已知定点A(0,1),B(0,-1),C(1,0),动点p满足向量AP*向量BP=K*向量PC^2
问题描述：
已知定点A(0,1),B(0,-1),C(1,0),动点p满足向量AP*向量BP=K*向量PC^2
求动点p轨迹方程,并说明方程表示的曲线类型.
答
设P(x,y),则：AP=(x,y)-(0,1)=(x,y-1),BP=(x,y)-(0,-1)=(x,y+1)
PC=(1,0)-(x,y)=(1-x,-y),而：AP dot BP=(x,y-1) dot (x,y+1)=x^2+y^2-1=k((1-x)^2+y^2)
k=1时,上式为：x=1,此时P点的轨迹是一条直线：x=1
k≠1时,即：(k-1)x^2+(k-1)y^2-2kx+k+1=0,即：x^2-2k/(k-1)+k^2/(k-1)^2+y^2=k^2/(k-1)^2-(k+1)/(k-1)
即：(x-k/(k-1))^2+y^2=1/(k-1)^2,当k>1时,P点的轨迹是圆,圆心(k/(k-1),0),半径：1/(k-1)
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