11. (20 points) As shown in Figure 1, for the ellipse $\Gamma: \frac{x^{2}}{4}+y^{2}=1$, two chords $A B$ and $C D$ satisfy $\overrightarrow{D C}=2 \overrightarrow{A B}$. Let the intersection of lines $A D$ and $B C$ be point $P$.
(1) Find the maximum value of $|A B|+|C D|$; 360227734
(2) If point $P$ lies on the parabola $y=\frac{1}{4} x^{2}-3$, find the maximum area of quadrilateral $A B C D$.
11. (1) Let $C(2 \cos \alpha, \sin \alpha)$, 
$D(2 \cos \beta, \sin \beta)$.
Then $|O C|^{2}=4 \cos ^{2} \alpha+\sin ^{2} \alpha$
$\leqslant 4 \cos ^{2} \alpha+4 \sin ^{2} \alpha=4$,
$|O D|^{2}=4 \cos ^{2} \beta+\sin ^{2} \beta$
$\leqslant 4 \cos ^{2} \beta+4 \sin ^{2} \beta=4$,
so $|O C| \leqslant 2,|O D| \leqslant 2$.
Notice that,
$|C D| \leqslant|O C|+|O D| \leqslant 4, Q$ : 3602277
$|A B|+|C D|=\frac{3}{2}|C D| \leqslant 6$,
When taking points $C(-2,0), D(2,0)$, the above equalities hold.
Therefore, the maximum value of $|A B|+|C D|$ is 6.
(2) Perform the coordinate transformation $\left\{\begin{array}{l}x^{\prime}=\frac{1}{2} x, \\ y^{\prime}=y,\end{array}\right.$ to get Figure 7.
Thus, the new parabola equation is $y=x^{2}-3$.
Let the area of quadrilateral $A B C D$ be $S$.
Take the midpoint $H$ of $C^{\prime} D^{\prime}$, and draw $O M \perp A^{\prime} D^{\prime}$ at point $M$.
Let $A^{\prime} M=D^{\prime} M=b, A^{\prime} P^{\prime}=2 b, O P^{\prime}=a$,
$P^{\prime}\left(t, t^{2}-3\right)$.
Then $a^{2}=t^{2}+\left(t^{2}-3\right)^{2}=t^{4}-5 t^{2}+9$
$$
=\left(t^{2}-\frac{5}{2}\right)^{2}+\frac{11}{4} \geqslant \frac{11}{4} \text {, }
$$

and $a=O P^{\prime}<O A^{\prime}+A^{\prime} D^{\prime}=1+A^{\prime} D^{\prime}<3$.
Thus, $a \in\left[\frac{\sqrt{11}}{2}, 3\right)$.
By $O M=\sqrt{1-b^{2}}$, in the right triangle $\triangle O M P^{\prime}$, we have
$1-b^{2}+9 b^{2}=a^{2}$
$\Rightarrow 8 b^{2}=a^{2}-1$.
Let $O H=y$. Then
$H D^{\prime}=\sqrt{1-y^{2}}$,

and
$P^{\prime} M \cdot P^{\prime} D^{\prime}=P^{\prime} O \cdot P^{\prime} H$
$\Rightarrow 12 b^{2}=a(a+y) \Rightarrow y=\frac{a^{2}-3}{2 a}$.
By $S=\frac{3}{4} S_{\triangle P C D}=\frac{3}{2} S_{\triangle P C^{\prime} D^{\prime}}$
$=\frac{3}{2} D^{\prime} H \cdot P^{\prime} H$
$=\frac{3}{2} \sqrt{1-\frac{\left(a^{2}-3\right)^{2}}{4 a^{2}}} \cdot \frac{3\left(a^{2}-1\right)}{2 a}$
$=\frac{9}{4} \sqrt{\left(1-\frac{\left(a^{2}-3\right)^{2}}{4 a^{2}}\right) \frac{\left(a^{2}-1\right)^{2}}{a^{2}}}$.
Let $f(x)=\left(1-\frac{(x-3)^{2}}{4 x}\right) \frac{(x-1)^{2}}{x}$
$=\frac{(9-x)(x-1)^{3}}{4 x^{2}}$,

where $x=a^{2} \in\left[\frac{11}{4}, 9\right)$.
By $f^{\prime}(x)=-\frac{(x-1)^{2}\left(x^{2}-4 x-9\right)}{2 x^{3}}$, we know that the maximum value of $f(x)$ is attained at $x=2+\sqrt{13} \in\left[\frac{11}{4}, 9\right)$, at which point,
$$
\begin{aligned}
S & =\frac{9}{4} \sqrt{f(2+\sqrt{13})} \\
& =\frac{9}{4} \times \frac{1+\sqrt{13}}{2(2+\sqrt{13})} \sqrt{(7-\sqrt{13})(1+\sqrt{13})} \\
& =\frac{(3 \sqrt{13}-6) \sqrt{2+2 \sqrt{13}}}{4} .
\end{aligned}
$$