已知抛物线y²=2px(p>0)异于顶点的两动点a,b满足ab为直径的圆过原点,证明直线AB过定

Question

，并求出该定点坐标

Answer 1

y^2=2px
A(2pa^2,2pa),B(2pb^2,2pb)
k(OA)*k(OB)=-1
[(2pa)/(2pa^2)]*[(2pb)/(2pb^2)]=-1
a=-1/b
k(AB)=(2pa-2pb)/(2pa^2-2pb^2)=1/(a+b)
AB:
y-2pa=(x-2pa^2)/(a+b)
(a+b)y-2pa*(a+b)=x-2pa^2
(b-1/b)y-2pb=x
b=1,x=y=2p
a=-1,x=2p,y=-2p
AB过定点(2p,2p)或过定点(2p,-2p)