21. As shown in the figure, curve $C_{1}$ is part of an ellipse centered at the origin $O$ with foci $F_{1}, F_{2}$, and curve $C_{2}$ is part of a parabola with vertex at $O$ and focus at $F_{2}(1,0)$. Point $A\left(\frac{3}{2}, \sqrt{6}\right)$ is the intersection of curves $C_{1}$ and $C_{2}$.
(1) Find the equations of the ellipse and the parabola on which curves $C_{1}$ and $C_{2}$ lie;
(2) Draw a line through $F_{2}$ that is not perpendicular to the $x$-axis, intersecting curves $C_{1}$ and $C_{2}$ at points $B, C, D, E$ in sequence.
(i) Find the range of the area of $\triangle C D F_{1}$;
(ii) If $G$ is the midpoint of $C D$ and $H$ is the midpoint of $B E$, is $\frac{|B E| \cdot\left|G F_{2}\right|}{|C D| \cdot\left|H F_{2}\right|}$ a constant? If so, find the constant; if not, explain why.
(1) According to the problem, for the ellipse we have $\left\{\begin{array}{l}a^{2}-b^{2}=1, \\ \frac{9}{4 a^{2}}+\frac{6}{b^{2}}=1\end{array} \Rightarrow\left\{\begin{array}{l}a^{2}=9, \\ b^{2}=8 .\end{array}\right.\right.$

Thus, the equation of the ellipse is $\frac{x^{2}}{9}+\frac{y^{2}}{8}=1$, and the equation of the parabola is $y^{2}=4 x$.
(2) (i ) Let the line $l: x=n y+1$, and combine $\left\{\begin{array}{l}x=n y+1, \\ y^{2}=4 x\end{array} \Rightarrow y^{2}-4 n y-4=0\right.$.

From $k_{F_{2} A}=2 \sqrt{6} \Rightarrow 0<n^{2}<\frac{1}{24}$,
Let $C\left(x_{1}, y_{1}\right), D\left(x_{2}, y_{2}\right)$, then $S_{\triangle C D F_{1}}=\left|y_{1}-y_{2}\right|=\sqrt{16 n^{2}+16}=4 \sqrt{n^{2}+1} \in\left(4, \frac{5 \sqrt{6}}{3}\right)$.
(ii ) Combine $\left\{\begin{array}{l}x=n y+1, \\ 8 x^{2}+9 y^{2}=72\end{array} \Rightarrow\left(8 n^{2}+9\right) y^{2}+16 n y-64=0\right.$.
Let $B\left(x_{3}, y_{3}\right), E\left(x_{4}, y_{4}\right)$, then $\left|y_{3}-y_{4}\right|=\frac{\sqrt{16^{2} n^{2}+4 \cdot 64\left(8 n^{2}+9\right)}}{8 n^{2}+9}=\frac{48 \sqrt{n^{2}+1}}{8 n^{2}+9}$.
Also, $y_{G}=2 n, y_{H}=\frac{-8 n}{8 n^{2}+9}$,
Therefore, $\frac{|B E| \cdot\left|G F_{2}\right|}{|C D| \cdot\left|H F_{2}\right|}=\frac{\left|y_{3}-y_{4}\right| \cdot\left|y_{G}\right|}{\left|y_{1}-y_{2}\right| \cdot\left|y_{H}\right|}=\frac{\frac{48 \sqrt{n^{2}+1}}{8 n^{2}+9} \cdot|2 n|}{4 \sqrt{n^{2}+1} \cdot \frac{|-8 n|}{8 n^{2}+9}}=\frac{48 \cdot 2}{4 \cdot 8}=3$ is a constant.