9. (16 points) Given the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, points $A, B$ are the left vertex and the upper vertex of the ellipse $E$, respectively. The line $A B$ is separated from the circle $C: x^{2}+y^{2}=\frac{c^{2}}{4}$ (where $c$ is the semi-focal distance of the ellipse). $P$ is a moving point on the line $A B$, and two tangents are drawn from point $P$ to the circle $C$, with the points of tangency being $M, N$. If there exists a point $P$ such that $\triangle P M N$ is an equilateral triangle, try to find the range of the eccentricity $e$ of the ellipse.
9. Given that the line $AB$ is separated from the circle $C: x^{2}+y^{2}=\frac{c^{2}}{4}$, we have
$$
\frac{a b}{\sqrt{a^{2}+b^{2}}}>\frac{c}{2},
$$

which implies
$$
4 a^{2} b^{2}>c^{2}\left(a^{2}+b^{2}\right) \Rightarrow 4 a^{2}\left(a^{2}-c^{2}\right)>c^{2}\left(2 a^{2}-c^{2}\right),
$$

leading to
$$
e^{4}-6 e^{2}+4>0 \text {. }
$$

Given $0<e<1$, we solve to get
$$
0<e^{2}<3-\sqrt{5} \text {. }
$$

Connecting $O M, O N, O P$. If there exists a point $P$ such that $\triangle P M N$ is an equilateral triangle, then in the right triangle $\triangle O P N$, $O P=2 O N=2 r=c$, so the distance from point $O$ to the line $AB$ is no greater than $c$, i.e.,
$$
\frac{a b}{\sqrt{a^{2}+b^{2}}} \leqslant c,
$$

which implies
$$
a^{2} b^{2} \leqslant c^{2}\left(a^{2}+b^{2}\right) \Rightarrow a^{2}\left(a^{2}-c^{2}\right) \leqslant c^{2}\left(2 a^{2}-c^{2}\right),
$$

leading to
$$
e^{4}-3 e^{2}+1 \leqslant 0 .
$$

Given $0<e<1$, we solve to get
$$
\frac{3-\sqrt{5}}{2} \leqslant e^{2}<1
$$

From equations (1) and (2), we have
$$
\frac{3-\sqrt{5}}{2} \leqslant e^{2}<3-\sqrt{5} \Rightarrow \frac{\sqrt{5}-1}{2} \leqslant e<\frac{\sqrt{10}-\sqrt{2}}{2} .
$$