题目内容
已知椭圆
+
=1(a>b>0)的右焦点为F2(3,0),离心率为e.
(Ⅰ)若e=
,求椭圆的方程;
(Ⅱ)设直线y=kx与椭圆相交于A,B两点,若
•
=0,且
<e≤
,求k的取值范围.
| x2 |
| a2 |
| y2 |
| b2 |
(Ⅰ)若e=
| ||
| 2 |
(Ⅱ)设直线y=kx与椭圆相交于A,B两点,若
| AF2 |
| BF2 |
| ||
| 2 |
| ||
| 2 |
(I)由题得:c=3,
=
?a=2
,b=
.
故椭圆方程为
+
=1;
(II)由
得(b2+a2k2)x2-a2b2=0,
设A(x1,y1),B(x2,y2),∴x1+x2=0,x1x2=
,又
=(3-x1,-y1),
=(3-x2,-y2),∴
•
=(1+k2)x1x2+9=0,即
+9=0,
∴k2=
=-1-
,
∵
<e≤
,
∴2
≤a≤3
,12≤a2≤18,
∴k2≥
,即 k∈(-∞,-
]∪[
,+∞).
| c |
| a |
| ||
| 2 |
| 3 |
| 3 |
故椭圆方程为
| x2 |
| 12 |
| y2 |
| 3 |
(II)由
|
设A(x1,y1),B(x2,y2),∴x1+x2=0,x1x2=
| -a2b2 |
| b2+a2k2 |
| AF2 |
| BF2 |
| AF2 |
| BF2 |
| -a2(a2-9)(1+k2) |
| a2k2+(a2-9) |
∴k2=
| a4-18a2+81 |
| -a4+18a2 |
| 81 |
| -a4+18a2 |
∵
| ||
| 2 |
| ||
| 2 |
∴2
| 3 |
| 2 |
∴k2≥
| 1 |
| 8 |
| ||
| 4 |
| ||
| 4 |
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