(15) From any point \( P \) on the line \( l: \frac{x}{12}+\frac{y}{8}=1 \), draw tangents \( PA \) and \( PB \) to the ellipse \( C: \frac{x^{2}}{24}+\frac{y^{2}}{16}=1 \), with points of tangency \( A \) and \( B \), respectively. Find the locus of the midpoint \( M \) of segment \( AB \).
(15) Let a point on $l$ be $P\left(x_{0}, y_{0}\right)$, the coordinates of the tangent points be $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$, and the midpoint of segment $A B$ be $M(x, y)$, so $x=\frac{x_{1}+x_{2}}{2}, y=$ $\frac{y_{1}+y_{2}}{2}$. Then the equation of the tangent line $P A$ is $\frac{x_{1} x}{24}+\frac{y_{1} y}{16}=1$, and the equation of the tangent line $P B$ is $\frac{x_{2} x}{24}+$ $\frac{y_{2} y}{16}=1$.
Substituting $P\left(x_{0}, y_{0}\right)$ into these equations, we get $\left\{\begin{array}{l}\frac{x_{1} x_{0}}{24}+\frac{y_{1} y_{0}}{16}=1, \\ \frac{x_{2} x_{0}}{24}+\frac{y_{2} y_{0}}{16}=1 .\end{array}\right.$
Therefore, the equation of the chord $A B$ is $\frac{x_{0} x}{24}+\frac{y_{0} y}{16}=1$.
(1) When $y_{0} \neq 0, x_{0} \neq 0$, we have $k_{A B}=-\frac{2 x_{0}}{3 y_{0}}$, so $\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=$ $-\frac{2 x_{0}}{3 y_{0}}$.
Since points $A$ and $B$ are on the ellipse, we have $\left\{\begin{array}{l}\frac{x_{1}^{2}}{24}+\frac{y_{1}^{2}}{16}=1, \\ \frac{x_{2}^{2}}{24}+\frac{y_{2}^{2}}{16}=1,\end{array}\right.$

Subtracting these equations, we get
$$
\frac{y_{1}-y_{2}}{x_{1}-x_{2}} \cdot \frac{y_{1}+y_{2}}{x_{1}+x_{2}}=-\frac{2}{3},
$$

Thus,
$$
-\frac{2 x_{0}}{3 y_{0}} \cdot \frac{y_{1}+y_{2}}{x_{1}+x_{2}}=-\frac{2}{3},
$$

Therefore,
$$
\frac{\frac{y_{1}+y_{2}}{2}}{\frac{x_{1}+x_{2}}{2}}=\frac{y_{0}}{x_{0}},
$$

which means $k_{O M}=k_{O P}$. Hence, the intersection point of line $O P$ and the chord $A B$ is the midpoint $M(x, y)$ of segment $A B$.
(2) When $y_{0}=0$ or $x_{0}=0$, the conclusion also holds.
Solving the system of equations $\left\{\begin{array}{l}\frac{x_{0} x}{24}+\frac{y_{0} y}{16}=1, \\ y=\frac{y_{0} x}{x_{0}}, \\ \frac{x_{0}}{12}+\frac{y_{0}}{8}=1\end{array} \Rightarrow \frac{x^{2}}{24}+\frac{y^{2}}{16}=\frac{x}{12}+\frac{y}{8}\right.$,

we get
$$
\frac{(x-1)^{2}}{\frac{5}{2}}+\frac{(y-1)^{2}}{\frac{5}{3}}=1 .
$$

Therefore, the locus of point $M$ is an ellipse centered at $(1,1)$, with semi-major and semi-minor axes of $\frac{\sqrt{10}}{2}$ and $\frac{\sqrt{15}}{3}$, respectively, and the major axis parallel to the $x$-axis (excluding the origin).
(Supplied by Cao Xiangjiang, Xinjiang Experimental High School)