Question

一动点至A(2,3)的距离等於其至B(3,1)的距离2倍,求这点的轨迹方程.

Answer 1

设此点为P(x,y)

PA^2=(x-2)^2+(y-3)^2

PB^2=(x-3)^2+(y-1)^2

PA=2PB

那么:(x-2)^2+(y-3)^2=4[(x-3)^2+(y-1)^2]

化简一下就可以了.