1)y^2=2pxF,A,B所在直线y=kx+b 解得y=kx-pk/2联立方程整理得(k^2)x^2-(k^2*p+2p)x+(p^2*k^2)/4=0韦达定理x1x2=c/a=(p^2)/4x1+x2=-b/a=p(2+1/k^2)2)|AF|=x1+p/2,|BF|=x2+p/2,代入,原式=2/p(空间不够过程略)...