The question asks to determine or prove the locus of a point defined by a geometric condition involving a circle (e.g., midpoint of a chord, midpoint of a segment, ratio of distances).
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 $P _ { 1 } , P _ { 2 }$ .
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 $P _ { 1 } ( - 6 , - 4 ) , P _ { 2 } ( 8,3 )$ .
3. A moving point $P$ has equal distance to the point $F ( 2,0 )$ and to the line $x + 2 = 0$. Then the locus equation of $P$ is $y ^ { 2 } = 8 x$. Analysis: This examines the definition and standard equation of a parabola. By the definition, the locus of $P$ is a parabola with focus $F ( 2,0 )$. Since $p = 2$, its equation is $y ^ { 2 } = 8 x$.
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.
6. Consider the family of circles $x 2 + y 2 = r 2,2 < r < 5$. If in the first quadrant, the common tangent to a circle of this family and the ellipse $4 \times 2 + 25 y 2 = 100$ meets the coordinate axes at A and B , then find the equation of the locus of the mid point of AB .
9. Let Cl and C 2 be two circles with C 2 lying inside Cl . A circle C lying inside C1touches C1 internally and C2 externally. Identify the locus of the centre of C.
12. Let $A B$ be a chord of the circle $x 2 + y 2 = r 2$ subtending a right angle at the centre. Then the locus of the centroid of the triangle PAB as P moves on the circle is: (A) A parabola (B) A circle (C) An ellipse (D) A pair of straight lines
13. At any point $P$ on the parabola $y ^ { 2 } - 2 y - 4 x + 5 = 0$, a tangent is drawn which meets the directrix at $Q$. Find the locus of point R which divides QP externally in the ratio $\frac { 1 } { 2 } : 1$. Sol. Any point on the parabola is $\mathrm { P } \left( 1 + \mathrm { t } ^ { 2 } , 1 + 2 \mathrm { t } \right)$. The equation of the tangent at P is $\mathrm { t } ( \mathrm { y } - 1 ) = \mathrm { x } - 1 + \mathrm { t } ^ { 2 }$ which meets the directrix $\mathrm { x } = 0$ at $\mathrm { Q } \left( 0,1 + \mathrm { t } - \frac { 1 } { \mathrm { t } } \right)$. Let R be $( \mathrm { h } , \mathrm { k } )$. Since it divides QP externally in the ratio $\frac { 1 } { 2 } : 1 , \mathrm { Q }$ is the mid point of RP $\Rightarrow 0 = \frac { \mathrm { h } + 1 + \mathrm { t } ^ { 2 } } { 2 }$ or $\mathrm { t } ^ { 2 } = - ( \mathrm { h } + 1 )$ and $1 + \mathrm { t } - \frac { 1 } { \mathrm { t } } = \frac { \mathrm { k } + 1 + 2 \mathrm { t } } { 2 }$ or $\mathrm { t } = \frac { 2 } { 1 - \mathrm { k } }$ So that $\frac { 4 } { ( 1 - k ) ^ { 2 } } + ( h + 1 ) = 0$ Or $( k - 1 ) ^ { 2 } ( h + 1 ) + 4 = 0$. Hence locus is $( y - 1 ) ^ { 2 } ( x + 1 ) + 4 = 0$.
13. At any point $P$ on the parabola $y ^ { 2 } - 2 y - 4 x + 5 = 0$, a tangent is drawn which meets the directrix at $Q$. Find the locus of point R which divides QP externally in the ratio $\frac { 1 } { 2 } : 1$. Sol. Any point on the parabola is $\mathrm { P } \left( 1 + \mathrm { t } ^ { 2 } , 1 + 2 \mathrm { t } \right)$. The equation of the tangent at P is $\mathrm { t } ( \mathrm { y } - 1 ) = \mathrm { x } - 1 + \mathrm { t } ^ { 2 }$ which meets the directrix $\mathrm { x } = 0$ at $\mathrm { Q } \left( 0,1 + \mathrm { t } - \frac { 1 } { \mathrm { t } } \right)$. Let R be $( \mathrm { h } , \mathrm { k } )$. Since it divides QP externally in the ratio $\frac { 1 } { 2 } : 1 , \mathrm { Q }$ is the mid point of RP $\Rightarrow 0 = \frac { \mathrm { h } + 1 + \mathrm { t } ^ { 2 } } { 2 }$ or $\mathrm { t } ^ { 2 } = - ( \mathrm { h } + 1 )$ and $1 + \mathrm { t } - \frac { 1 } { \mathrm { t } } = \frac { \mathrm { k } + 1 + 2 \mathrm { t } } { 2 }$ or $\mathrm { t } = \frac { 2 } { 1 - \mathrm { k } }$ So that $\frac { 4 } { ( 1 - k ) ^ { 2 } } + ( h + 1 ) = 0$ Or $( k - 1 ) ^ { 2 } ( h + 1 ) + 4 = 0$. Hence locus is $( y - 1 ) ^ { 2 } ( x + 1 ) + 4 = 0$.
10. Tangents are drawn from any point on the hyperbola $x ^ { 2 } / 9 - y ^ { 2 } / 4 = 1$ to the circle $x ^ { 2 } + y ^ { 2 } = 9$. Find the locus of mid-point of the chord of contact.
28. A circle touches the line L and the circle $\mathrm { C } _ { 1 }$ externally such that both the circles are on the same side of the line, then the locus of centre of the circle is (A) ellipse (B) hyperbola (C) parabola (D) parts of straight line Sol. (C) Let C be the centre of the required circle. Now draw a line parallel to L at a distance of $\mathrm { r } _ { 1 }$ (radius of $\mathrm { C } _ { 1 }$ ) from it. Now $\mathrm { CP } _ { 1 } = \mathrm { AC } \Rightarrow \mathrm { C }$ lies on a parabola. [Figure]