双曲线X^2/4-Y^2=1,双曲线上有一点P,F1,F2为焦点,∠PF1F2为直角,求△PF1F2的面积

Question

双曲线X^2/4-Y^2=1,双曲线上有一点P,F1,F2为焦点,∠PF1F2为直角,求△PF1F2的面积

Answer 1

X^2/4-Y^2=1
a^2=4,b^2=1,c^2=4+1=5
a=2,b=1,c=根号5.
PF2-PF1=2a=4
又PF2^2=PF1^2+F1F2^2
(4+PF1)^2=PF1^2+(2根号5)^2
16+8PF1+PF1^2=PF1^2+20
PF1=1/2
所以,S(PF1F2)=1/2F1F2*PF1=1/2*2根号5*1/2=根号5/2