Question

设过抛物线y2=4x的焦点F的直线交抛物线于A、B两点,且AB中点为M,则点M的轨迹方程是

Answer 1

F(1,0)

M(x,y),yA+yB=2y

k(AB)=(yA-yB)=y/(x-1)

(yA)^2-(yB)^2=4(xA-xB)

(yA+yB)*(yA-yB)/(xA-xB)=4

2y*[y/(x-1)]=4

y^2=2(x-1)