题目内容
离心率e=![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172348240199522/SYS201311031723482401995004_ST/0.png)
【答案】分析:根据离心率和准线方程求得a和c,则b可得,则椭圆的方程可得.
解答:解:由e=
=
,
=3,
求得a=
,c=
,
∴b=
=
=
,
∴椭圆的方程为:
+
=1.
故答案为:
+
=1.
点评:本题主要考查了椭圆的标准方程,椭圆的简单性质.考查了学生分析问题和解决问题的能力.
解答:解:由e=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172348240199522/SYS201311031723482401995004_DA/0.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172348240199522/SYS201311031723482401995004_DA/1.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172348240199522/SYS201311031723482401995004_DA/2.png)
求得a=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172348240199522/SYS201311031723482401995004_DA/3.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172348240199522/SYS201311031723482401995004_DA/4.png)
∴b=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172348240199522/SYS201311031723482401995004_DA/5.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172348240199522/SYS201311031723482401995004_DA/6.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172348240199522/SYS201311031723482401995004_DA/7.png)
∴椭圆的方程为:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172348240199522/SYS201311031723482401995004_DA/8.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172348240199522/SYS201311031723482401995004_DA/9.png)
故答案为:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172348240199522/SYS201311031723482401995004_DA/10.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103172348240199522/SYS201311031723482401995004_DA/11.png)
点评:本题主要考查了椭圆的标准方程,椭圆的简单性质.考查了学生分析问题和解决问题的能力.
![](http://thumb.zyjl.cn/images/loading.gif)
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