9. For the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the right vertex is $A$, the upper vertex is $B$, and the left focus is $F$. If $\angle A B F=$ $90^{\circ}$, then the eccentricity of the ellipse is $\qquad$.
9. $\frac{\sqrt{5}-1}{2}$.

Given $\angle A B F=90^{\circ}$, according to the projection theorem, we have $a c=b^{2}=a^{2}-c^{2}$,

i.e., $\left(\frac{c}{a}\right)^{2}+\left(\frac{c}{a}\right)-1=0$.
Thus, $e=\frac{c}{a}=\frac{-1 \pm \sqrt{5}}{2}$ (the negative value is discarded).