21. (12 points) As shown in Figure 4, curve $C_{1}$ is part of an ellipse centered at the origin $O$ with foci $F_{1}$ and $F_{2}$. 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 parabola on which curves $C_{1}$ and $C_{2}$ lie.
(2) Draw a line through point $F_{2}$ that is not perpendicular to the $x$-axis, intersecting curves $C_{1}$ and $C_{2}$ at points $B$, $C$, $D$, and $E$.
(i) Determine the range of the area of $\triangle C D F_{1}$.
(ii) If $G$ and $H$ are the midpoints of $C D$ and $B E$ respectively, is $\frac{|B E|\left|G F_{2}\right|}{|C D|\left|H F_{2}\right|}$ a constant? If so, find the constant; if not, explain why.
21. (1) Let the equation of the parabola be $y^{2}=2 p x$. Given that $F_{2}(1,0)$ is the focus, we have
$$
p=2 \Rightarrow y^{2}=4 x \text {. }
$$

Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}-1}=1$.
Substituting the point $\left(\frac{3}{2}, \sqrt{6}\right)$ into the above equation gives $a^{2}=9$.
Thus, the equation of the ellipse is $\frac{x^{2}}{9}+\frac{y^{2}}{8}=1$.
(2) (i) Let the equation of the line be $x=m y+1$. Then
$$
\begin{array}{l}
m \in\left(-\frac{\sqrt{6}}{12}, 0\right) \cup\left(0, \frac{\sqrt{6}}{12}\right) . \\
\text { By }\left\{\begin{array}{l}
y^{2}=4 x, \\
x=m y+1
\end{array}\right. \\
\Rightarrow y^{2}-4 m y-4=0 \\
\Rightarrow y_{1}+y_{2}=4 m, y_{1} y_{2}=-4 .
\end{array}
$$

Let $C\left(x_{1}, y_{1}\right), D\left(x_{2}, y_{2}\right)$. Then
$$
\begin{array}{l}
S_{\triangle F_{1} C D}=\frac{1}{2} \times 2\left|y_{1}-y_{2}\right| \\
=\sqrt{\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}} \\
=4 \sqrt{m^{2}+1} .
\end{array}
$$

Since $m^{2} \in\left(0, \frac{1}{24}\right)$, we have
$S_{\triangle F_{1} C D} \in\left(4, \frac{5 \sqrt{6}}{3}\right)$.
(ii) Let $B\left(x_{3}, y_{3}\right), E\left(x_{4}, y_{4}\right)$.

By $\left\{\begin{array}{l}\frac{x^{2}}{9}+\frac{y^{2}}{8}=1, \\ x=m y+1\end{array}\right.$,
$$
\begin{array}{l}
\Rightarrow\left(8 m^{2}+9\right) y^{2}+16 m y-64=0 \\
\Rightarrow y_{3}+y_{4}=-\frac{16 m}{8 m^{2}+9}, y_{3} y_{4}=-\frac{64}{8 m^{2}+9} \\
\Rightarrow\left|y_{3}-y_{4}\right|=\sqrt{\frac{256 \times 9\left(m^{2}+1\right)}{\left(8 m^{2}+9\right)^{2}}} \\
\quad=\frac{48 \sqrt{m^{2}+1}}{8 m^{2}+9} \\
\Rightarrow \frac{|B E|\left|G F_{2}\right|}{|C D|\left|H F_{2}\right|}=\frac{\left|y_{3}-y_{4}\right|}{\left|y_{1}-y_{2}\right|} \cdot \frac{\left|y_{1}+y_{2}\right|}{\left|y_{3}+y_{4}\right|}=3 .
\end{array}
$$