Question

定点A(3,1)动点B在圆x^2+y^2=4上,P在线段AB上,BP:PA=1:2,求点P轨迹方程（x-1）^2+(y-

Answer 1

P(x,y)

(x-xB)/(3-x)=BP:PA=1/2,xB=(3x-3)/2

(y-yB)/(1-y)=BP:PA=1/2,yB=(3y-1)/2

(xB)^2+(yB)^2=4

[(3x-3)/2]^2+[3y-1)/2]^2=4

(x-1)^2+(y-1/3)^2=16/9,

x定义域[-1/3,7/3]