题目内容
已知M是椭圆
+
=1上一点,F1,F2是椭圆的两个焦点,I是△MF1F2的内心,延长MI交F1F2于N,则
等于______.
x2 |
9 |
y2 |
5 |
|MI| |
|NI| |
如图,连接IF1,IF2.在△MF1I中,F1I是∠MF1N的角平分线,
根据三角形内角平分线性质定理,
=
,同理可得
=
,
∴
=
=
;
根据等比定理
=
=
=
=
.
故答案为:
.
根据三角形内角平分线性质定理,
|MI| |
|NI| |
|MF1| |
|F1N| |
|MI| |
|NI| |
|MF2| |
|F2N| |
∴
|MI| |
|NI| |
|MF1| |
|F1N| |
|MF2| |
|F2N| |
根据等比定理
|MI| |
|NI| |
|MF1|+|MF2| |
|F1N|+|F2N| |
2a |
2c |
2×3 | ||
2×
|
3 |
2 |
故答案为:
3 |
2 |
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