10. (20 points) An ellipse is constructed with the side $AB$ of the equilateral $\triangle ABC$ as its major axis. If the ellipse intersects the circumcircle of $\triangle ABC$ at exactly two points, find the range of the ellipse's eccentricity $e$.

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10. As shown in Figure 9, let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, with the semi-focal distance being $c, A(-a, 0)$, and $B(a, 0)$.
Assume the ellipse and the circumcircle of $\triangle A B C$ have a common point $P(x, y) (y > 0)$ other than points $A$ and $B$.
Since $A, C, B, P$ are concyclic, we have $\angle A P B = 180^{\circ} - \angle C = 120^{\circ}$.
Thus, $-\sqrt{3} = \tan \angle A P B = \frac{\frac{y}{x-a} - \frac{y}{x+a}}{1 + \frac{y}{x-a} \cdot \frac{y}{x+a}}$.
Simplifying, we get $\sqrt{3}\left(x^{2} + y^{2} - a^{2}\right) + 2 a y = 0$.
Since point $P$ lies on the ellipse, we have $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$.
Eliminating $x$ from the two equations, we get
$$
\sqrt{3} y^{2}\left(b^{2} - a^{2}\right) + 2 a b^{2} y = 0 \text{. }
$$

Since $0 < y \leqslant b$, we have
$$
\begin{array}{l}
y = \frac{2 a b^{2}}{\sqrt{3}\left(a^{2} - b^{2}\right)} = \frac{2 a\left(a^{2} - c^{2}\right)}{\sqrt{3} c^{2}} \leqslant b \\
\Rightarrow \frac{2\left(1 - e^{2}\right)}{\sqrt{3} e^{2}} \leqslant \sqrt{1 - e^{2}} \\
\Rightarrow 4\left(1 - e^{2}\right) \leqslant 3 e^{4} \Rightarrow \frac{\sqrt{6}}{3} \leqslant e < 1 .
\end{array}
$$

However, according to the problem, a point $P$ satisfying the above conditions does not exist, so the range of the eccentricity $e$ of the ellipse is $\left(0, \frac{\sqrt{6}}{3}\right)$.