题目内容
(本小题满分12分)
如下图,O1(– 2,0),O2(2,0),圆O1与圆O2的半径都是1,
![]()
(1)
过动点P分别作圆O1、圆O2的切线PM、PN(M、N分别为切点),使得
.求动点P的轨迹方程;
(2)
若直线
交圆O2于A、B,又点C(3,1),当m取何值时,△ABC的面积最大?
【答案】
(1)![]()
(2)m = – 1或 – 3
【解析】解:(1) ∵ ![]()
∴ ![]()
设P(x,y),则![]()
整理得
,即为所求·············································· 6分
(2) (方法一) ∵ ![]()
∴ AB∥O2C
∴
S△ABC =![]()
(d为O2到AB的距离),而![]()
∴
![]()
当且仅当
时,取等号
而![]()
∴
时,△O2AB的面积最大.···························· 12分
![]()
![]()
C到AB的距离![]()
∴
![]()
![]()
当
或– 3时,S取得最大值,而m = – 1,– 3满足①式
∴ m = – 1或 – 3··························································· 12分
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