Question

斜率为2的直线与等轴双曲线x^2-y^2=12相交于两点A、B,求AB中点的轨迹方程.

Answer 1

设A,B坐标分别是:(x1,y1),(x2,y2),AB中点P的坐标是(x,y)

所以有:x1+x2=2x,y1+y2=2y

又:x1^2-y1^2=12,x2^2-y2^2=12

二式相减得:(x1+x2)(x1-x2)-(y1+y2)(y1-y2)=0

所以,AB的斜率是:K=(y1-y2)/(x1-x2)=(x1+x2)/(y1+y2)=x/y

又:K=2

所以,x/y=2

即轨迹方程是:x=2y.