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,O为坐标原点,OP向量=1/2(OA向量+O

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)