2. For the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, in what range of the eccentricity $e$ does the ellipse always have a point $P$ such that the product of the distances from point $P$ to the two foci equals the square of the focal distance?
2. Let $F_{1}, F_{2}$ be the two foci of the ellipse, then $\left|F_{1} F_{2}\right|=2 c$, and by the given condition $\left|P F_{1}\right| \cdot\left|P F_{2}\right|=4 c^{2}$. By property 8, we know $4 c+$ $|O P|^{2}=b^{2}+a^{2}$, i.e., $|O P|^{2}=b^{2}+a^{2}-4 c^{2}$. Also, $b^{2} \leqslant|O P|^{2} \leqslant a^{2}$, i.e., $b^{2} \leqslant b^{2}-4 c^{2}+a^{2} \leqslant a^{2}$, from which we have $4 c^{2}$ $\leqslant a^{2}$ and $a^{2} \leqslant 5 c^{2}$, hence $e \leqslant \frac{1}{2}$ and $e \geqslant \frac{\sqrt{5}}{5}$, i.e., the range of $e$ is $\left[\frac{\sqrt{5}}{5}, \frac{1}{2}\right]$.