11. Let point $P$ be a moving point on the circle $C_{1}: x^{2}+y^{2}=2$, and draw a perpendicular from point $P$ to the $x$-axis, with the foot of the perpendicular being $Q$. Point $M$ satisfies $\sqrt{2} M Q=\stackrel{\text { mu }}{P}$. (1) Find the equation of the trajectory $C_{2}$ of $M$ (2) Draw two tangents from a point $T$ on the line $x=2$ to the circle $C_{1}$, with the points of tangency being $A$ and $B$. If the line $AB$ intersects the curve $C_{2}$ from (1) at points $C$ and $D$, find the range of $\frac{|C D|}{|A B|}$.
11. Analysis: (1) Let point $P\left(x_{p}, y_{p}\right)$, then $M\left(x_{p}, \frac{\sqrt{2}}{2} y_{p}\right)$, and from $x_{p}^{2}+y_{p}^{2}=2$ we know that the trajectory equation of point $M$ is: $\frac{x^{2}}{2}+y^{2}=1$.
(2) Let $T(2, t), A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$, then the equation of the chord of contact $AB$ is: $y=-\frac{2 x-2}{t}$, substituting into the equation of $C_{1}$ we get: $\left(t^{2}+4\right) x^{2}-8 x+4-2 t^{2}=0$, by Vieta's formulas: $x_{1}+x_{2}=\frac{8}{t^{2}+4}, x_{1} x_{2}=\frac{4-2 t^{2}}{t^{2}+4}$. Combining the equation of line $AB$ with the trajectory equation of point $M$ we get: $\left(t^{2}+8\right) x^{2}-16 x+8-2 t^{2}=0$, by Vieta's formulas: $x_{3}+x_{4}=\frac{16}{t^{2}+8}, x_{3} x_{4}=\frac{8-2 t^{2}}{t^{2}+8}$.
Thus, $\frac{|C D|}{|A B|}=\sqrt{\frac{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}{\left(x_{3}+x_{4}\right)^{2}-4 x_{3} x_{4}}}=\frac{\left(t^{2}+4\right) \sqrt{t^{2}+4}}{\left(t^{2}+8\right) \sqrt{t^{2}+2}}$, let $s=t^{2}, s \in[0,+\infty)$, $f(s)=\frac{(s+4) \sqrt{s+4}}{(s+8) \sqrt{s+2}}$, then $f^{\prime}(s)=\frac{\sqrt{(s+4)(s+2)}\left(\frac{3}{2}(s+8)-(s+4)-\frac{(s+4)(s+8)}{2 s+4}\right)}{(s+8) \sqrt{s+2}}$.

Since $\frac{3}{2}(s+8)-(s+4)-\frac{(s+4)(s+8)}{2 s+4}=\frac{(s+16)(s+2)-(s+4)(s+8)}{2 s+4}=\frac{6 s}{2 s+4} \geq 0$ always holds, hence $f(s) \in\left[\frac{\sqrt{2}}{2}, 1\right)$, i.e., $\frac{|C D|}{|A B|} \in\left[\frac{\sqrt{2}}{2}, 1\right)$.