5. Given the ellipse $\frac{x^{2}}{24}+\frac{y^{2}}{16}=1$, the line $l: \frac{x}{12}+\frac{y}{8}=$ $1, P$ is a point on $l$, the ray $O P$ intersects the ellipse at point $R$, and point $Q$ is on the ray $O P$ such that $|O Q| \cdot|O P|=|O R|^{2}$. When point $P$ moves along the line $l$, find the equation of the trajectory of point $Q$, and describe what the trajectory is.

When point $P$ moves along the line $l$, find the equation of the trajectory of point $Q$, and describe what the trajectory is.
(In the polar coordinate system, the equations of the line $l$ and the ellipse are
$$
\rho=\frac{24}{2 \cos \theta+3 \sin \theta}, \rho^{2}=\frac{48}{2+\sin ^{2} \theta} .
$$

Let $Q(\rho, \theta) 、 R\left(\rho_{1}, \theta\right) 、 P\left(\rho_{2}, \theta\right)$. Then
$$
\rho=|O Q| \neq 0 \text { and } \rho \rho_{2}=\rho_{1}^{2} \text {. }
$$

From this, the equation of the trajectory of point $Q$ is
$$
\rho\left(2+\sin ^{2} \theta\right)-2(2 \cos \theta+3 \sin \theta)=0(\rho \neq 0) \text {. }
$$

Converting to the corresponding Cartesian coordinate system and completing the square, we get
$$
\frac{(x-1)^{2}}{\frac{5}{2}}+\frac{(y-1)^{2}}{\frac{5}{3}}=1 \text {. }
$$

Therefore, its trajectory is an ellipse (excluding the origin), 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 being $\frac{\sqrt{10}}{2}$ and the semi-minor axis being $\frac{\sqrt{15}}{3}$.)