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.