11. Given the ellipse $C_{1}: \frac{x^{2}}{24}+\frac{y^{2}}{b^{2}}=1(0<b<2 \sqrt{6})$ with its right focus $F_{1}$ coinciding with the focus of the parabola $C_{2}: y^{2}=4 p x\left(p \in \mathbf{Z}_{+}\right)$, a line $l$ passing through $F_{1}$ with a positive integer slope intersects $C_{1}$ at points $A, B$ and $C_{2}$ at points $C, D$. If $13|A B|=\sqrt{6}|C D|$, find the values of $b$ and $p$.
Solve:
As shown in the figure, $p=c=\sqrt{24-b^{2}}$, let $l: x=n y+p$, where
$n=\frac{1}{k}$. By solving the system $\left\{\begin{array}{l}x=n y+p, \\ y^{2}=4 p x\end{array} \Rightarrow y^{2}-4 p n y-4 p^{2}=0\right.$
$\Rightarrow|C D|=\sqrt{n^{2}+1} \cdot \sqrt{16 p^{2} n^{2}+16 p^{2}}=4 p\left(n^{2}+1\right) ;$
By solving the system $\left\{\begin{array}{l}x=n y+c, \\ \frac{x^{2}}{24}+\frac{y^{2}}{b^{2}}=1\end{array} \Rightarrow b^{2}(n y+c)^{2}+24 y^{2}=24 b^{2}\right.$
$$
\begin{array}{l}
\Rightarrow\left(b^{2} n^{2}+24\right) y^{2}+2 b^{2} n c y+b^{2} c^{2}-24 b^{2}=0 \\
\Rightarrow|A B|=\sqrt{n^{2}+1} \cdot \frac{\sqrt{96 b^{2}\left(b^{2} n^{2}+24-c^{2}\right)}}{b^{2} n^{2}+24}=\frac{4 \sqrt{6} b^{2}\left(n^{2}+1\right)}{b^{2} n^{2}+24} .
\end{array}
$$

Thus, $\frac{13 \cdot 4 \sqrt{6} b^{2}\left(n^{2}+1\right)}{b^{2} n^{2}+24}=4 \sqrt{6} p\left(n^{2}+1\right) \Rightarrow \frac{13 b^{2}}{b^{2} n^{2}+24}=p \Rightarrow \frac{13 b^{2} k^{2}}{b^{2}+24 k^{2}}=p$ $\Rightarrow 13 k^{2}\left(24-p^{2}\right)=p\left(24-p^{2}+24 k^{2}\right)$. Since $p=c<\sqrt{24}<5$, only $p=4, k=2$ satisfy the condition. Therefore, $p=4, b=2 \sqrt{2}$.