过P(5.4)作直线与椭圆x2/5+y2/4=1交于AB两点,则AB中点M的轨迹方程?
问题描述:
过P(5.4)作直线与椭圆x2/5+y2/4=1交于AB两点,则AB中点M的轨迹方程?
答
设A(x1,y1),B(x2,y2),M(x0,y0)PAB直线方程为:y-4=k(x-5)因AB在椭圆上:4x1²+5y1²=204x2²+5y2²=20两式相减:4(x1+x2)(x1-x2)=-5(y1+y2)(y1-y2)4(x1+x2)/(y1+y2)=-5(y1-y2)/(x1-x2)=-5K4x0/y0=-5...