已知M,N为两个定点,|MN|=6,且动点P满足向量PM*向量PN=6,求点P的轨迹方程

设M(-3,0),N(3,0),P(x,y)
PM=((-3-x,-y),PN=(3-x,-y),
向量PM*向量PN=6,
向量PM*向量PN
=(-3-x)(3-x)+(-y)^2
=x^2-9+y^2
所以 x^2-9+y^2=6
即 x^2+y^2=15
故点P的轨迹方程为：x^2+y^2=15.