- 设F1和F2是双曲线(x^2)/4
- 设和F2是双曲线(x^2)/4-y^2=1的两个焦点，点P在双曲线上，且满足∠F1PF2=90度，则ΔF1PF2的面积等于
- 有一个规律，ΔF1PF2的面积等于 b^2cot(∠F1PF2/2)
[证明：
(F1F2)^2=(PF1)^2+(PF2)^2-2(PF1)(PF2)cos∠F1PF2
(2c)^2=(PF1-PF2)^2+2(PF1)(PF2)(1-cos∠F1PF2)
4c^2=4a^2+2(PF1)(PF2)(1-cos∠F1PF2)
2b^2=(PF1)(PF2)(1-cos∠F1PF2)
(PF1)(PF2)=2b^2/(1-cos∠F1PF2)
面积=[(PF1)(PF2)sin∠F1PF2]/2=b^2(sin∠F1PF2)/(1-cos∠F1PF2)
=b^2/[(1-cos∠F1PF2)/sin∠F1PF2]
=b^2/tan∠F1PF2
=b^2cot(∠F1PF2/2)
同理可得椭圆的为b^2tan(∠F1PF2/2)]
所以答案为1