设AB的方程为x=ky+b,代入y^2=2px中,得,y^2+2pky-2pb=0
y1+y2=2pk,y1y2= -2pb
设A(x1,y1),B(x2,y2).PA⊥PB,kpa*kpb= -1
(y1-y0)/(x1-x0)*(y2-y0)/(x2-x0)= -1,(1)
代入,(1)x1=y1^2/2p,x0=y0^2/2p,x2=y0^2/2p得
(y1+y0)(y2+y0)= -4p^2
y1y2+(y1+y2)y0+y0^2+4p^2=0
-2pb+2pky0+2px0+4p^2=0
b=ky0+x0+2p
AB的方程为x=ky+b=ky+ky0+x0+2p=k(y+y0)+x0+2p
x=k(y+y0)+x0+2p
所以它一定经过（2p+X0,-Y0）