(12) Given point $F$ is the right focus of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the eccentricity of the ellipse is $\frac{\sqrt{3}}{2}$, a line $l$ passing through point $F$ intersects the ellipse at points $A, B$ (point $A$ is above the $x$-axis), and $\overrightarrow{A F}=3 \overrightarrow{F B}$, then the slope of line $l$ is $\qquad$
Solution: $-\sqrt{2}$.
The polar equation with the pole at the right focus is $\rho=\frac{e p}{1+e \cos \theta}$,
so $|A F|=\frac{e p}{1+e \cos \theta},|B F|=\frac{e p}{1-e \cos \theta}$,
thus $\frac{e p}{1+e \cos \theta}=\frac{3 e p}{1-e \cos \theta}$, which gives $\cos \theta=-\frac{\sqrt{3}}{3}, \tan \theta=-\sqrt{2}$,
therefore the slope of the line $l$ is $-\sqrt{2}$.