Three. (20 points) Let the left and right foci of an ellipse be $F_{1}$ and $F_{2}$, respectively, and the left directrix be $l$. Point $P$ is on the ellipse. Draw $P Q \perp$ $l$, with $Q$ being the foot of the perpendicular. Try to determine: For what kind of ellipse does there exist such a point $P$ that $P Q F_{1} F_{2}$ forms a parallelogram? Explain your reasoning (the answer should be expressed as an equation or inequality involving the eccentricity $e$).
Three, as shown in Figure 5, let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, and point $P$ is above the $x$-axis, with coordinates $\left(x_{P}, y_{P}\right)$, $c^{2}=a^{2}-b^{2}(c>0)$, then $F_{1}(-c, 0)$, $F_{2}(c, 0)$, and the left directrix $l: x=-\frac{a^{2}}{c}$.

Since $P Q F_{1} F_{2}$ is a parallelogram, $P Q \cong F_{1} F_{2}$. Therefore,
$$
x_{P}-\left(-\frac{a^{2}}{c}\right)=c-(-c),
$$

which means $x_{P}=2 c-\frac{a^{2}}{c}=\frac{2 c^{2}-a^{2}}{c}$.
Since $\frac{x_{P}^{2}}{a^{2}}+\frac{y_{P}^{2}}{b^{2}}=1$, the necessary and sufficient condition for such a point $P$ to exist is $x_{P}^{2} \leqslant a^{2}$, i.e., $\left(\frac{2 c^{2}-a^{2}}{c}\right)^{2} \leqslant a^{2}$, or $\left(2 e-\frac{1}{e}\right)^{2} \leqslant 1$ (where the eccentricity $e=\frac{c}{a}$), $-1 \leqslant 2 e-\frac{1}{e} \leqslant 1, -e \leqslant 2 e^{2}-1 \leqslant e$. Therefore,
$$
\left\{\begin{array}{l}
2 e^{2}+e-1 \geqslant 0, \\
2 e^{2}-e-1 \leqslant 0, \\
0<e<1 .
\end{array} \text { Solving, we get } \frac{1}{2} \leqslant e<1 .\right.
$$

In summary, such a point $P$ exists if and only if the eccentricity $e$ of the ellipse satisfies the condition $\frac{1}{2} \leqslant e<1$.