Example 5 Given the ellipse $C: \frac{x^{2}}{24}+\frac{y^{2}}{16}=1$, the line $l: \frac{x}{12}+\frac{y}{8}=1$, and point $P$ on $l$, the ray $O P$ intersects the ellipse at point $R$. Point $Q$ is on $O P$ such that $|O Q| \cdot|O P|=|O R|^{2}$. Find the equation of the trajectory of point $Q$ as point $P$ moves along $l$; and identify what kind of curve the trajectory is.
Solve: In the given coordinate system, establish a polar coordinate system with the origin $O$ as the pole and the positive half-axis of the $x$-axis as the polar axis. Let $\angle P Q x=\theta$, $|O R|=\rho_{1},|O P|=\rho_{2},|O Q|=\rho$. $R$ is on the ellipse $\frac{x^{2}}{24}+\frac{y^{2}}{16}=1$, so $\frac{\rho_{1}^{2} \cos ^{2} \theta}{24}+\frac{\rho_{1}^{2} \sin ^{2} \theta}{16}=1$. Therefore, $\frac{1}{\rho_{1}^{2}}=\frac{\cos ^{2} \theta}{24}+\frac{\sin ^{2} \theta}{16}$. (*) Since $P$ is on the line $\frac{x}{12}+\frac{y}{8}=1$, we have $\frac{\rho_{2} \cos \theta}{12}+\frac{\rho_{2} \sin \theta}{8}=1$. Therefore, $\frac{1}{\rho_{2}}=\frac{\cos \theta}{12}+\frac{\sin \theta}{8}$. From $|O Q| \cdot|O P|=|O R|^{2}$, we get $\rho_{2} \cdot \rho=\rho_{1}^{2}$, so $\rho \cdot \frac{1}{\rho_{1}^{2}}=\frac{1}{\rho_{2}}$, thus $\frac{\rho^{2}}{\rho_{1}^{2}}=\frac{\rho}{\rho_{2}}$. Substituting (*) into this, we get: $\frac{x^{2}}{24}+\frac{y^{2}}{16}=\frac{x}{12}+\frac{y}{8}$, so $2 x^{2}+3 y^{2}-4 x-6 y=0, 2(x-1)^{2}+3(y-1)^{2}=5$. Therefore, the required trajectory equation is $\frac{(x-1)^{2}}{\frac{5}{2}}+\frac{(y-1)^{2}}{\frac{5}{3}}=1$, which is an ellipse centered at $(1,1)$, with the semi-major axis length $\frac{\sqrt{10}}{2}$, the semi-minor axis length $\frac{\sqrt{15}}{3}$, and the foci on the line $y=1$.