2. Given the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{8}=1$, with foci $F_{1}, F_{2}$, and $P$ as any point on the ellipse (but $P$ is not on the $x$-axis), the incenter of $\triangle P F_{1} F_{2}$ is $I$. A line parallel to the $x$-axis through $I$ intersects $P F_{1}, P F_{2}$ at $A$,
$B$. Then $\frac{S_{\triangle P A B}}{S_{\triangle P r_{1} r_{2}}}=$ $\qquad$ .
2. $\frac{9}{16}$. Given, $a=3, b^{2}=8, c=1$, then
$$
\left|P F_{1}\right|+\left|P F_{2}\right|=2 a=6,\left|F_{1} F_{2}\right|=2 c=2 \text {. }
$$
The perimeter of $\triangle P F_{1} F_{2}$ is 8.
Let the inradius of $\triangle P F_{1} F_{2}$ be $r$, and point $P(x, y)$.
By $\triangle P A B \backsim \triangle P F_{1} F_{2}$, we get
$$
\begin{array}{c}
\frac{S_{\triangle P A B}}{S_{\triangle P F_{1} F_{2}}}=\left(\frac{y-r}{y}\right)^{2} . \\
\text { Also, } S_{\triangle P F_{1} F_{2}}=\frac{1}{2}\left|F_{1} F_{2}\right||y|=|y| .
\end{array}
$$

Let $p$ be the semiperimeter of $\triangle P F_{1} F_{2}$, then
$$
S_{\triangle P F_{1} F_{2}}=p r=4 r,
$$

Thus, $|y|=4 r$, which means
$$
\frac{S_{\triangle P A B}}{S_{\triangle P F_{1} F_{2}}}=\frac{9}{16} .
$$