题目内容
过圆x2+y2=1上一点P作圆的切线与x轴和y轴分别交于A,B两点,O是坐标原点,则|![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_ST/0.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_ST/1.png)
【答案】分析:设∠OBP=α,由O<α<
,∠OAP=
-α,知|
+2
|=|(
,
)|然后利用向量的模以及基本不等式求出表达式的最小值即可.
解答:解:设∠OAP=α,
∵O<α<
,∠OBP=
-α,
,
,
∴|
+2
|=|(
,
)|=
=![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/15.png)
=![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/16.png)
=![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/17.png)
=3,当且仅当tan2
时,表达式取得最小值.
故答案为:3.
点评:本题考查直线和圆的方程的应用,是基础题.解题时要认真审题,仔细解答,注意合理地运用均值不等式进行解题.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/0.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/1.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/2.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/3.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/4.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/5.png)
解答:解:设∠OAP=α,
∵O<α<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/6.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/7.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/8.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/9.png)
∴|
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/10.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/11.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/12.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/13.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/14.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/15.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/16.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/17.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/18.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103101406991146919/SYS201311031014069911469011_DA/19.png)
故答案为:3.
点评:本题考查直线和圆的方程的应用,是基础题.解题时要认真审题,仔细解答,注意合理地运用均值不等式进行解题.
![](http://thumb.zyjl.cn/images/loading.gif)
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