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Let |AF2|=t and |AB|=3x, then |BF 1|=4x, |AF 1|=5x,

According to the definition of hyperbola, | af1| | af2 | = | bf2 | | BF1| = 2a.

That is, 5x-t=(3x+t)-4x=2a, and the solution is t=3x and a = x.

∵| AB |: | BF1|: | AF1| = 3: 4: 5, so △ABF 1 is Rt△ with B as the right angle.

∴ cos ∠ BAF1= | ab || af1| = 35, and cos∠F2AF 1=-35 can be obtained.

△F2AF 1，| f 1 F2 | 2 = | af 1 | 2+| af2 | 2 | af 1 |？ |AF2|cos∠F2AF 1

=25x2+9x2-2×5x×3x×(-35)=52x2, then |F 1F2|=2 13x.

So the eccentricity of this hyperbola is e=2c2a=2 13x2x= 13.

So choose: a