13 The line $l$ is the left directrix of the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$, $F_{1}$ is the left focus, and the moving point $P$ is on $l$. The third vertex of the equilateral triangle with side $P F_{1}$ is $M$, and the points $P$, $F_{1}$, and $M$ are arranged counterclockwise. Find the equation of the trajectory of point $M$.
(13) Let $l$ intersect the $x$-axis at point $P_{0}$. When point $P$ is at point $P_{0}$, point $M$ is at point $M_{0}$. When point $P$ is above point $P_{0}$, point $M$ is in the upper right of point $M_{0}$. Since $\angle P F_{1} M = \angle P_{0} F_{1} M_{0}, \angle P F_{1} P_{0} = \angle M_{0} F_{1} M, P_{0} F_{1} = M_{0} F_{1}, P F_{1} = M F_{1}$, we have $\triangle P F_{1} P_{0} \cong \triangle M F_{1} M_{0}$, which implies $\angle P P_{0} F_{1} = \angle M M_{0} F_{1} = 90^{\circ}$, i.e., $M M_{0} \perp M_{0} F_{1}$. When point $P$ is below point $P_{0}$, point $M$ is in the lower left of point $M_{0}$. Similarly, we can prove $M M_{0} \perp M_{0} F_{1}$. Given $F_{1}(-1,0), M_{0}\left(-\frac{5}{2}, \frac{3 \sqrt{3}}{2}\right), k_{M_{0} F_{1}} = -\sqrt{3}$, the equation of the trajectory of $M$ is
$$
y - \frac{3 \sqrt{3}}{2} = \frac{\sqrt{3}}{3}\left(x + \frac{5}{2}\right),
$$

which simplifies to $x - \sqrt{3} y + 7 = 0$.