2. For the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the left and right foci are $F_{1}, F_{2}$ respectively. If there are exactly 6 different points $P$ on the ellipse such that $\triangle F_{1} F_{2} P$ is an isosceles triangle, then the range of the eccentricity of the ellipse $C$ is $\qquad$.
2. $\left(\frac{1}{3}, \frac{1}{2}\right) \cup\left(\frac{1}{2}, 1\right)$.

According to the problem, we know
$$
\left\{\begin{array}{l}
4 c>2 a-2 c, \\
2 c \neq a
\end{array} \Rightarrow \frac{1}{3}<e<1 \text { and } e \neq \frac{1}{2} .\right.
$$

Therefore, the range of the eccentricity is $\left(\frac{1}{3}, \frac{1}{2}\right) \cup\left(\frac{1}{2}, 1\right)$.