求证:AB=2p/sin²a焦点F坐标(0.5p,0),设直线L过F,则直线L方程为y=k(x-0.5p)联立y²=2px得k²x²-(pk²+2p)x+p²k²/4=0由韦达定理得x1+x2=p+2p/k²AB=x1+0.5p+x2+0.5p=x1+x2+p=2p+2...