Question

设椭圆的方程为X平方+Y平方/4=1，过M（0,1）的直线交椭圆于AB两点

设椭圆的方程为X平方+Y平方/4=1，过M（0,1）的直线交椭圆于AB两点，O为坐标原点，OP向量=1/2(OA向量+OB向量)，当L绕点M旋转时，求动点P的轨迹方程

Answer 1

E: x^2+y^2/4 = 1 (1)
M(0,1)
OP = (1/2)(OA+OB)
L: passing through M(0,1)
y = mx +c
1= c
ie
L: y = mx +1 (2)
Sub (2) into (1)
x^2 + (mx+1)^2/4 =1
4x^2 + (mx+1)^2 = 4
(4+m^2)x^2 + 2mx -3 =0
Let P be (x,y)
then
2x = -2m/(4+m^2) (3)
from (2)
y = mx+1
x = (y-1)/m (4)
Sub (4) into (1)
(y-1)^2/m^2 + y^2/4 = 1
4(y-1)^2 + m^2y^2 = 4m^2
(4+m^2)y^2 - 8y + 4(1-m^2) =0
then
2y = 8/(4+m^2)
4+m^2 = 4/y
m = √[4(1-y)/y] (5)
Sub (5) into (3)
2x = -2m/(4+m^2)
x = -√[4(1-y)/y]/ (4/y)
x^2 = [4(1-y)/y] / [4/y]^2
= y(1-y)/4
4x^2 = y(1-y)

P的轨迹方程:
4x^2 = y(1-y)