(1) 求点 $P$ 到椭圆上点的距离的最大值;
(2) 求 $|C D|$ 的最小值.

|【小问 1 详解】

$$|P Q|^2=12 \cos ^2 \theta+(1-\sin \theta)^2=13-11 \sin ^2 \theta-2 \sin \theta=-11\left(\sin \theta+\frac{1}{11}\right)^2+\frac{144}{11} \leq \frac{144}{11}$$
, 当且仅当 $\sin \theta=-\frac{1}{11}$ 时取等号, 故 $|P Q|$ 的最大值是 $\frac{12 \sqrt{11}}{11}$.
【小问 2 详解】

$\left(k^2+\frac{1}{12}\right) x^2+k x-\frac{3}{4}=0$, 设 $A\left(x_1, y_1\right), B\left(x_2, y_2\right)$, 所以 $\left\{\begin{array}{l}x_1+x_2=-\frac{k}{k^2+\frac{1}{12}} \\ x_1 x_2=-\frac{3}{4\left(k^2+\frac{1}{12}\right)}\end{array}\right.$

\begin{aligned} & |C D|=\sqrt{1+\frac{1}{4}}\left|x_C-x_D\right|=\frac{\sqrt{5}}{2}\left|\frac{4 x_1}{(2 k+1) x_1-1}-\frac{4 x_2}{(2 k+1) x_2-1}\right| \\ & =2 \sqrt{5}\left|\frac{x_1-x_2}{\left[(2 k+1) x_1-1\right]\left[(2 k+1) x_2-1\right]}\right|=2 \sqrt{5}\left|\frac{x_1-x_2}{(2 k+1)^2 x_1 x_2-(2 k+1)\left(x_1+x_2\right)+1}\right| \\ & =\frac{3 \sqrt{5}}{2} \cdot \frac{\sqrt{16 k^2+1}}{|3 k+1|}=\frac{6 \sqrt{5}}{5} \cdot \frac{\sqrt{16 k^2+1} \sqrt{\frac{9}{16}+1}}{|3 k+1|} \geq \frac{6 \sqrt{5}}{5} \times \frac{\sqrt{\left(4 k \times \frac{3}{4}+1 \times 1\right)^2}}{|3 k+1|}=\frac{6 \sqrt{5}}{5}, \end{aligned}

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