记A点坐标(XA,YA),B点坐标(XB,YB)
设直线I,KY=X-2P -------(1)
则X=KY+2P ----------(2)
(2)代入抛物线方程,有Y^2-2PKY-4P^2=0 ===>YAYB=-4P^2
(1)==>K^2*Y^2=X^2-2PX+4P^2,将抛物线方程代入有X^2-2P(K^2+1)X+4P^2
====> XAXB=4P^2=-YAYB
OA^2=XA^2+YA^2
OB^2=YA^2+YB^2
AB^2=(XB-XA)^2+(YB-YA)^2=XA^2+XB^2-2XAXB+YA^2+YB^2-2YAYB=OA^2+OB^2
所以角AOB为直角