不妨設(shè)P點(diǎn)在雙曲線右支,則由橢圓及雙曲線定義得:
|PF1|+|PF2|=2a①,|PF1|-|PF2|=2m②,
由①②得:|PF1|=a+m,|PF2|=a-m,
又a²-b²=m²+n²=c²,
∴cos∠F1PF2=(|PF1|²+|PF2|²-|F1F2|²)/2|PF1|·lPF2l
=[(a+m)²+(a-m)²-(2c)²]/2(a+m)(a-m)
=[(a²-c²)+(m²-c²)]/(a²-m²)
=(b²-n²)/(b²+n²),sin∠F1PF2=2bn/(b²+n²),
∴tan½∠F1PF2=(1-cos∠F1PF2)/sin∠F1PF=n/b,
∴∠F1PF2=2arc tan(n/b),
ΔF1PF2面積=½lPF1l·lPF2l·sin∠F1PF2
=½(a+m)(a-m)·2bn/(b²+n²)
=bn(a²-m²)/(a²-m²)=bn.