Question

过点(-2,0)的直线l交双曲线X^2-Y^2=1于A,B两点,求线段AB的中点M的轨迹方程

Answer 1

设A(x1,y1),B(x2,y2),M(x,y),直线的斜率为K.则有x1^2-y1^2=1,x2^2-y2^2=1.两式相减有：(y1-y2)/(x1-x2)=(x1+x2)/(y1+y2).又点M是线段AB的中点,故x1+x2=2x,y1+y2=2y,所以K=2x/(2y)=x/y.又点M在直线上,所以y-0=(x/y)(x+2),即(x+1)^2-y^2=1.当直线的斜率不存在时,点M(-2,0)在曲线(x+1)^2-y^2=1上,所以点M的轨迹方程为(x+1)^2-y^2=1.