設(shè)拋物線方程為y^2=2px,則焦點(diǎn)P為（p/2,0),直線AB方程為：y=k(x-p/2),
令點(diǎn)A（x1,y1),B(x2,y2）,y1^2=2px1,y2^2=2px2,兩式相減得
（y1-y2)(y1+y2)=2p(x1-x2),又y1=k(x1-p/2),y2=k(x2-p/2),故y1-y2=k(x1-x2)
故y1+y2=2p/k,又y1+y2=k(x1+x2)-kp,故x1+x2=p+2p/k^2,
對(duì)y1+y2=2p/k兩邊平方,得y1^2+y2^2+2y1*y2=（2p/k)^2,
既2px1+2px2+2y1y2=(2p/k)^2,將x1+x2=p+2p/k^2代入得y1*y2=-p^2,
MN^2=(y1+y2)^2=(2p/k)^2,MO^2=p^2/4+y1^2,NO^2=p^2/4+y2^2
由余弦定理：cos