9 Given the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$ with its left and right foci being $F_{1}$ and $F_{2}$, respectively, and point $P$ lies on the line $l: x-\sqrt{3} y+8+2 \sqrt{3}=0$. When $\angle F_{1} P F_{2}$ takes its maximum value, the ratio $\frac{\left|P F_{1}\right|}{\left|P F_{2}\right|}$ is $\qquad$
From plane geometry, we know that to maximize $\angle F_{1} P F_{2}$, the circle passing through points $F_{1}, F_{2}, P$ must be tangent to line $l$ at point $P$. Suppose line $l$ intersects the $x$-axis at $A(-8-2 \sqrt{3}, 0)$, then $\angle A P F_{1}=\angle A F_{2} P$, i.e., $\triangle A P F_{1} \sim \triangle A F_{2} P$, which means
$$
\frac{\left|P F_{1}\right|}{\left|P F_{2}\right|}=\frac{|A P|}{\left|A F_{2}\right|},
$$

By the power of a point theorem,
$$
|A P|^{2}=\left|A F_{1}\right| \cdot\left|A F_{2}\right| \text {, }
$$

Given $F_{1}(-2 \sqrt{3}, 0), F_{2}(2 \sqrt{3}, 0), A(-8-2 \sqrt{3}, 0)$, we have $\left|A F_{1}\right|=8,\left|A F_{2}\right|=8+4 \sqrt{3}$. Substituting into (1) and (2), we get
$$
\frac{\left|P F_{1}\right|}{\left|\tilde{P F}_{2}\right|}=\sqrt{\frac{\left|A F_{1}\right|}{\left|A F_{2}\right|}}=\sqrt{\frac{8}{8+4 \sqrt{3}}}=\sqrt{4-2 \sqrt{3}}=\sqrt{3}-1 .
$$