9. As shown in the figure, $ABCD$ is a rectangle, $AB=8, BC=4, M, N$ are the midpoints of $AB, CD$ respectively. A moving line $l$ is used as a fold to flip the part of the rectangle below it, so that after each flip, point $M$ lands on side $CD$, denoted as $M'$. Draw $M'P$ perpendicular to $CD$ intersecting line $l$ at point $P$. Let the trajectory of point $P$ be curve $E$.
(1) Establish an appropriate Cartesian coordinate system and find the equation of curve $E$;
(2) $F$ is a point on $MN$, $\overrightarrow{FN}=-3 \overrightarrow{FM}$, a line through point $F$ intersects curve $E$ at points $S, T$, and $\overrightarrow{SF}=\lambda \overrightarrow{FT}$, find the range of the real number $\lambda$.
(1) Connect $P M$. Since $l$ is the perpendicular bisector of segment $M M^{\prime}$, we have
$|P M|=\left|P M^{\prime}\right|$, which means the distance from point $P$ to point $M$ is equal to the distance from point $P$ to line $C D$. Therefore, the locus of point $P$ is a segment of a parabola with point $M$ as the focus and line $C D$ as the directrix.
Taking the midpoint $O$ of $M N$ as the origin, a line parallel to $C D$ as the $x$-axis, and line $M N$ as the $y$-axis, we establish a Cartesian coordinate system as shown in the figure. Since $M(0,-2), A(-4,-2), B(4,-2)$, the equation of curve $E$ is $x^{2}=-8 y(-4 \leqslant x \leqslant 4)$.
(2) $F(0,-1)$, let $S T: y=k x-1\left(-\frac{1}{4} \leqslant k \leqslant \frac{1}{4}\right), S\left(x_{1}, y_{1}\right), T\left(x_{2}, y_{2}\right)$, and solve the system of equations

\[
\left\{\begin{array}{l}
y=k x-1, \\
x^{2}=-8 y
\end{array} \Rightarrow x^{2}+8 k x-8=0\right.
\]

Then $\lambda=-\frac{x_{1}}{x_{2}} \Rightarrow \lambda+\frac{1}{\lambda}=-\frac{x_{1}^{2}+x_{2}^{2}}{x_{1} x_{2}}$ $=\frac{64 k^{2}+16}{8}=8 k^{2}+2 \in\left[2, \frac{5}{2}\right]$. Therefore, $\lambda+\frac{1}{\lambda} \leqslant \frac{5}{2} \Rightarrow \lambda \in\left[\frac{1}{2}, 2\right]$.