设AB方程为：y=k(x-2)+1 则：x^2/9+[k(x-2)+1]^2/4=1 (4+9k^2)x^2+18k(1-2k)x+9(1-2k)^2-36=0 x1+x2=18k(2k-1)/(4+9k^2) y1+y2=k(x1+x2)+2(1-2k) =18k^2(2k-1)/(4+9k^2)+2(1-2k) =-8(2k-1)/(4+9k^2) 设AB中点M坐标为(x,y) 则：x=(x1+x2)/2=4k(2k-1)/(4+9k^2) y=(y1+y2)/2=-4(2k-1)/(4+9k^2) 所以,x=-ky 把：k=-x/y代入：y=k(x-2)+1得：y=-x(x-2)/y+1 y^2-y+x(x-2)=0 这就是线段AB中点M的轨迹方程