As shown in the figure, let | f1b | = k (k >; 0), then | af 1 | = 3 | f 1b | = 3k.

∴|AB|=4k According to the properties of ellipse, we get:

|AF2|=2a？ 3k，|BF2|=2a？ k

∫cos∠AF2B = 3/5，

In △ABF2, obtained by cosine theorem,

|AB|？ =|AF2|？ +|BF2|2|AF2|？ |BF2|cos∠AF2B

That is (4k)? =(2a？ 3k)？ +(2a？ k)6/5(2a？ 3k)(2a？ k)，

Simplify to get (a+k)(a? 3k)=0, and a+k >; 0, so a=3k,

∴|af2|=|af 1|=a=3k,|bf2|=5k，

∴|BF2|？ =|AF2|？ +|AB|？ ,

∴AF 1⊥AF2，

△ AF 1F2 is an isosceles right triangle.

∴|AF2|？ +|AF 1|？ =|F 1F2|？ That is, one? +a？ =(2c)？

∴c=√2/2a，

Eccentricity of ellipse e = c/a = ∴ 2/2.