根据已知得,PA⊥AO,AM⊥PO,
∴|MO|*|PO|=|OA|²=1.
设P(X0,Y0),M(x,y),
则√(x²+y²()*√(X0²+Y0²)=1
即x²+y²=1/(X0²+Y0²),
∵P在椭圆(x²/9)+(y²/4)=1上,
∴(x0²/9)+(y0²/4)=1
∴x²+y²=[(X0²/9)+(Y0²/4)]/(X0²+Y0²)(把“1”换成“(x0²/9)+(y0²/4)”得到的)
∵O、M、P三点共线,
∴y/x=Y0/X0
∴变形为x²+y²=[(x²/9)+(y²/4)]/(x²+y²),
当P(0,±2),M(0,±1/2)时,y/x没有意义,但是带入也满足
∴最终得M的轨迹方程为(x²+y²)²=x²/9+y²/4
最后一步变形看不懂的话,我把它写出来：
是根据y/x=Y0/X0得到的
[(X0²/9)+(Y0²/4)]/(X0²+Y0²)
=X0²/[9(X0²+Y0²)]+Y0²/[4(X0²+Y0²)]
=X0²/[9(X0²+Y0²)]+Y0²/[4(X0²+Y0²)]
= 1/[9(1 + Y0²/X0²)] + 1/[4(X0²/Y0² + 1)](上下同时约去分子)
= 1/[9(1 + y²/x²)] + 1/[4(x²/y² + 1)]
= x²/[9(x²+y²)]+y²/[4(x²+y²)]
= (4x²+9y²)/[36(x²+y²)]
∴x²+y²=(4x²+9y²)/[36(x²+y²)]
36(x²+y²)²=(4x²+9y²)
(x²+y²)²=x²/9+y²/4