are contained in $R$ and have centers on the $x$-axis. Let $C$ be the circle that has largest radius among the circles in $\mathcal { F }$. Let $( \alpha , \beta )$ be a point where the circle $C$ meets the curve $y ^ { 2 } = 4 - x$. The radius of the circle $C$ is $\_\_\_\_$.

Consider the region $R = \left\{ ( x , y ) \in \mathbb { R } \times \mathbb { R } : x \geq 0 \right.$ and $\left. y ^ { 2 } \leq 4 - x \right\}$. Let $\mathcal { F }$ be the family of all circles that are contained in $R$ and have centers on the $x$-axis. Let $C$ be the circle that has largest radius among the circles in $\mathcal { F }$. Let $( \alpha , \beta )$ be a point where the circle $C$ meets the curve $y ^ { 2 } = 4 - x$. The value of $\alpha$ is $\_\_\_\_$.

$P$ and $Q$ are two distinct points on the parabola, $y ^ { 2 } = 4 x$, with parameters $t$ and $t _ { 1 }$, respectively. If the normal at $P$ passes through $Q$, then the minimum value of $t _ { 1 } ^ { 2 }$, is
(1) 8
(2) 4
(3) 6
(4) 2

The radius of a circle, having minimum area, which touches the curve $y = 4 - x^2$ and the lines $y = |x|$ is:
(1) $2(\sqrt{2} + 1)$
(2) $2(\sqrt{2} - 1)$
(3) $4(\sqrt{2} - 1)$
(4) $4(\sqrt{2} + 1)$

If a point $P ( 0 , - 2 )$ and $Q$ is any point on the circle, $x ^ { 2 } + y ^ { 2 } - 5 x - y + 5 = 0$, then the maximum value of $( P Q ) ^ { 2 }$ is
(1) $8 + 5 \sqrt { 3 }$
(2) $\frac { 47 + 10 \sqrt { 6 } } { 2 }$
(3) $14 + 5 \sqrt { 3 }$
(4) $\frac { 25 + \sqrt { 6 } } { 2 }$

Let $PQ$ be a diameter of the circle $x ^ { 2 } + y ^ { 2 } = 9$. If $\alpha$ and $\beta$ are the lengths of the perpendiculars from $P$ and $Q$ on the straight line, $x + y = 2$ respectively, then the maximum value of $\alpha \beta$ is $\_\_\_\_$

Let $A ( 1,4 )$ and $B ( 1 , - 5 )$ be two points. Let $P$ be a point on the circle $( ( x - 1 ) ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 1$, such that $( P A ) ^ { 2 } + ( P B ) ^ { 2 }$ have maximum value, then the points, $P , A$ and $B$ lie on
(1) a hyperbola
(2) a straight line
(3) an ellipse
(4) a parabola

Let $r _ { 1 }$ and $r _ { 2 }$ be the radii of the largest and smallest circles, respectively, which pass through the point $( - 4,1 )$ and having their centres on the circumference of the circle $x ^ { 2 } + y ^ { 2 } + 2 x + 4 y - 4 = 0$. If $\frac { r _ { 1 } } { r _ { 2 } } = a + b \sqrt { 2 }$, then $a + b$ is equal to:
(1) 3
(2) 11
(3) 5
(4) 7

Let $P$ and $Q$ be any points on the curves $( x - 1 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 1$ and $y = x ^ { 2 }$, respectively. The distance between $P$ and $Q$ is minimum for some value of the abscissa of $P$ in the interval
(1) $\left( 0 , \frac { 1 } { 4 } \right)$
(2) $\left( \frac { 1 } { 2 } , \frac { 3 } { 4 } \right)$
(3) $\left( \frac { 1 } { 4 } , \frac { 1 } { 2 } \right)$
(4) $\left( \frac { 3 } { 4 } , 1 \right)$

Let a variable line passing through the centre of the circle $x^2 + y^2 - 16x - 4y = 0$, meet the positive coordinate axes at the point $A$ and $B$. Then the minimum value of $OA + OB$, where $O$ is the origin, is equal to
(1) 12
(2) 18
(3) 20
(4) 24

If the shortest distance of the parabola $y ^ { 2 } = 4 x$ from the centre of the circle $x ^ { 2 } + y ^ { 2 } - 4 x - 16 y + 64 = 0$ is $d$ , then $\mathrm { d } ^ { 2 }$ is equal to:
(1) 16
(2) 24
(3) 20
(4) 36

Consider two straight lines
$$y=1, \quad y=-1$$
and the point $\mathrm{A}(0,3)$ in the $xy$-plane. Take a point P on the straight line $y=1$ and a point Q on the straight line $y=-1$ such that
$$\angle\mathrm{PAQ}=90^\circ.$$
Let the two points P and Q move preserving the above conditions. We are to find the minimum value of the length of the line segment PQ.
First, denote the coordinates of P by $(\alpha,1)$ and the coordinates of Q by $(\beta,-1)$. Then the condition $\angle\mathrm{PAQ}=90^\circ$ is reduced to the conditions $\alpha\neq 0$, $\beta\neq 0$ and
$$\alpha\beta = \mathbf{AB}.$$
Since we know that $\alpha$ and $\beta$ have opposite signs, let us assume that $\alpha<0<\beta$.
Then we have
$$\begin{aligned} \mathrm{PQ}^2 &= (\beta-\alpha)^2+\mathbf{C} \\ &= \alpha^2+\beta^2+\mathbf{DE} \\ &\geqq 2|\alpha\beta|+\mathbf{DE} = \mathbf{FG}. \end{aligned}$$
So we have
$$\mathrm{PQ}\geqq\mathbf{H}.$$
Hence, when
$$\alpha=\mathbf{IJ}\sqrt{\mathbf{K}} \text{ and } \beta=\mathbf{L}\sqrt{\mathbf{M}},$$
PQ takes the minimum value $\mathbf{H}$.

On a coordinate plane, there is a circle with radius 7 and center $O$. It is known that points $A, B$ are on the circle and $\overline{AB} = 8$. Then the dot product $\overrightarrow{OA} \cdot \overrightarrow{OB} =$ (14--1) (14--2).

In the Cartesian coordinate plane, two circles with one centered at $(12,0)$ and the other centered at $(0,9)$ intersect only at point $(4,6)$.
What is the distance between the points on these circles that are closest to the origin?
A) $\sqrt { 5 }$ B) $\sqrt { 10 }$ C) $\sqrt { 13 }$ D) $2 \sqrt { 5 }$ E) $2 \sqrt { 10 }$