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?
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?
(1)
(2)m =" –" 1或– 3
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
∴
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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