9. Given $F_{1}, F_{2}$ are the left and right foci of the ellipse $\Gamma: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, $\left|F_{1} F_{2}\right|=2$. $A$ is the right endpoint of $\Gamma$, and the line $l$ passes through point $A$ and is perpendicular to the $x$-axis. $P$ 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 $\qquad$.
As shown in the figure, by Miller's theorem, when the circle passing through points $F_{1}, F_{2}$ is tangent to the line $l$ at point $P$, $\angle F_{1} P F_{2}$ is maximized. At this time, by the sine rule, we get $2 R=\frac{\left|F_{1} F_{2}\right|}{\sin \angle F_{1} P F_{2}}=2 \sqrt{2} \Rightarrow R=\sqrt{2}$.
Obviously, the center $C$ is on the $y$-axis, and the $y$-coordinate of point $C$ is the same as that of point $P$. By $x_{F_{2}}^{2}+y_{C}^{2}=2 \Rightarrow y_{C}= \pm 1$, so $P(\sqrt{2}, \pm 1)$.