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, 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 $\qquad$.
9. $\sqrt{3}-1$.

From plane geometry, to maximize $\angle F_{1} P F_{2}$, the circle passing through points $F_{1}$, $F_{2}$, and $P$ must be tangent to line $l$ at point $P$. Suppose line $l$ intersects the $x$-axis at point $A(-8-2 \sqrt{3}, 0)$, then $\angle A P F_{1} = \angle A F_{2} P$, i.e., $\triangle A P F_{1} \backsim \triangle A F_{2} P$.
Thus, $\frac{\left|P F_{1}\right|}{\left|P F_{2}\right|} = \frac{|A P|}{\left|A F_{2}\right|}$.
By the power of a point theorem, we have
$$
|A P|^{2} = \left|A F_{1}\right| \cdot \left|A F_{2}\right| \text{. }
$$

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