题目内容
如图,以AB为直径的圆有一内接梯形
,且
.若双曲线
以A、B为焦点,且过C、D两点,则当梯形的周长最大时,双曲线的离心率为( ).
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232239289004994.png)
A、
B、
C、2 D、![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823223928963461.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823223928838534.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823223928853586.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823223928885343.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408232239289004994.png)
A、
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823223928931346.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823223928947409.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823223928963461.png)
B
解:设∠BAC=θ,作CE⊥AB于点E,
则BC=2Rsinθ,EB=BCcos(90°-θ)=2Rsin2θ,有CD=2R-4Rsin2θ,
梯形的周长l=AB+2BC+CD=2R+4Rsinθ+2R-4Rsin2=-4R(sinθ-
)2+5R.
当sinθ=
,即θ=30°时,l有最大值5R,这时,BC=R,AC=
R,a="1" 2 (AC-BC)="1" 2 (
-1)R,e=
=
+1.
故答案为
+1
则BC=2Rsinθ,EB=BCcos(90°-θ)=2Rsin2θ,有CD=2R-4Rsin2θ,
梯形的周长l=AB+2BC+CD=2R+4Rsinθ+2R-4Rsin2=-4R(sinθ-
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823223928978343.png)
当sinθ=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823223928978343.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823223928931346.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823223928931346.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823223929056362.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823223928931346.png)
故答案为
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823223928931346.png)
![](http://thumb.zyjl.cn/images/loading.gif)
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