19. (This question is worth 14 points) Given the ellipse $C : \frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > b > 0 )$ with eccentricity $\frac { \sqrt { 2 } } { 2 }$, point $P ( 0,1 )$ and point $A ( m , n ) ( m \neq 0 )$ are both on the ellipse $C$. The line $P A$ intersects the $x$-axis at point $M$. (I) Find the equation of ellipse $C$ and find the coordinates of point $M$ (expressed in terms of $m , n$); (II) Let $O$ be the origin. Point $B$ is symmetric to point $A$ with respect to the $x$-axis. The line $P B$ intersects the $x$-axis at point $N$. Question: Does there exist a point $Q$ on the $y$-axis such that $\angle O Q M = \angle O N Q$? If it exists, find the coordinates of point $Q$; if it does not exist, explain the reason.

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

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]

Given curve $C : x ^ { 2 } + y ^ { 2 } = 16 ( y > 0 )$, from any point $P$ on $C$, draw a perpendicular segment $P P ^ { \prime }$ to the $x$-axis, where $P ^ { \prime }$ is the foot of the perpendicular. The locus of the midpoint of segment $P P ^ { \prime }$ is
A. $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 1 \quad ( y > 0 )$
B. $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 8 } = 1 \quad ( y > 0 )$
C. $\frac { y ^ { 2 } } { 16 } + \frac { x ^ { 2 } } { 4 } = 1 \quad ( y > 0 )$
D. $\frac { y ^ { 2 } } { 16 } + \frac { x ^ { 2 } } { 8 } = 1 \quad ( y > 0 )$

Tangents are drawn to a given circle from a point on a given straight line, which does not meet the given circle. Prove that the locus of the mid-point of the chord joining the two points of contact of the tangents with the circle is a circle.

A triangle $ABC$ has a fixed base $BC$. If $AB : AC = 1 : 2$, then the locus of the vertex $A$ is
(A) a circle whose centre is the midpoint of $BC$
(B) a circle whose centre is on the line $BC$ but not the midpoint of $BC$
(C) a straight line
(D) none of the above.

Let $P$ be a variable point on a circle $C$ and $Q$ be a fixed point outside $C$. If $R$ is the mid-point of the line segment $PQ$, then the locus of $R$ is
(A) a circle
(B) an ellipse
(C) a line segment
(D) segment of a parabola

Let $P$ be a variable point on a circle $C$ and $Q$ be a fixed point outside $C$. If $R$ is the mid-point of the line segment $P Q$, then the locus of $R$ is
(a) a circle
(b) an ellipse
(c) a line segment
(d) segment of a parabola.

Let $P$ be a variable point on a circle $C$ and $Q$ be a fixed point outside $C$. If $R$ is the mid-point of the line segment $P Q$, then the locus of $R$ is
(a) a circle
(b) an ellipse
(c) a line segment
(d) segment of a parabola.

A triangle $ABC$ has a fixed base $BC$. If $AB : AC = 1 : 2$, then the locus of the vertex $A$ is
(A) a circle whose centre is the midpoint of $BC$
(B) a circle whose centre is on the line $BC$ but not the midpoint of $BC$
(C) a straight line
(D) none of the above

Let $P$ be a variable point on a circle $C$ and $Q$ be a fixed point outside $C$. If $R$ is the mid-point of the line segment $PQ$, then the locus of $R$ is
(A) a circle
(B) an ellipse
(C) a line segment
(D) segment of a parabola

A triangle $A B C$ has a fixed base $B C$. If $A B : A C = 1 : 2$, then the locus of the vertex $A$ is
(A) a circle whose centre is the midpoint of $B C$
(B) a circle whose centre is on the line $B C$ but not the midpoint of $B C$
(C) a straight line
(D) none of the above

Let $P$ be a variable point on a circle $C$ and $Q$ be a fixed point outside $C$. If $R$ is the mid-point of the line segment $P Q$, then the locus of $R$ is
(A) a circle
(B) an ellipse
(C) a line segment
(D) segment of a parabola

Let $C$ be a circle of area $A$ with centre at $O$. Let $P$ be a moving point such that its distance from $O$ is always twice the length of a tangent drawn from $P$ to the circle. Then the point $P$ must move along
(A) the sides of a square centred at $O$, with area $\frac{4}{3}A$.
(B) the sides of an equilateral triangle centred at $O$, with area $4A$.
(C) a circle centred at $O$, with area $\frac{4}{3}A$.
(D) a circle centred at $O$, with area $4A$.

