问题
解答题
已知定点A(0,1),B(0,-1),C(1,0).动点P满足:
(1)求动点P的轨迹方程,并说明方程表示的曲线; (2)当k=2时,求|2
|
答案
(1)设动点的坐标为P(x,y),则
=(x,y-1),AP
=(x,y+1),BP
=(1-x,-y)PC
∵
•AP
=k|BP
|2,∴x2+y2-1=k[(x-1)2+y2]即(1-k)x2+(1-k)y2+2kx-k-1=0.PC
若k=1,则方程为x=1,表示过点(1,0)是平行于y轴的直线.
若k≠1,则方程化为:(x+
)2+y2=(k 1-k
)2,1 1-k
表示以(-
,0)为圆心,以k 1-k
为半径的圆.1 |1-k|
(2)当k=2时,方程化为(x-2)2+y2=1.
∵2
+AP
=2(x,y-1)+(x,y+1)=(3x,3y-1),BP
∴|2
+AP
|=BP
.又x2+y2=4x-3,9x2+9y2-6y+1
∴|2
+AP
|=BP
∵(x-2)2+y2=1,∴令x=2+cosθ,y=sinθ.36x-6y-26
则36x-6y-26=36cosθ-6sinθ+46=6
cos(θ+φ)+46∈[46-637
,46+637
],37
∴|2
+AP
|max=BP
=3+46+6 37
,|237
+AP
|min=BP
=46-6 37
-3.37