在平面上,已知定点A,B且AB=6,如果动点P到点A的距离和到点B的距离之比为2：1,求动点P的轨迹方程.

取A,B的坐标分别为A(-3,0),B(3,0)
P(x,y)
|OA| :|OB| = 2:1
|OA|^2 :|OB|^2 = 4:1
|OA|^2 = 4|OB|^2
|OA|^2 = (x+3)^2 + y^2
|OB|^2 = (x-3)^2 + y^2
(x+3)^2 + y^2 = 4[(x-3)^2 + y^2 ]
(x-5)^2 + y^2 = 16