设P(X0，Y0)是抛物线y^2=2px(p>0)上一定点，A,B是抛物线上两点，且PA垂直PB，求证：AB过点（2p+X0,-Y0）

Question

如题所述

Answer 1

设P(x0,y0)是抛物线y2=2px上一定点.A,B是抛物线上两点,且满足PA⊥PB,则AB过点(2p+x0,-y0).
证明设A(x1,y1),B(x2,y2)是抛物线上两点,
因为PA⊥PB,所以(x0-x1)(x0-x2)+(y0-y1)(y0-y2)=0,
即(y0^2-y1^2)(y0^2-y2^2)+4p(y0-y1)(y0-y2)=0
因为y0,y1,y2互不相等,所以(y0+y1)(y0+y2)=-4p,
y1y2=-4p-(y1+y2)y0-2px0.
(1)另一方面当x1≠x2,KAB=xy11--yx22=y12+py2,则AB所在直线方程为:y-y1=y12+py2(x-x1),化简并整理得:y=y12+pxy2+y1y1+y2y2,(2)(1)代(2)得:y=y11+y2(2px-4p-2px0)-y0,
此时直线AB恒过定点(2p+x0,-y0),

Answer 2

设AB的方程为x=ky+b，代入y^2=2px中，得，y^2+2pky-2pb=0
y1+y2=2pk,y1y2= -2pb
设A(x1,y1),B(x2,y2).PA⊥PB,kpa*kpb= -1
(y1-y0)/(x1-x0)*(y2-y0)/(x2-x0)= -1,(1)
代入,(1)x1=y1^2/2p,x0=y0^2/2p,x2=y0^2/2p得
(y1+y0)(y2+y0)= -4p^2
y1y2+(y1+y2)y0+y0^2+4p^2=0
-2pb+2pky0+2px0+4p^2=0
b=ky0+x0+2p
AB的方程为x=ky+b=ky+ky0+x0+2p=k(y+y0)+x0+2p
x=k(y+y0)+x0+2p
所以它一定经过（2p+X0,-Y0）本回答被网友采纳