Question

过a（0 -1）做直线l叫抛物线y^2=4x与bc两点 求bc中点p的轨迹方程

Answer 1

直线L过(0,-1),交抛物线y^2=4x于B、C两点,设BC中点P(x,y),则

kBC=(yB-yC)/(xB-xC)=(y+1)/x

yB+yC=2y

(yB)^2=4(xB).(1)

(yC)^2=4(xC).(2)

(1)-(2):

(yB+yC)*(yB-yC)=4(xB-xC)

(yB+yC)*(yB-yC)/(xB-xC)=4

2y*(y+1)/x=4

y^2+y=2x

BC中点P的轨迹方程是抛物线：

(y+1/2)^2=2(x+1/8)