Example 2 Given the ellipse $C: \frac{x^{2}}{24}+\frac{y^{2}}{16}=1$, the line $l: \frac{x}{12}+\frac{y}{8}=1$, and a point $P$ on $l$, the ray $O P$ intersects the ellipse $C$ at $R$. Also, point $Q$ is on $O P$ such that $|O Q| \cdot|O P|=|O R|^{2}$. Find the locus of point $Q$ as $P$ moves along $l$, and identify what kind of curve the locus is.
Let $Q(x, y), P\left(x_{P}, y_{P}\right), R\left(x_{R}, y_{R}\right)$. Since $|O Q| \cdot|O P|=|O R|^{2}$, we have $\frac{|O P|}{|O Q|} = \frac{|O R|^{2}}{|O Q|^{2}} \stackrel{\text { set as }}{=} t$, then $x_{R}^{2}=t x^{2}, y_{R}^{2}=t y^{2}, x_{P}=t x, y_{P}=t y$. Substituting into the equations of $C$ and $l$ yields $\frac{t x^{2}}{24}+\frac{t y^{2}}{16}=1=\frac{t x}{12}+\frac{t y}{8}$. Removing $t$ and rearranging gives $\frac{(x-1)^{2}}{\frac{5}{2}}+\frac{(y-1)^{2}}{\frac{5}{3}}=1$. Therefore, the required trajectory is an ellipse, with its center at $(1,1)$, and its axes of symmetry parallel to the $x$-axis and $y$-axis, respectively, with the semi-major axis $a=\frac{\sqrt{10}}{2}$ and the semi-minor axis $b=\frac{\sqrt{15}}{3}$.