AB的中点P的轨迹方程是2(Y-5)^2+X^2=2.
设过点M(0,3)的直线为y=kx+3,
将y=kx+3代入椭圆方程x^2/4+y^2/3=1得
(4k^2+3)x^2+24kx+24=0
x2,x2是方程的根,则有
x1+x2=-24k/(4k^2+3)
y1=kx1+3,y2=kx2+3
设AB的中点P的坐标为P(X,Y),则
X=(x1+x2)/2=-24k/(4k^2+3)
Y=(y1+y2)/2=k(x1+x2)/2+3=-12k^2/(2k^2+1)+4
X=4k/(2k^2+1)
Y-4=4k^2/(2k^2+1)
Y-4=kX,k=(Y-4)/X
将k=(Y-4)/X代入X=4k/(2k^2+1)
2(Y-4)^2-4(Y-4)+X^2=0
2(Y-5)^2+X^2=2
AB的中点P的轨迹方程是2(Y-5)^2+X^2=2,轨迹仍是一个椭圆.