Question

已知A、B是抛物线y^2=4x上两点,0是抛物线的顶点,且OA垂直OB,求证：直线AB过定点

Answer 1

设A(x1,y1),B(x2,y2)
则4x1=y1²,4x2=y2²轿信
向量OA=(x1,y1),向量OB=(x2,y2)
OA*OB=x1x2+y1y2=0
16x1x2=y1²y2²
x1x2=(y1²y2²)/16
代入x1x2+y1y2=0
(y1²y2²)/16+y1y2=0
y1y2≠0
所以,有：y1y2/16+1=0
得：y1y2=-16
则x1x2=16
A(y1²/4,y1),B(y2²/4,y2)
由两点式写出方程：(y-y1)/(y1-y2)=(x-y1²/4)/(y1²/4-y2²/4)
(y-y1)/(y1-y2)=(4x-y1²)/(y1+y2)(y1-y2)
y-y1=(4x-y1²)/(y1+y2)
(y-y1)(y1+y2)=4x-y1²
(y1+y2)y-y1²闭带轮-y1y2=4x-y1²
-4x+(y1+y2)y-y1y2=0
把y1y2=-16代入,得AB的直线方程为：-4x+(y1+y2)y+16=0
要过定点,则与y1+y2无行贺关,所以：y=0,-4x+16=0
得：x=4,y=0
所以,直线AB过定点(4,0)

Answer 2

简单孙搜计算一下，镇凯租详情如图御兆所示