9. Given $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\Gamma$:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$

with $\left|F_{1} F_{2}\right|=2$, and $A$ is the right endpoint of $\Gamma$. The line $l$ passes through point $A$ and is perpendicular to the $x$-axis, and $P$ is a moving point on line $l$. If the maximum value of $\angle F_{1} P F_{2}$ is $\frac{\pi}{4}$, then the coordinates of point $P$ at this time are . $\qquad$
9. $(\sqrt{2}, 1)$ or $(\sqrt{2},-1)$.

Let $\angle F_{1} P F_{2}=\theta, \angle P F_{1} A=\alpha, \angle P F_{2} A=\beta$.
Then $\theta=\beta-\alpha$.
Let $P A=|x|$. Then
$$
\begin{array}{l}
\tan \alpha=\frac{x}{a+c}, \tan \beta=\frac{x}{a-c}, \\
\tan \theta=\tan (\beta-\alpha)=\frac{\frac{x}{a-c}-\frac{x}{a+c}}{1+\frac{x}{a-c} \cdot \frac{x}{a+c}} \\
=\frac{2 c x}{b^{2}+x^{2}} \leqslant \frac{2 c x}{2 b x}=\frac{c}{b} .
\end{array}
$$

When and only when $x=b$, the equality holds.
Given that the maximum value of $\angle F_{1} P F_{2}$ is $\frac{\pi}{4}$, we have
$$
\frac{c}{b}=\tan \frac{\pi}{4}=1 \Rightarrow b=c \text {. }
$$

Also, $\left|F_{1} F_{2}\right|=2$, then $c=1, b=1, a=\sqrt{2}$. Therefore, when $\angle F_{1} P F_{2}$ is maximized, $|x|=b$, i.e., $P(\sqrt{2}, 1)$ or $(\sqrt{2},-1)$.