10. Curve $C_{1}$ is a part of an ellipse with the origin $O$ as the center and $F_{1}, F_{2}$ as the foci, and curve $C_{2}$ is a part of a parabola with $O$ as the vertex and $F_{2}$ as the focus. Point $A$ is the intersection of curves $C_{1}$ and $C_{2}$, and $\angle A F_{2} F_{1}$ is an obtuse angle. If $\left|A F_{1}\right|=\frac{7}{2}$, $\left|A F_{2}\right|=\frac{5}{2}$, (I) find the equations of the ellipse and parabola on which curves $C_{1}$ and $C_{2}$ lie;
(II) Draw a line through $F_{2}$ that is not perpendicular to the $x$-axis, intersecting curves $C_{1}$ and $C_{2}$ sequentially at points $B, C, D, E$. If $G$ is the midpoint of $CD$ and $H$ is the midpoint of $BE$, 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 the reason.
10. Analysis ~(I) Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, then $2 a=\left|A F_{1}\right|+\left|A F_{2}\right|=\frac{7}{2}+\frac{5}{2}=6$, which gives $a=3$. Let $A(x, y), F_{1}(-c, 0), F_{2}(c, 0)$, then $(x+c)^{2}+y^{2}=\left(\frac{7}{2}\right)^{2}$, $(x-c)^{2}+y^{2}=\left(\frac{5}{2}\right)^{2}$. Subtracting the two equations gives $x c=\frac{3}{2}$. By the definition of the parabola, $\left|A F_{2}\right|=x+c=\frac{5}{2}$, so $c=1, x=\frac{3}{2}$ or $x=1, c=\frac{3}{2}$ (discard).

Thus, the equation of the ellipse is $\frac{x^{2}}{9}+\frac{y^{2}}{8}=1$, and the equation of the parabola is $y^{2}=4 x$.
(II) Let $B\left(x_{1}, y_{1}\right), E\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$. Substituting the line $y=k(x-1)$ into $\frac{x^{2}}{9}+\frac{y^{2}}{8}=1$,

we get: $8\left(\frac{y}{k}+1\right)^{2}+9 y^{2}-72=0$,
which simplifies to $\left(8+9 k^{2}\right) y^{2}+16 k y-64 k^{2}=0$,

then $y_{1}+y_{2}=-\frac{16 k}{8+9 k^{2}}, y_{1} y_{2}=-\frac{64 k^{2}}{8+9 k^{2}}$.

Similarly, substituting $y=k(x-1)$ into $y^{2}=4 x$ gives: $k y^{2}-4 y-4 k=0$,
then $y_{3}+y_{4}=\frac{4}{k}, y_{3} y_{4}=-4$,
so $\frac{|B E| \cdot\left|G F_{2}\right|}{|C D| \cdot\left|H F_{2}\right|}=\frac{\left|y_{1}-y_{2}\right|}{\left|y_{3}-y_{4}\right|} \cdot \frac{\frac{1}{2}\left|y_{3}+y_{4}\right|}{\frac{1}{2}\left|y_{1}+y_{2}\right|}$
$=\sqrt{\frac{\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}}{\left(y_{1}+y_{2}\right)^{2}} \cdot \frac{\left(y_{3}+y_{4}\right)^{2}}{\left(y_{3}+y_{4}\right)^{2}-4 y_{3} y_{4}}}=3$ is a constant.