则|PF1|=a+ex1,|PF2|=a-ex1.
在△PF1F2中,由余弦定理得 cos120°=−
| 1 |
| 2 |
| (a+ex1)2+(a−ex1)2−4c2 |
| 2(a+ex1)(a−ex1) |
解得 x12=
| 4c2−3a2 |
| e2 |
∵x12∈(0,a2],
∴0≤
| 4c2−3a2 |
| e2 |
即4c2-3a2≥0.且e2<1
∴e=
| c |
| a |
| ||
| 2 |
故椭圆离心率的取范围是 e∈[
| ||
| 2 |
故选C
| 1 |
| 2 |
| ||
| 3 |
| ||
| 2 |
| ||
| 2 |
| 1 |
| 2 |
| (a+ex1)2+(a−ex1)2−4c2 |
| 2(a+ex1)(a−ex1) |
| 4c2−3a2 |
| e2 |
| 4c2−3a2 |
| e2 |
| c |
| a |
| ||
| 2 |
| ||
| 2 |