设AF1=AB=m,则BF1=√2m,AF2=m-2a,局扒祥BF2=√2m-2a,∴AB=AF2+BF2=m,解得4a=√2m,AF2=(1-√2/2)m,△AF1F2为Rt三角桐搏形,∴F1F2^2=AF1^2+AF2^2∴4c^2=(5/2-√2)m^2,此竖又4a^2=1/2m^2,∴e^2=5-2√2选择B