A,B是抛物线y^2=2px(p>0)上的两点,满足OA垂直OB,求证直线AB恒过一定点

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=(y1-y2)/(x1-x2)*(x-x1)
因为 y1^2=2px1,y2^2=2px2两式相减
y1^2-y^2=2p(x1-x2)
(y1+y2)(y1-y2)=2p(x1-x2)
(y1-y2)/(x1-x2)=2p/(y1+y2)
故y-y1=2p/(y1+y2)*(x-x1)

又y1*y2=-4P^2,y1^2=2px1,y2^2=2px2
(y-y1)(y1+y2)=2p*(x-x1)
yy1+yy2-y1^2-y1y2=2px-2px1
yy1+yy2-2px1+4p^2=2px-2px1
yy1+yy2=2px-4p^2
故(y2+y1)*y=2p*(x-2p)
x=2p时，y恒为0
所以直线AB过定点(2p,0)

设AB直线为：y=kx+a
代入抛物线方程消去y
k²x²+2akx+a²-2px=0
则(x1+x2)/2=(p-ak)/k²,x1*x2=a²/k²
则(y1+y2)/2=p/k
AB²=(y1-y2)²+(x1-x2)²=(x1-x2)²(k²+1)
因(x1-x2)²=(x1+x2)²-4x1*x2
AB²={[2(p-ak)/k²]²-4a²/k²}(k²+1)
=4(p²-2akp)(k²+1)/k^4
因为OA垂直OB，则AB中点到O的距离=AB/2
则(p/k)²+[(p-ak)/k²]²=(p²-2akp)(k²+1)/k^4
=>a=-2kp
则直线方程为y=kx-2kp