** Given a line segment $l(l \geqslant 1)$, whose ends move on the parabola $y=x^{2}$, try to find:
(1) the equation of the trajectory of the midpoint $M$,
(2) the coordinates of the lowest point $M$ with respect to the $x$-axis.
(1) Choose the angle $\theta$ between $P_{1} P_{2}$ and the $x$-axis as the parameter. Let the coordinates of $P_{1}$ and $P_{2}$ be $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$, then $y_{1}=x_{1}^{2}$, $y_{2}=x_{2}^{2}, x_{2}-x_{1}=l \cos \theta, x_{2}^{2}-x_{1}^{2}=l \sin \theta$. Solving these, we get $x_{1}=\frac{1}{2}(\tan \theta-l \cos \theta), x_{2}=\frac{1}{2}(\tan \theta+l \cos \theta)$.
Let the coordinates of the midpoint $M$ be $(X, Y)$, then
$$
\left\{\begin{array}{l}
X=\frac{1}{2}\left(x_{1}+x_{2}\right)=\frac{1}{2} \tan \theta, \\
Y=\frac{1}{2}\left(y_{1}+y_{2}\right)=\frac{1}{4}\left(\tan ^{2} \theta+l^{2} \cos ^{2} \theta\right)
\end{array}\right.
$$

which is the parametric equation of the trajectory of $M$.
(2) Since
$$
\begin{aligned}
Y & =\frac{1}{4}\left(\tan ^{2} \theta+l^{2} \cos ^{2} \theta\right)=\frac{1}{4}\left(\frac{l^{2} \cos ^{4} \theta-\cos ^{2} \theta+1}{\cos ^{2} \theta}\right) \\
& =\frac{1}{4}\left[\frac{\left(l \cos ^{2} \theta-1\right)^{2}}{\cos ^{2} \theta}+2 l-1\right],
\end{aligned}
$$

$Y$ has a minimum value when $l \cos ^{2} \theta-1=0$, i.e., $\cos \theta= \pm \frac{1}{\sqrt{l}}$. The minimum value of $Y$ is $Y_{\text {min }}=\frac{1}{4}(2 l-1)$, at which time
$$
X= \pm \frac{\sqrt{\frac{1}{\cos ^{2} \theta}-1}}{2}= \pm \frac{1}{2} \sqrt{l-1} \text {, }
$$

Thus, the coordinates of the lowest point $M$ from the $x$-axis are $\left( \pm \frac{1}{2} \sqrt{l-1}, \frac{1}{4}(2 l-1)\right)$.