斜率为-2的椭圆x2+2y2=2的动弦中点轨迹方程是.

问题描述:

斜率为-2的椭圆x2+2y2=2的动弦中点轨迹方程是.

设直线方程为:y=-2x+m;
设直线与椭圆交点分别为A,B,设A(x1,y1) B(x2,y2
又因为x12+2y12=2       (1)
x22+2y22=2             (2)
(1)-(2)得:x12-x22=2y22-2y12
(x1+x2)(x1-x2)=-2(y1+y2)(y1-y2
k=-2=-

x1+x2
2(y1+y2

设中点为P(x,y)
所以2=
x
2y

x-4y=0