9. Given the ellipse $\frac{x^{2}}{1 \mathrm{~h}}+\frac{y^{2}}{4}=1$ with its left and right foci being $F_{1}$ and $F_{2}$, respectively, point $P$ lies on the line $l: x-\sqrt{3} y+8$ $+2 \sqrt{3}=0$. When $\angle F_{1} P F_{2}$ is maximized, the ratio $\frac{\left|P F_{1}\right|}{\left|P F_{2}\right|}$ is $\qquad$.
$F_{1}(-2 \sqrt{3}, 0), F_{2}(2 \sqrt{3}, 0)$. Let $\bigodot M$ be the circle passing through $F_{1}, F_{2}$ and tangent to line $l$, with the point of tangency being $P$. Line $l$ intersects the $x$-axis at point $N(-8-2 \sqrt{3}, 0)$, then $|N P|^{2}=\left|N F_{1}\right| \cdot\left|N F_{2}\right|=8(8+4 \sqrt{3})$ $=16(4+2 \sqrt{3}) \Rightarrow|N P|=4(\sqrt{3}+1)$. Therefore, $\triangle N F_{1} P \sim \triangle N P F_{2} \Rightarrow \frac{\left|P F_{1}\right|}{\left|P F_{2}\right|}=\frac{\left|N F_{1}\right|}{|N P|}$ $=\frac{8}{4(\sqrt{3}+1)}=\sqrt{3}-1$.