求椭圆方程x^2/2+y^2=1中过定点P(0,2)的弦AB中点M的轨迹方程

直线AB:y=kx+2椭圆方程与y=kx+2联立得到(1+2k^2)x^2+8kx+6=0x1+x2=-8k/(1+2k^2)A(x1,y1),B(x2,y2)弦AB的中点M(x,y),x=(x1+x2)/2,y=(y1+y2)/2y1=kx1+2,y2=kx2+2x=-4k/(1+2k^2)y=k(x1+x2)/2+2=2/(1+2k^2) (1)x/y=-2k,k...