直线m:x=ky+p,(1)
代入y^2=2px,得
y^2-2kpy-2p^2=0,
设A(x1,y1),B(x2,y2),则
y1+y2=2kp,y1y2=-2p^2,
由(1)(x1+p)(x2+p)=(ky1+2p)(ky2+2p)
=k^2*y1y2+2kp(y1+y2)+4p^2
=-2k^2*p^2+4k^2*p^2+4p^2
=(2k^2+4)p^2,
NA*NB=(x1+p,y1)*(x2+p,y2)
=(x1+p)(x2+p)+y1y2
=(2k^2+2)p^2,
当k=0时它取最小值2p^2.