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

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)

设I:y=kx-1
交点满足
y=kx-1
y^2=4x
(kx-1)^2=4x
k^2*x^2-(2k+4)x+1=0
所以x1+x2=(2k+4)/k^2
y1+y2=(kx1-1)+(kx2-1)=k(x1+x2)-1
=(k+4)/k
所以p点坐标为((k+2)/k^2,(k+4)/2k)
p的轨迹方程为4y^2-1=8x