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PMPN(MN分别为切点),使得.求动点P的轨迹方程;

(2)    若直线交圆O2AB,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) ∵

Pxy),则

整理得,即为所求·············································· 6分

        (2) (方法一) ∵

ABO2C

SABC =

(dO2AB的距离),而

当且仅当时,取等号

时,△O2AB的面积最大.···························· 12分

   

CAB的距离

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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