Question

过抛物线y=ax^2(a＞0)的顶点做两条垂直的弦OA,OB,

Answer 1

1）因为OA⊥OB,且A,B在曲线上,则可设A(k/a,(1/a)k^2),B(-1/(ak),1/(ak^2)),AB中点P(x,y)

则有KAB=[(1/a)k^2-1/(ak^2)]/[k/a+1/(ak)]=k-1/k,且过A点

于是AB方程可写为y=(k-1/k)(x-k/a)+k^2/a,整理得y=kx-x/k+1/a,即ky=k^2x-x+k/a

亦即k(y-1/a)=x(k^2-1),显然过定点(0,1/a),命题得证.

2）易得中点坐标

x=[k/a-1/(ak)]/2=[k-1/k]/(2a)

y=[(1/a)k^2+(1/a)(1/k^2)]/2=(k^2+1/k^2)/(2a)

有x^2=(k^2+1/k^2)/(4a^2)-1/(2a^2)

于是消去K得AB中点P的轨迹方程

x^2=y/(2a)-1/(2a^2),a>0