11. (20 points) The ellipse $C_{1}$ and the hyperbola $C_{2}$ have common foci $( \pm c, 0)(c>0), C_{1}$ and $C_{2}$ have an eccentricity difference of no more than 1, and $C_{2}$ has an asymptote with a slope of no less than $\frac{4}{3}$. $C_{1}$ and $C_{2}$ intersect the positive x-axis at points $A$ and $B$, respectively, and the intersection point of the two curves in the first quadrant is $D$. Question: Does $\triangle A B D$ have a maximum area? If so, find the maximum area and provide the equations of $C_{1}$ and $C_{2}$; if not, explain the reason.
11. Let the eccentricities of the ellipse $C_{1}$ and the hyperbola $C_{2}$ be $e_{1}$ and $e_{2}$, respectively.
Then the equations of the ellipse $C_{1}$ and the hyperbola $C_{2}$ are
$\left(1-e_{1}^{2}\right) x^{2}+y^{2}=\frac{1-e_{1}^{2}}{e_{1}^{2}} c^{2}$,
$\left(1-e_{2}^{2}\right) x^{2}+y^{2}=\frac{1-e_{2}^{2}}{e_{2}^{2}} c^{2}$.
Let $D\left(x_{0}, y_{0}\right)$. By solving (1) and (2) simultaneously, we get
$y_{0}=\frac{c}{e_{1} e_{2}} \sqrt{\left(1-e_{1}^{2}\right)\left(e_{2}^{2}-1\right)}$.
Then $S_{\triangle A B C}=\frac{1}{2}\left(\frac{c}{e_{1}}-\frac{c}{e_{2}}\right) y_{0}$
$=\frac{1}{2} c^{2} \frac{e_{2}-e_{1}}{e_{1}^{2} e_{2}^{2}} \sqrt{\left(1-e_{1}^{2}\right)\left(e_{2}^{2}-1\right)}$.
Let $f\left(e_{1}, e_{2}\right)=\frac{e_{2}-e_{1}}{e_{1}^{2} e_{2}^{2}} \sqrt{\left(1-e_{1}^{2}\right)\left(e_{2}^{2}-1\right)}$.
From $e_{2}-e_{1} \leqslant 1 \Rightarrow e_{1} \geqslant e_{2}-1$
$\Rightarrow e_{1} \in\left[e_{2}-1,1\right)$.
From the slope of one asymptote of the hyperbola $C_{2}$ being no less than $\frac{4}{3}$, we know $e_{2} \geqslant \frac{5}{3} \Rightarrow e_{2} \in\left[\frac{5}{3}, 2\right)$.
Fixing $e_{2}$ in $f\left(e_{1}, e_{2}\right)$ and considering it as a function of $e_{1}$, it is easy to see that as $e_{1}$ increases, $e_{2}-e_{1}$ decreases, $\frac{1}{e_{1}^{2} e_{2}^{2}}$ decreases, and $\sqrt{\left(1-e_{1}^{2}\right)\left(e_{2}^{2}-1\right)}$ decreases. Therefore, $f\left(e_{1}, e_{2}\right)$ is decreasing on $e_{1} \in\left[e_{2}-1,1\right)$ when $e_{2}$ is fixed.
Thus, $f\left(e_{1}, e_{2}\right) \leqslant f\left(e_{2}-1, e_{2}\right)$
$$
\begin{array}{l}
=\frac{1}{\left(e_{2}-1\right)^{2} e_{2}^{2}} \sqrt{\left(2 e_{2}-e_{2}^{2}\right)\left(e_{2}^{2}-1\right)} \\
=\sqrt{\frac{\left(2-e_{2}\right)\left(e_{2}+1\right)}{e_{2}^{3}\left(e_{2}-1\right)^{3}}} .
\end{array}
$$

Notice that, $\left(2-e_{2}\right)\left(e_{2}+1\right)=-\left(e_{2}-\frac{1}{2}\right)^{2}+\frac{9}{4}$.
Hence, $\left(2-e_{2}\right)\left(e_{2}+1\right)$ is monotonically decreasing on $\left[\frac{5}{3}, 2\right)$.
Also, $\frac{1}{e_{2}^{3}\left(e_{2}-1\right)^{3}}$ is clearly monotonically decreasing on $\left[\frac{5}{3}, 2\right)$.
Thus, $\sqrt{\frac{\left(2-e_{2}\right)\left(e_{2}+1\right)}{e_{2}^{3}\left(e_{2}-1\right)^{3}}}$
$\leqslant \sqrt{\frac{\left(2-\frac{5}{3}\right)\left(\frac{5}{3}+1\right)}{\left(\frac{5}{3}\right)^{3}\left(\frac{2}{3}\right)^{3}}}=\frac{9 \sqrt{5}}{25}$.
Therefore, $f\left(e_{1}, e_{2}\right) \leqslant \frac{9 \sqrt{5}}{25} \Rightarrow S_{\triangle A B C} \leqslant \frac{9 \sqrt{5}}{50} c^{2}$.
Hence, the maximum value of $S_{\triangle A B C}$ is $\frac{9 \sqrt{5}}{50} c^{2}$, and it is achieved when $e_{2}=\frac{5}{3}, e_{1}=\frac{2}{3}$.
At this point, the equation of the ellipse $C_{1}$ is $\frac{x^{2}}{\frac{9}{4} c^{2}}+\frac{y^{2}}{\frac{5}{4} c^{2}}=1$, and the equation of the hyperbola $C_{2}$ is $\frac{x^{2}}{\frac{9}{25} c^{2}}-\frac{y^{2}}{\frac{16}{25} c^{2}}=1$.