设A,B为抛物线y^2=2px(p>0)上的两点,满足OA垂直OB(O为原点),证明直线AB经过定点

设A(X1,Y1),B(X2,Y2)则 y1^2=2px1,y2^2=2px2
∠AOB=90
(y1*y2)/(x1*x2)=-1 即y1*y2=-4P^2
由直线AB得:y-y1=(y2-y1)/(x2-x1)*(x-x1)
即y-y1=2p/(y2+y1)*(x-x1)因为 y1^2=2px1,y2^2=2px2和y1*y2=-4P^2
故:(y2+y1)*y=2p*(x-2p)
所以直线AB过定点(2p,0)