设中点M(x,y),A(x1,y1),B(x2,y2)
x1+x2=2x
y1+y2=2y
直线AB斜率k=y/(x-1)=(y2-y1)/(x2-x1) （x≠1）
把A、B代入椭圆,得
x1²/9+y1²/4=1.①
x2²/9+y2²/4=1.②
①-②得
(x1²-x2²)/9+(y1²-y2²)/4=0
(x1+x2)(x1-x2)/9+(y1+y2)(y1-y2)/4=0
(x1+x2)/9+(y1+y2)(y1-y2)/[4(x1-x2)]=0
2x/9 + (2y/4)*[y/(x-1)]=0
4(x-0.5)²+9y²=1
当直线AB没有斜率时,AB中点就是点(1,0),代入也满足.
所以AB 中点M的轨迹方程是
4(x-0.5)²+9y²=1