由AF1•F1F2=O,得AF1⊥F1F2,
所以⊿F1AF2是Rt⊿,cos∠F1AF2=|AF1|/|AF2|
从而 AF1•AF2=|AF1|•|AF2|cos∠F1AF2=|AF1|²
由条件知AF1•AF2=c²
所以 |AF1|²=c²,|AF1|=c,
|AF2|²=|AF1|²+|F1F2|²=5c²,|AF2|=√5•c
又|AF1|+|AF2|=2a,即(1+√5)c=2a,e=2/(1+√5)=(√5-1)/2
由AF1•F1F2=O,得AF1⊥F1F2,
所以⊿F1AF2是Rt⊿,cos∠F1AF2=|AF1|/|AF2|
从而 AF1•AF2=|AF1|•|AF2|cos∠F1AF2=|AF1|²
由条件知AF1•AF2=c²
所以 |AF1|²=c²,|AF1|=c,
|AF2|²=|AF1|²+|F1F2|²=5c²,|AF2|=√5•c
又|AF1|+|AF2|=2a,即(1+√5)c=2a,e=2/(1+√5)=(√5-1)/2