Question: Let the two endpoints $A$ and $B$ of a line segment $AB$ of length $l (l \geqslant 1)$ move on the parabola $y=x^{2}$, and let $M(x, y)$ be the midpoint of $AB$. Find the coordinates of point $M$ when it is closest to the $x$-axis.
Solution: Let the focus of the parabola $y=x^{2}$ be $F\left(0, \frac{1}{4}\right)$,

and the directrix $l: y=-\frac{1}{4}$.
To find the coordinates of point $M$ when it is closest to the $x$-axis, this is equivalent to finding the coordinates of point $M$ when it is closest to the directrix $l$.

Construct $Y_{1} \perp l, A A_{1} \perp l, B B_{1} \perp l$, and connect $A A^{\prime} 、 B F$, then
$$
\begin{aligned}
& \left|M M_{1}\right|=\frac{1}{2}\left(\left|A A_{1}\right|+\left|B B_{1}\right|\right) \\
= & \frac{1}{2}(|F A|+|F B|) \geqslant \frac{1}{2}|A B| . \\
\therefore \quad & \left|M_{1}\right| \mathrm{m}_{1 \mathrm{n}}=\frac{1}{2} l .
\end{aligned}
$$

Thus, $y_{\mathrm{min}}=\frac{1}{2} l-\frac{1}{4}=\frac{2 l-1}{4}$.
When $y=\frac{2 l-1}{4}$, $x= \pm \frac{\sqrt{2 l-1}}{2}$.
Therefore, the coordinates of point $M$ when it is closest to the $x$-axis are
$$
\left( \pm \frac{\sqrt{2 l-1}}{2}, \frac{2 l-1}{4}\right) .
$$