13.7.3 * $M$ is a moving point on the circle $x^{2}+y^{2}-6 x-8 y=0$, $O$ is the origin, $N$ is a point on the ray $O M$, $|O M| \cdot|O N|=150$. Find the equation of the locus of point $N$.

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Let the parametric equation of the ray $O M$ be $\left\{\begin{array}{l}x=t \cos \alpha, \\ y=t \sin \alpha,\end{array}\right.$ (where $t$ is the parameter). Substituting into the equation $x^{2}+y^{2}-6 x-8 y=0$, we get $t_{1}=6 \cos \alpha+8 \sin \alpha$. Let the parameter corresponding to point $N$ be $t_{2}$. Given that $|O M| \cdot|O N|=150$ and $N$ lies on the ray $O M$, we know $t_{1} \cdot t_{2}=150$, so $t_{2}=\frac{150}{t_{1}}$. Therefore, the parametric equation of the trajectory of point $N$ is
$$
\left\{\begin{array}{l}
x=t_{2} \cos \alpha=\frac{150 \cos \alpha}{6 \cos \alpha+8 \sin \alpha}, \\
y=t_{2} \sin \alpha=\frac{150 \sin \alpha}{6 \cos \alpha+8 \sin \alpha} .
\end{array}\right.
$$

Eliminating $\alpha$ yields $3 x+4 y-75=0$, which is the required trajectory equation.