13.7.7 ** As shown in the figure, $F$ is a fixed point, $l$ is a fixed line, and the distance from point $F$ to line $l$ is $p$. Point $M$ slides on line $l$, and the moving point $N$ is on the extension of $M F$, satisfying the condition,
$\frac{|F N|}{|M N|}=\frac{1}{|M F|}$. Find the locus of the moving point $N$.
Draw $F K \perp l$, intersecting $l$ at $K$, and establish a polar coordinate system with $F$ as the pole and the reverse extension of $F K$ as the polar axis. Let the coordinates of $N$ be ( $\rho$, $\theta$ ), and assume $\rho>0, \cos \theta>0$. Then $|M F|-\frac{\rho}{\cos \theta}$, $|F N|=\rho$. Therefore, $|M N|=|M F|+|F N|=\frac{p}{\cos \theta}+\rho$. By the problem, $|M F| \cdot|F N|=$ $|M N|$, so $N$ satisfies $\frac{p}{\cos \theta} \cdot \rho=\frac{p}{\cos \theta}+\rho$. Thus, the polar equation of point $N$ is $\rho=$ $\frac{1}{1-\frac{1}{p} \cos \theta}(01$, i.e., $01$ when, the required trajectory is the part of the ellipse to the right of the line $l^{\prime}$.