Example 7 Given $\overrightarrow{O A}=(2,0), \overrightarrow{O C}=\overrightarrow{A B}=(0, 1)$, a moving point $M$ is at a distance $d$ from the fixed line $y=1$, and $\overrightarrow{O M} \cdot \overrightarrow{A M}=k\left(\overrightarrow{C M} \cdot \overrightarrow{B M}-d^{2}\right)$, where $O$ is the origin, and $k$ is a parameter.
(1) Find the trajectory of the moving point $M$;
(2) When $k=\frac{1}{2}$, find the maximum and minimum values of $|\overrightarrow{O M}+2 \overrightarrow{A M}|$; $\square$
(3) If the trajectory of the moving point $M$ is a conic section, find the eccentricity.
Solution: (1) Let $M(x, y)$, then $\overrightarrow{O A}=(2,0), \overrightarrow{O C}=$
$$
\begin{array}{l}
\overrightarrow{A B}=(0,1), \therefore C(0,1), B(2,1), \overrightarrow{O M}=(x, y), \\
\overrightarrow{A M}=(x-2, y), \overrightarrow{C M}=(x, y-1), \overrightarrow{B M}= \\
(x-2, y-1), d=|y-1| .
\end{array}
$$

From $\overrightarrow{O M} \cdot \overrightarrow{A M}=k\left(\overrightarrow{C M} \cdot \overrightarrow{B M}-d^{2}\right)$ we get:
$(1-k) x^{2}+2(k-1) x+y^{2}=0$ as the equation of the trajectory of $M$.

When $k=1$, $y=0$, the trajectory is a straight line; when $k=0$, the trajectory is a circle; when $k>1$, the trajectory is a hyperbola; when $k<0$ or $0<k<1$, the trajectory of $M$ is an ellipse.
(2) When $k=\frac{1}{2}$, the trajectory equation of the moving point $M$ is $(x-1)^{2}+2 y^{2}=1$,

Thus $|\overrightarrow{O M}+2 \overrightarrow{A M}|^{2}=(3 x-4)^{2}+9 y^{2}=\frac{9}{2}$ $\left(x-\frac{5}{3}\right)^{2}+\frac{7}{2}$.
From $(x-1)^{2}+2 y^{2}=1$ we get $0 \leqslant x \leqslant 2$,
$\therefore$ When $x=\frac{5}{3}$, $|\overrightarrow{O M}+2 \overrightarrow{A M}|$ reaches its minimum value $\frac{\sqrt{14}}{2}$; when $x=0$, $|\overrightarrow{O M}+2 \overrightarrow{A M}|$ reaches its maximum value 4.
(3) Since $\frac{\sqrt{3}}{3} \leqslant e \leqslant \frac{\sqrt{2}}{2}$, the trajectory is an ellipse, with the equation $\frac{(x-1)^{2}}{1}+\frac{y^{2}}{1-k}=1$.

When $0<k<1$, $a^{2}=1, b^{2}=1-k, c^{2}=k, e^{2}=\frac{c^{2}}{a^{2}}=k$, thus $\frac{1}{3} \leqslant k \leqslant \frac{1}{2}$;

When $k<0$, $a^{2}=1-k, b^{2}=1, c^{2}=-k, e^{2}=\frac{c^{2}}{a^{2}}=\frac{-k}{1-k}$, thus $-1 \leqslant k \leqslant-\frac{1}{2}$.
In summary, $k \in\left[-1,-\frac{1}{2}\right] \cup\left[\frac{1}{3}, \frac{1}{2}\right]$.