It is known that P is any point on the circle x square plus y square = 4, passing through the point P to make the x-axis PQ. (1) find the trajectory equation of the midpoint m of the line PQ Problem (2) taking point a (- 2,0) as the right angle vertex, we can find the area of the isosceles right triangle ABC which is connected to the locus of point M

Let the midpoint coordinate be (x1, Y1), then x = x1, y = 2y1. Take it into the circle equation to get 4Y1 ^ 2 + X1 ^ 2 = 4, that is, 4Y ^ 2 + x ^ 2 = 4

Let p be the moving point on the square of the circle (x + 1) + the square of y = 25. Let Q be the projection of P on the x-axis, and m be the midpoint of the line PQ. When P moves on the circle, the trajectory of M is obtained

Let the coordinates of point m be (x, y)
Then the coordinates of P are (x, 2Y)
P follows the equation of a circle on a circle
(x+1)²+(2y)²=25
(x+1)²+4y²=25
(x+1)²/25+y²/(25/4)=1
So the trajectory of M is centered on (- 1,0)
Ellipse with 5 / 2 as major axis and 1 / 5 as minor axis

Take any point P in the circle x2 + y2 = 4, pass through the point P as the vertical line segment PD of X axis, and D is the vertical foot. When the point P moves on the circle, the locus of the midpoint m of the segment PD is ()
A. Ellipse B. hyperbola C. parabola D. circle

Let m (x, y), from the topic D (x, 0), P (x, Y1) ∵ m be the midpoint of the line PD, ∵ Y1 + 0 = 2Y, Y1 = 2Y. And ∵ P (x, Y1) on the circle x2 + y2 = 4, ∵ x2 + Y12 = 4, ∵ x2 + 4y2 = 4, that is, x24 + y2 = 1

1. Point P is a moving point on the circle x ^ 2 + y ^ 2 = 16, PQ is perpendicular to the x-axis, and the perpendicular foot is Q. find the trajectory equation of the midpoint m of PQ in the vertical section
2. Find the locus of the point whose distance from the fixed point F (2,2) is equal to the distance from the fixed line x + y = 1

1.x^2+4y^2=16
2.x^2+y^2-6x-6y-2xy+15=0

Given the fixed point P (1,0), the moving point q is on the circle C: (x + 1) ^ 2 + y ^ 2 = 16, and the perpendicular of PQ intersects the point m, then the trajectory equation of the moving point m is——

Should the vertical bisector of PQ intersect CQ at point m?
Since the vertical bisector of PQ intersects CQ at m, so | MP | = | MQ |
So | MC | + | MP | = | MC | + | MQ | = | MQ | = 4 > | CP |
It can be seen from the definition of ellipse that the trajectory of point m is based on C (- 1,0)
An ellipse with point P (1,0) as the focus and 4 as the major axis,
The equation is: x ^ 2 / 4 + y ^ 2 / 3 = 1

Through the circle, through any point P on the X + y square of the circle, make a vertical line of the X axis, the perpendicular foot is Q, and find the ordinary square of the locus of the midpoint of the line PQ

The length of the major axis is the diameter of the original circle, and the minor axis is the ellipse with the radius of the original circle. If the radius of the original circle is r, the trajectory equation is x + 4 times y = R. is it correct?

Given that P1 (2, - 1), P2 (- 1,3), P is on the straight line p1p2, and the vector | p1p | = 2 / 3 | PP2 |

Because p1p and PP2 are collinear,
So (1) if p1p = 2 / 3 * PP2,
Then op-op1 = 2 / 3 * (op2-op),
The solution is op = 3 / 5 * OP1 + 2 / 5 * op2 = (6 / 5, - 3 / 5) + (- 2 / 5,6 / 5) = (4 / 5,3 / 5);
(2) If p1p = - 2 / 3 * PP2,
Then op-op1 = - 2 / 3 * (op2-op),
The solution is op = 3op1-2op2 = (6, - 3) + (2, - 6) = (8, - 9);
So the P coordinates are (4 / 5,3 / 5) or (8, - 9)

1. Given that P1 (6, - 3), P2 (- 3,8), P is on the extension line of line p1p2, and the vector | p1p | = 2 | PP2 |

Because | p1p | = 2 | PP2 |, we know that P is in the middle of P1, P2 or on the side of P2 on the extension line of p1p2
① Between two points: point P is the trisection point of p1p2, the abscissa of point P is 1 / 3 (6 - (- 3)) + (- 3) = 0, and the ordinate is 1 / 3 ((- 3) - 8) + 8 = 13 / 3
② On the P2 side: P2 is the midpoint of PP1, so the abscissa of P is 2 * (- 3) - 6 = - 12, and the ordinate is 2 * 8 - (- 3) = 19

Given that P is on p1p2 and vector p1p = vector λ PP2, how to find the coordinates of point P to op = OP1 + p1p = OP1 + input PP2 = OP1 + input (op2-op)?

OP = OP1 + p1p = OP1 + input PP2 = OP1 + input (op2-op)
That is op = OP1 + in (op2-op) = OP1 + in op2 in Op
(1 + in) op = OP1 + in op2
OP = OP1 / (1 + in) + op2 / (1 + in)

If we know P1 (2, - 1), P2 (0, 5) and point P is on the extension line of p1p2, | p1p | = 2 | PP2 |, then the coordinates of P ()
A. （2，-7）B. （43，3）C. （23，3）D. （-2，11）

Let P (x, y) be the midpoint of PP1, then ∵ P1 (2, - 1), P2 (0, 5), ∵ 0 = x + 2, 10 = Y-1 ∵ x = - 2, y = 11 ∵ P (- 2, 11)