解析:∵A(-a,0),F(-c,0),B1(0,b),B2(0,-b).
=(a,-b),=(c,b).
∵⊥,∴·=ac-b2=0. ∴b2=ac,即c2-ac-a2=0,e2-e-1=0.
又e>1, ∴e=.