Question

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

Answer 1

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)