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Ellipse problem in senior high school mathematics
Solution:

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.