问题
解答题
P为椭圆
(1)求△F1PF2的面积; (2)求P点的坐标. |
答案
∵a=5,b=3
∴c=4(1)
设|PF1|=t1,|PF2|=t2,
则t1+t2=10①t12+t22-2t1t2•cos60°=82②,
由①2-②得t1t2=12,
∴S△F1PF2=
1 |
2 |
1 |
2 |
| ||
2 |
3 |
(2)设P(x,y),由S△F1PF2=
1 |
2 |
3 |
∴|y|=
3
| ||
4 |
3
| ||
4 |
3
| ||
4 |
5
| ||
4 |
5
| ||
4 |
3
| ||
4 |
5
| ||
4 |
3
| ||
4 |
5
| ||
4 |
3
| ||
4 |
5
| ||
4 |
3
| ||
4 |