A triangle $A B C$ has a fixed base $B C$. If $A B : A C = 1 : 2$, then the locus of the vertex $A$ is
(a) a circle whose centre is the midpoint of $B C$.
(b) a circle whose centre is on the line $B C$ but not the midpoint of $B C$.
(c) a straight line.
(d) none of the above.

Tangents are drawn from the point $P ( 3,4 )$ to the ellipse $\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 4 } = 1$ touching the ellipse at points A and B.
The equation of the locus of the point whose distances from the point $P$ and the line AB are equal, is
A) $9 x ^ { 2 } + y ^ { 2 } - 6 x y - 54 x - 62 y + 241 = 0$
B) $x ^ { 2 } + 9 y ^ { 2 } + 6 x y - 54 x + 62 y - 241 = 0$
C) $9 x ^ { 2 } + 9 y ^ { 2 } - 6 x y - 54 x - 62 y - 241 = 0$
D) $x ^ { 2 } + y ^ { 2 } - 2 x y + 27 x + 31 y - 120 = 0$

Let $a, r, s, t$ be nonzero real numbers. Let $P(at^2, 2at)$, $Q$, $R(ar^2, 2ar)$ and $S(as^2, 2as)$ be distinct points on the parabola $y^2 = 4ax$. Suppose that $PQ$ is the focal chord and lines $QR$ and $PK$ are parallel, where $K$ is the point $(2a, 0)$.
The value of $r$ is
(A) $-\frac{1}{t}$
(B) $\frac{t^2+1}{t}$
(C) $\frac{1}{t}$
(D) $\frac{t^2-1}{t}$

Let $RS$ be the diameter of the circle $x^2 + y^2 = 1$, where $S$ is the point $(1,0)$. Let $P$ be a variable point (other than $R$ and $S$) on the circle and tangents to the circle at $S$ and $P$ meet at the point $Q$. The normal to the circle at $P$ intersects a line drawn through $Q$ parallel to $RS$ at point $E$. Then the locus of $E$ passes through the point(s)
(A) $\left(\frac{1}{3}, \frac{1}{\sqrt{3}}\right)$
(B) $\left(\frac{1}{4}, \frac{1}{2}\right)$
(C) $\left(\frac{1}{3}, -\frac{1}{\sqrt{3}}\right)$
(D) $\left(\frac{1}{4}, -\frac{1}{2}\right)$

Let $T$ be the line passing through the points $P ( - 2,7 )$ and $Q ( 2 , - 5 )$. Let $F _ { 1 }$ be the set of all pairs of circles ( $S _ { 1 } , S _ { 2 }$ ) such that $T$ is tangent to $S _ { 1 }$ at $P$ and tangent to $S _ { 2 }$ at $Q$, and also such that $S _ { 1 }$ and $S _ { 2 }$ touch each other at a point, say, $M$. Let $E _ { 1 }$ be the set representing the locus of $M$ as the pair ( $S _ { 1 } , S _ { 2 }$ ) varies in $F _ { 1 }$. Let the set of all straight line segments joining a pair of distinct points of $E _ { 1 }$ and passing through the point $R ( 1,1 )$ be $F _ { 2 }$. Let $E _ { 2 }$ be the set of the mid-points of the line segments in the set $F _ { 2 }$. Then, which of the following statement(s) is (are) TRUE?
(A) The point $( - 2,7 )$ lies in $E _ { 1 }$
(B) The point $\left( \frac { 4 } { 5 } , \frac { 7 } { 5 } \right)$ does NOT lie in $E _ { 2 }$
(C) The point $\left( \frac { 1 } { 2 } , 1 \right)$ lies in $E _ { 2 }$
(D) The point $\left( 0 , \frac { 3 } { 2 } \right)$ does NOT lie in $E _ { 1 }$

