由题意可知过焦点的直线方程为y=x? p 2 ,联立有 y2=2px y=x? p 2 ?x2?3px+ p2 4 =0,∴x1+x2=3p,x1x2= p2 4 ∴|x1-x2|= (x1+x2)2?4x1x2 = (3p)2?4× p2 4 又|AB|= (1+12) (3p)2?4× p2