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In the given ellipse, $a=5$ and $b=3,$ so $c=\sqrt{a^2-b^2}=4.$ We can take $F=(4,0).$

Let $A=(x,y).$ Then $\frac{x^2}{25}+\frac{y^2}{9}=1$ and $$(x-4{)}^2+y^2={\left(\frac{3}{2}\right)}^2=\frac{9}{4}.$$Solving for $y^2$ in $\frac{x^2}{25}+\frac{y^2}{9}=1,$ we get $$y^2=\frac{225-9x^2}{25}.$$Substituting, we get $$(x-4{)}^2+\frac{225-9x^2}{25}=\frac{9}{4}.$$This simplifies to $64x^2-800x+2275=0,$ which factors as $(8x-65)(8x-35)=0.$ Since $x\le 5,$ $x=\frac{35}{8}.$ Then $$\frac{(35/8{)}^2}{25}+\frac{y^2}{9}=1.$$This leads to $y^2=\frac{135}{64},$ so $y=\frac{\sqrt{135}}{8}=\pm \frac{3\sqrt{15}}{8}.$ We can take $y=\frac{3\sqrt{15}}{8}.$

Thus, the slope of line $AB$ is $$\frac{\frac{3\sqrt{15}}{8}}{\frac{35}{8}-4}=\sqrt{15},$$so its equation is $$y=\sqrt{15}(x-4).$$To find $B,$ we substitute into the equation of the ellipse, to get $$\frac{x^2}{25}+\frac{15(x-4{)}^2}{9}=1.$$This simplifies to $128x^2-1000x+1925=0.$ We could try factoring it, but we know that $x=\frac{35}{8}$ is a solution (because we are solving for the intersection of the line and the ellipse, and $A$ is an intersection point.) Hence, by Vieta's formulas, the other solution is $$x=\frac{1000}{128}-\frac{35}{8}=\frac{55}{16}.$$Then $y=\sqrt{15}(x-4)=-\frac{9\sqrt{15}}{16}.$ Hence, $$BF=\sqrt{{\left(\frac{55}{16}-4\right)}^2+{\left(-\frac{9\sqrt{15}}{16}\right)}^2}=\boxed{\frac{9}{4}}.$$