16. Let $P$ be a moving point on the circle $C_{1}: x^{2}+y^{2}=2$, and let $Q$ be the foot of the perpendicular from $P$ to the $x$-axis. Point $M$ satisfies $\sqrt{2} \overrightarrow{M Q}=\overrightarrow{P Q}$.
(1) Find the equation of the trajectory $C_{2}$ of point $M$;
(2) Draw two tangents from a point $T$ on the line $x=2$ to the circle $C_{1}$, and let the points of tangency be $A$ and $B$. If the line $AB$ intersects the curve $C_{2}$ from (1) at points $C$ and $D$, find the range of $\frac{|C D|}{|A B|}$.
16. (1) Let point $M(x, y)$.

From $\sqrt{2} \overrightarrow{M Q}=\overrightarrow{P Q}$, we know point $P(x, \sqrt{2} y)$.
Since point $P$ is on the circle $C_{1}: x^{2}+y^{2}=2$, thus, $x^{2}+2 y^{2}=2$,
which means the trajectory equation of point $M$ is $\frac{x^{2}}{2}+y^{2}=1$.
(2) Let point $T(2, t)$. Then the equation of the chord of contact $A B$ is $2 x+t y=2$.
Let points $C\left(x_{1}, y_{1}\right), D\left(x_{2}, y_{2}\right)$.
Then the distance from the center $O$ to $A B$ is $d=\frac{2}{\sqrt{4+t^{2}}}$.
Thus, $|A B|=2 \sqrt{r^{2}-d^{2}}=2 \sqrt{\frac{2 t^{2}+4}{t^{2}+4}}$.
Also, $\left\{\begin{array}{l}2 x+t y=2, \\ \frac{x^{2}}{2}+y^{2}=1,\end{array}\right.$ gives $\left(t^{2}+8\right) y^{2}-4 t y$ then $y_{1}+y_{2}=\frac{4 t}{t^{2}+8}, y_{1} y_{2}=-\frac{4}{t^{2}+8}$.
Thus, $|C D|=\sqrt{1+\frac{t^{2}}{4}}\left|y_{1}-y_{2}\right|$
$=\frac{2 \sqrt{t^{2}+4} \cdot \sqrt{2 t^{2}+8}}{t^{2}+8}$.
Therefore, $\frac{|A B|}{|C D|}=\frac{\left(t^{2}+8\right) \sqrt{t^{2}+2}}{\left(t^{2}+4\right) \sqrt{t^{2}+4}}$.
Let $t^{2}+4=s$. Then $s \geqslant 4$. Thus,
$\frac{|A B|}{|C D|}=\sqrt{\frac{s^{3}+6 s^{2}-32}{s^{3}}}=\sqrt{1+\frac{6}{s}-\frac{32}{s^{3}}}$.
Let $\frac{1}{s}=m\left(m \in\left(0, \frac{1}{4}\right]\right)$. Thus,
$\frac{|A B|}{|C D|}=\sqrt{1+6 m-32 m^{3}}$.
Let $f(m)=1+6 m-32 m^{3}$. Then
$f^{\prime}(m)=6-96 m^{2}$.
Let $f^{\prime}(m)=0 \Rightarrow m=\frac{1}{4}$.
Thus, $f(m)$ is monotonically increasing on $\left(0, \frac{1}{4}\right]$. Hence, $f(m) \in$ $(1, \sqrt{2}]$, which means the range of $\frac{|C D|}{|A B|}$ is $\left[\frac{\sqrt{2}}{2}, 1\right)$.