设∠PF1F2=β ∠PF2F1=γ
由正弦定理得|PF1|/sinγ=|PF2|/sinβ=|F1F2|/sin(β+γ)
∴sin(β+γ)/(sinβ+sinγ)=|F1F2|/(|PF1|+|PF2|)=2c/2a=c/a
∴c/a={2sin[(β+γ)/2]cos[(β+γ)/2]}/{2sin[(β+γ)/2]cos[(β-γ)/2]}
化简得:
c/a=cos[(β+γ)/2]/cos[(β-γ)/2]
由余弦定理:|PF1|²+|PF2|²-2|PF2||PF1|cosα=|F1F2|²
∴(|PF1|+|PF2|)²-2(1+cosα)|PF2||PF1|=|F1F2|²
即|PF2||PF1|=4(a²-c²)/[2(1+cosα)]=2b²/(1+cosα)
∴S=1/2|PF2||PF1|sinα=1/2*[2b²/(1+cosα)]sinα=b²tan(α/2)