9. (16 points) Given the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{4}=1$. Find the real number pair $(a, b)$, such that for any line $l_{a}$ with slope $a$ intersecting the ellipse at points $A$ and $B$, and the line $x=b$ intersecting the upper half of the ellipse at point $P$, the triangle $\triangle P A B$ always has its incenter on the line $x=b$.

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$$
\begin{array}{l}
\text { Let points } A\left(x_{A}, y_{A}\right), B\left(x_{B}, y_{B}\right), P\left(x_{P}, y_{P}\right), \text { and } l_{a}: y=a x+m.
\text { Substituting the equation of line } l_{a} \text { into the ellipse equation, we get } \\
x^{2}+9(a x+m)^{2}-36=0 \\
\Rightarrow\left(9 a^{2}+1\right) x^{2}+18 a m x+9 m^{2}-36=0 \\
\Rightarrow x_{A}+x_{B}=\frac{-18 a m}{9 a^{2}+1}, x_{A} x_{B}=\frac{9 m^{2}-36}{9 a^{2}+1} .
\end{array}
$$
$$
\begin{array}{l}
\text { Given } k_{P A}+k_{P B}=0 \\
\Rightarrow\left(y_{P}-y_{A}\right)\left(x_{P}-x_{B}\right)+\left(y_{P}-y_{B}\right)\left(x_{P}-x_{A}\right)=0 \\
\Rightarrow 2 x_{P} y_{P}-x_{P}\left(y_{A}+y_{B}\right)-y_{P}\left(x_{A}+x_{B}\right)+ \\
y_{A} x_{B}+y_{B} x_{A}=0 \\
\Rightarrow 2 x_{P} y_{P}-x_{P}\left[a\left(x_{A}+x_{B}\right)+2 m\right]-y_{P}\left(x_{A}+x_{B}\right)+ \\
2 a x_{A} x_{B}+m\left(x_{A}+x_{B}\right)=0 \\
\Rightarrow 2 x_{P} y_{P}-x_{P}\left(2 m-\frac{18 a^{2} m}{9 a^{2}+1}\right)+y_{P} \cdot \frac{18 a m}{9 a^{2}+1}+ \\
2 a \cdot \frac{9 m^{2}-36}{9 a^{2}+1}-\frac{18 a m^{2}}{9 a^{2}+1}=0 \\
\Rightarrow 2 x_{P} y_{P}-x_{P} \cdot \frac{2 m}{9 a^{2}+1}+y_{P} \cdot \frac{18 a m}{9 a^{2}+1}-\frac{72 a}{9 a^{2}+1}=0 \\
\Rightarrow\left(2 x_{P} y_{P}-\frac{72 a}{9 a^{2}+1}\right)+\frac{m}{9 a^{2}+1}\left(18 a y_{P}-2 x_{P}\right)=0 \\
\Rightarrow\left\{\begin{array} { l } 
{ x _ { P } y _ { P } = \frac { 3 6 a } { 9 a ^ { 2 } + 1 } , } \\
{ x _ { P } = 9 a y _ { P } }
\end{array} \Rightarrow \left\{\begin{array}{l}
x_{P}=\frac{18 a}{\sqrt{9 a^{2}+1}}, \\
y_{P}=\frac{2}{\sqrt{9 a^{2}+1}} .
\end{array}\right.\right. \\
\end{array}
$$

Upon verification, point $P$ lies on the ellipse.
Thus, the solution is $(a, b)=\left(a, \frac{18 a}{\sqrt{9 a^{2}+1}}\right)(a \in \mathbf{R})$.