10. (20 points) Given $A_{1} 、 A_{2}$ are the endpoints of the major axis of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $P$ is any point on the ellipse, and the lines $P A_{1} 、 P A_{2}$ intersect the line $x=\frac{a^{2}}{c}$ at points $C$ and $D$ respectively. If there is a fixed point $M$ outside the ellipse such that $\angle C M D$ is always a right angle, find the coordinates of $M$.
10. As shown in Figure 4, it is clear that $A_{1}(-a, 0), A_{2}(a, 0)$.
Let $P(m, n), C\left(\frac{a^{2}}{c}, y_{1}\right), D\left(\frac{a^{2}}{c}, y_{2}\right)$. Then
$l_{P A_{1}}: y=\frac{n}{m+a}(x+a)$.
Let $x=\frac{a^{2}}{c}$, we get $y_{1}=\frac{n a(a+c)}{(m+a) c}$.
Similarly, $y_{2}=\frac{n a(a-c)}{(m-a) c}$.
From $\frac{m^{2}}{a^{2}}+\frac{n^{2}}{b^{2}}=1 \Rightarrow \frac{m^{2}-a^{2}}{a^{2}}=-\frac{n^{2}}{b^{2}}$
$\Rightarrow \frac{n^{2} a^{2}}{m^{2}-a^{2}}=-b^{2}$.
Thus, $y_{1} y_{2}=\frac{n^{2} a^{2}\left(a^{2}-c^{2}\right)}{\left(m^{2}-a^{2}\right) c^{2}}$
$$
=\frac{-b^{2}\left(a^{2}-c^{2}\right)}{c^{2}}=-\frac{b^{4}}{c^{2}}<0 \text {. }
$$

Therefore, points $C$ and $D$ are on opposite sides of the $x$-axis. Let $C D$ intersect the $x$-axis at point $E$, and take point $M$ on the $x$-axis such that $|E M|=\frac{b^{2}}{c}$.
Since $|E M|^{2}=\left|y_{1} y_{2}\right|=|E C||E D|$
$\Rightarrow \angle C M D=90^{\circ}$.
Point $M$ lies on the circle with diameter $C D$. Taking any two different positions of point $P$, we know that $M$ is the intersection of the two circles, and there are at most two $M$s, which are $(c, 0)$ or $\left(\frac{a^{2}+b^{2}}{c}, 0\right)$.

It is easy to see that $(c, 0)$ is inside the ellipse, so the only $M$ that meets the requirement has coordinates $\left(\frac{a^{2}+b^{2}}{c}, 0\right)$.