设P坐标为(x1,y1)
则:x1^2=2py1
过P点的切线方程为:x1x=p(y+y1)
F坐标:(0,p/2),FQ斜率=-p/x1
FQ方程:y=-px/x1+p/2
解方程组:
x1x=p(y+y1)
y=-px/x1+p/2
得:
x1=2px/(p-2y)
y1=(2x^2-yp+2y^2)/(p-2y)
代人x1^2=2py1得:
[2px/(p-2y)]^2=2p(2x^2-yp+2y^2)/(p-2y)
4x^2/(p-2y)=2y-p
2x=±(2y-p)
x+y-p/2=0,或,x-y+p/2=0
这就是点G的轨迹方程