十二月陶
花朵
共回答了15个问题采纳率:86.7% 举报
(1)设P(x,y),
AP =(x,y-1),
BP =(x,y+1) ,
PC =(1-x,-y) .
当k=1时,由
AP •
BP =k|
PC | 2 ,得x 2 +y 2 -1=(1-x) 2 +y 2 ,
整理得:x=1,表示过(1,0)且平行于y轴的直线;
当k≠1时,由
AP •
BP =k|
PC | 2 ,得x 2 +y 2 -1=k(1-x) 2 +ky 2 ,
整理得: (x+
k
1-k ) 2 + y 2 = (
1
1-k ) 2 ,表示以点 (-
k
1-k ,0) 为圆心,以
1
|1-k| 为半径的圆.
(2)当k=2时,方程化为(x-2) 2 +y 2 =1,即x 2 +y 2 =4x-3,
∵2
AP +
BP =(3x,3y-1) ,
∴ |2
AP +
BP |=
9 x 2 +9 y 2 -6y+1 ,又x 2 +y 2 =4x-3,
∴ |2
AP +
BP |=
36x-6y-26 =
6(6x-y)-26 .
问题归结为求6x-y的最值,令t=6x-y,
∵点P在圆(x-2) 2 +y 2 =1,圆心到直线t=6x-y的距离不大于圆的半径,
∴
|12-t|
37 ≤1 ,解得12-
37 ≤t≤12+
37 .
∴
37 -3≤|2
AP +
BP |≤12+
37 .
1年前
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