|AF1|-|AF2|=2a,|BF1|-|BF2|=2a两式相加得:|AF1|+|BF1|-|AB|=4a,
又在双曲线中,|AB|=2×
b2 |
a |
∴△ABF1周长为:|AF1|+|BF1|+|AB|=2|AB|+4a=4×
b2 |
a |
∵△ABF1内切圆的半径为a,
∴△ABF1面积为:S=
1 |
2 |
又S=
1 |
2 |
∴
1 |
2 |
b2 |
a |
1 |
2 |
即c2-a2=ac
解得:e=
c |
a |
1+
| ||
2 |
1+
| ||
2 |
故答案为:
1+
| ||
2 |
x2 |
a2 |
y2 |
b2 |
b2 |
a |
b2 |
a |
1 |
2 |
1 |
2 |
1 |
2 |
b2 |
a |
1 |
2 |
c |
a |
1+
| ||
2 |
1+
| ||
2 |
1+
| ||
2 |