设过左焦点的直线方程为x=my-1带入椭圆方程得(4m²+5)y²-8my-16=0 y1+y2=8m/(4m²+5)
x1+x2=-10/(4m²+5),即AB中点坐标为x=4m/(4m²+5),y=-5/(4m²+5),)
两式消掉M得 5y²+4x²-4x=0就是所求轨迹方程.
设过左焦点的直线方程为x=my-1带入椭圆方程得(4m²+5)y²-8my-16=0 y1+y2=8m/(4m²+5)
x1+x2=-10/(4m²+5),即AB中点坐标为x=4m/(4m²+5),y=-5/(4m²+5),)
两式消掉M得 5y²+4x²-4x=0就是所求轨迹方程.