Question

A.B是抛物线Y^2=2PX(P>0)上的两点,且OA垂直OB,求证直线AB过定点。

Answer 1

设A (2pm^2，2pm) ，B(2pn^2，2pn)
OA⊥OB
则(2pm^2)(2pn^2)+(2pm)(2pn)
mn=-1
直线方程为
(2pm-2pn)x+(2pn^2-2pm^2)+4(p^2)(m^2)n-2(p^2)m(n^2)=0
整理得x-(m+n）y-2p=0
显然，此直线经过定点（2p，0）

Answer 2

设A(y1²/2p, y1) B(y2²/2p, y2)
∵OA⊥OB
∴[y1/(y1²/2p)]*[y2/(y2²/2p]=-1
即y1*y2=-4p²
k=(y1-y2)/(y1²/2p-y2²/2p)=2p/(y1+y2)
AB: y-y1=[2p/(y1+y2)](x-y1²/2p)
当y=0时 x=-2p
∴AB过定点(-2p, 0)

Answer 3

可设A(2pa² 2pa) B(2pb² 2pb)
∵OA⊥OB
∴(1/a)*(1/b)=-1
ab=-1
易知,直线AB方程为
x=(a+b)y-2pab
即x=(a+b)y+2p
显然,当y=0时, x=2p
即动直线AB恒过定点(2p,0)