Question

已知两点A、B分别在双曲线X^2-3Y^2=1的两条渐近线上,且三角形AOB的面积为8,求线段AB中点M的轨迹方程.

Answer 1

设A(x1根号3,x1)

B(x2根号3,-x2)

x1x2根号3=8,3x1x2=8根号3

A B中点[(x1+x2)(根号3)/2,(x1-x2)/2]

[(x1+x2)(根号3)/2]^2=3(x1^2+x2^2+2x1x2)/4

[(x1-x2)/2]^2=(x1^2+x2^2-2x1x2)/4

所以[(x1+x2)(根号3)/2]^2-3[(x1-x2)/2]^2=(6x1x2+6x1x2)/4=3x1x2=8根号3

所以轨迹为x^2-3y^2=8根号3