Question

已知y^2=2px(p>0)上两点AB,OA=OB,且AOB垂心是抛物线焦点,求AB方程

Answer 1

设抛物线焦点F(p/2,0)

设A(y1^2/2p,y1),B(y2^2/2p,y2)

AOB垂心为F,则OF垂直AB且AF垂直OB

显然OF斜率为0,则AB垂直于x轴,y1^2/2p=y2^2/2p,则y2=-y1

则B(y1^2/2p,-y1),AB所在直线为x=y1^2/2p

则向量AF=(p/2-y1^2/2p,-y1),向量OB=(y1^2/2p,-y1)

AF垂直OB,则(p/2-y1^2/2p)*y1^2/2p+y1^2=0 解得y1^2=5p^2

得AB所在直线为x=y1^2/2p=(5/2)*p