已知椭圆中心在坐标原点O,焦点在x轴上,长轴长是短轴长的2倍,且经过点M(2,1),直线l平行OM,且与椭圆交于A、B两个不同的点.
(1)求椭圆方程;
(2)若∠AOB为钝角,求直线l在y轴上的截距m的取值范围;
(3)求证直线MA、MB与x轴围成的三角形总是等腰三角形.
【答案】
分析:(1)设椭圆方程
![](http://thumb2018.1010pic.com//pic6/res/gzsx/web/STSource/20131202111904488331202/SYS201312021119044883312020_DA/0.png)
,利用长轴长是短轴长的2倍,且经过点M(2,1),建立方程组,即可求得椭圆方程;
(2)设l方程与椭圆方程联立,利用韦达定理及∠AOB为钝角,结合向量知识,即可求直线l在y轴上的截距m的取值范围;
(3)依题即证k
AM+k
BM=0,利用韦达定理代入,即可证得结论.
解答:(1)解:设椭圆方程
![](http://thumb2018.1010pic.com//pic6/res/gzsx/web/STSource/20131202111904488331202/SYS201312021119044883312020_DA/1.png)
,依题意可得
![](http://thumb2018.1010pic.com//pic6/res/gzsx/web/STSource/20131202111904488331202/SYS201312021119044883312020_DA/2.png)
…2分
可得
![](http://thumb2018.1010pic.com//pic6/res/gzsx/web/STSource/20131202111904488331202/SYS201312021119044883312020_DA/3.png)
,所以椭圆方程为
![](http://thumb2018.1010pic.com//pic6/res/gzsx/web/STSource/20131202111904488331202/SYS201312021119044883312020_DA/4.png)
….4分
(2)解:设l方程为:
![](http://thumb2018.1010pic.com//pic6/res/gzsx/web/STSource/20131202111904488331202/SYS201312021119044883312020_DA/5.png)
,与椭圆方程联立得:x
2+2mx+2m
2-4=0
由韦达定理得:x
1+x
2=-2m,
![](http://thumb2018.1010pic.com//pic6/res/gzsx/web/STSource/20131202111904488331202/SYS201312021119044883312020_DA/6.png)
…6分
设A(x
1,y
1),B(x
2,y
2),
因为∠AOB为钝角,所以
![](http://thumb2018.1010pic.com//pic6/res/gzsx/web/STSource/20131202111904488331202/SYS201312021119044883312020_DA/7.png)
=
![](http://thumb2018.1010pic.com//pic6/res/gzsx/web/STSource/20131202111904488331202/SYS201312021119044883312020_DA/8.png)
=
![](http://thumb2018.1010pic.com//pic6/res/gzsx/web/STSource/20131202111904488331202/SYS201312021119044883312020_DA/9.png)
…7分
又直线l平行OM,∴
![](http://thumb2018.1010pic.com//pic6/res/gzsx/web/STSource/20131202111904488331202/SYS201312021119044883312020_DA/10.png)
….8分
(3)证明:依题即证k
AM+k
BM=0…9分
而
![](http://thumb2018.1010pic.com//pic6/res/gzsx/web/STSource/20131202111904488331202/SYS201312021119044883312020_DA/11.png)
..…10分
将
![](http://thumb2018.1010pic.com//pic6/res/gzsx/web/STSource/20131202111904488331202/SYS201312021119044883312020_DA/12.png)
,
![](http://thumb2018.1010pic.com//pic6/res/gzsx/web/STSource/20131202111904488331202/SYS201312021119044883312020_DA/13.png)
代入上式,得
![](http://thumb2018.1010pic.com//pic6/res/gzsx/web/STSource/20131202111904488331202/SYS201312021119044883312020_DA/14.png)
….12分
将(2)中韦达定理代入得,上式=
![](http://thumb2018.1010pic.com//pic6/res/gzsx/web/STSource/20131202111904488331202/SYS201312021119044883312020_DA/15.png)
=0
即证.…14分
点评:本题考查椭圆的标准方程,考查直线与椭圆的位置关系,考查向量知识的运用,考查学生的计算能力,属于中档题.