Question

已知椭圆方程为(x^2)/(4)+(y^2)/(3)=1,点P在第二象限,且∠PF1F2=120度,求△PF1F2的面积

Answer 1

a^2=4 b^2=3 c^2=1

PF1+PF2=2a=4

cos120=（PF1^2+4-PF2^2）/4PF1=-1/2

PF1^2+2PF1+4=PF2^2=(4-PF1^2)=16-8PF1+PF1^2

PF1=6/5

△PF1F2的面积=(1/2)*PF1*2*sin60=3根3/5