Let $S$ be the circle in the $x y$-plane defined by the equation $x ^ { 2 } + y ^ { 2 } = 4$. Let $E _ { 1 } E _ { 2 }$ and $F _ { 1 } F _ { 2 }$ be the chords of $S$ passing through the point $P _ { 0 } ( 1,1 )$ and parallel to the $x$-axis and the $y$-axis, respectively. Let $G _ { 1 } G _ { 2 }$ be the chord of $S$ passing through $P _ { 0 }$ and having slope $- 1$. Let the tangents to $S$ at $E _ { 1 }$ and $E _ { 2 }$ meet at $E _ { 3 }$, the tangents to $S$ at $F _ { 1 }$ and $F _ { 2 }$ meet at $F _ { 3 }$, and the tangents to $S$ at $G _ { 1 }$ and $G _ { 2 }$ meet at $G _ { 3 }$. Then, the points $E _ { 3 } , F _ { 3 }$, and $G _ { 3 }$ lie on the curve
(A) $x + y = 4$
(B) $( x - 4 ) ^ { 2 } + ( y - 4 ) ^ { 2 } = 16$
(C) $( x - 4 ) ( y - 4 ) = 4$
(D) $x y = 4$

Let $S$ be the circle in the $x y$-plane defined by the equation $x ^ { 2 } + y ^ { 2 } = 4$. Let $P$ be a point on the circle $S$ with both coordinates being positive. Let the tangent to $S$ at $P$ intersect the coordinate axes at the points $M$ and $N$. Then, the mid-point of the line segment $M N$ must lie on the curve
(A) $( x + y ) ^ { 2 } = 3 x y$
(B) $x ^ { 2 / 3 } + y ^ { 2 / 3 } = 2 ^ { 4 / 3 }$
(C) $x ^ { 2 } + y ^ { 2 } = 2 x y$
(D) $x ^ { 2 } + y ^ { 2 } = x ^ { 2 } y ^ { 2 }$

The locus of the foot of perpendicular drawn from the centre of the ellipse $x ^ { 2 } + 3 y ^ { 2 } = 6$ on any tangent to it is
(1) $\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } = 6 x ^ { 2 } + 2 y ^ { 2 }$
(2) $\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } = 6 x ^ { 2 } - 2 y ^ { 2 }$
(3) $\left( x ^ { 2 } - y ^ { 2 } \right) ^ { 2 } = 6 x ^ { 2 } + 2 y ^ { 2 }$
(4) $\left( x ^ { 2 } - y ^ { 2 } \right) ^ { 2 } = 6 x ^ { 2 } - 2 y ^ { 2 }$

Let $a$ and $b$ be any two numbers satisfying $\frac { 1 } { a ^ { 2 } } + \frac { 1 } { b ^ { 2 } } = \frac { 1 } { 4 }$. Then, the foot of perpendicular from the origin on the variable line $\frac { x } { a } + \frac { y } { b } = 1$ lies on:
(1) A circle of radius $= 2$
(2) A hyperbola with each semi-axis $= \sqrt { 2 }$.
(3) A hyperbola with each semi-axis $= 2$
(4) A circle of radius $= \sqrt { 2 }$

Locus of the image of the point $(2, 3)$ in the line $(2x - 3y + 4) + k(x - 2y + 3) = 0$, $k \in \mathbb{R}$, is a:
(1) straight line parallel to $x$-axis
(2) straight line parallel to $y$-axis
(3) circle of radius $\sqrt{2}$
(4) circle of radius $\sqrt{3}$

Let $O$ be the vertex and $Q$ be any point on the parabola, $x ^ { 2 } = 8 y$. If the point $P$ divides the line segment $OQ$ internally in the ratio $1 : 3$, then the locus of $P$ is
(1) $x ^ { 2 } = 2 y$
(2) $x ^ { 2 } = y$
(3) $y ^ { 2 } = x$
(4) $y ^ { 2 } = 2 x$