题目内容
如图,P是椭圆
+
=1(xy≠0)上的动点,F1、F2是椭圆的焦点,M是∠F1PF2的平分线上一点,且
•
=0.则|OM|的取值范围______.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140611/201406111400361965036.png)
x2 |
25 |
y2 |
16 |
F2M |
MP |
![](http://thumb.zyjl.cn/pic2/upload/papers/20140611/201406111400361965036.png)
∵
•
=0,∴
⊥
延长F2M交PF1于点N,可知△PNF2为等腰三角形,![](http://thumb.zyjl.cn/pic2/upload/papers/20140611/2014061114003883210045.png)
且M为F2M的中点,可得OM是△PF1F2的中位线
∴|OM|=
|NF1|=
(|PF1|-|PN|)
=
(|PF1|-|PF2|)=
(2a-2|PF2|)=a-|PF2|
∵a-c<|PF2|<a+c
∴0<|OM|<c=
=3
∴|OM|的取值范围是(0,3)
故答案为:(0,3)
F2M |
MP |
F2M |
MP |
延长F2M交PF1于点N,可知△PNF2为等腰三角形,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140611/2014061114003883210045.png)
且M为F2M的中点,可得OM是△PF1F2的中位线
∴|OM|=
1 |
2 |
1 |
2 |
=
1 |
2 |
1 |
2 |
∵a-c<|PF2|<a+c
∴0<|OM|<c=
a2-b2 |
∴|OM|的取值范围是(0,3)
故答案为:(0,3)
![](http://thumb.zyjl.cn/images/loading.gif)
练习册系列答案
相关题目