Example 9 Given the parabola $C_{1}: y^{2}=4 a x(a>0)$, the ellipse $C$ is centered at the origin, with the focus $F$ of the parabola $C_{1}$ as its right focus, and the ratio of the major axis to the minor axis is $\sqrt{2}$. A line $l$ with an inclination angle of $\frac{\pi}{4}$ is drawn through $F$, intersecting the ellipse $C$ at point $P$ (point $P$ is above the $x$-axis), and the parabola $C_{1}$ at $Q$ (point $Q$ is below the $x$-axis),
(1) Find the coordinates of $P$ and $Q$;
(2) Move point $Q$ upward along the line $l$ to $Q^{\prime}$, such that $\left|Q Q^{\prime}\right|=4 a$, find the equation of the hyperbola passing through $P$ and $Q^{\prime}$, with the center at the origin and the axes of symmetry being the coordinate axes;
(3) Let $A(t, 0)(t$ be a constant, and $t>4)$, when $a$ varies in the closed interval $[1,2]$, find the maximum value of the area of $\triangle A P Q$ and the corresponding value of $a$.
(1) According to the problem, the equation of the ellipse $C$ is $\frac{x^{2}}{2 a^{2}}+\frac{y^{2}}{a^{2}}=1$,
and the equation of the line $l$ is $y=x-a$,
solving these equations simultaneously yields $P\left(\frac{4}{3} a, \frac{1}{3} a\right), Q((3-2 \sqrt{2})$ $a,(2-2 \sqrt{2}) a)$.
(2) Let $Q^{\prime}(x, y)$, then $x>(3-2 \sqrt{2}) a$, and $\left|Q Q^{\prime}\right|=\sqrt{2}|x-(3-2 \sqrt{2}) a|=4 a \Rightarrow x=3 a$,
so $y=2 a$, i.e., $Q^{\prime}(3 a, 2 a)$.
Assume the equation of the hyperbola is $m x^{2}-n y^{2}=1$,
from $\left\{\begin{array}{l}9 m a^{2}-4 n a^{2}=1, \\ \frac{16}{9} m a^{2}-\frac{1}{9} n a^{2}=1,\end{array} \Rightarrow\left\{\begin{array}{l}m=\frac{7}{11 a^{2}}, \\ n=\frac{13}{11 a^{2}},\end{array}\right.\right.$
thus the required equation is $\frac{7 x^{2}}{11 a^{2}}-\frac{13 y^{2}}{11 a^{2}}=1$.
(3) $S_{\triangle A P Q}=S_{\triangle A P F}+S_{\triangle A Q F}=\frac{1}{2}|t-a|$ $\left|\frac{1}{3} a+(3-2 \sqrt{2}) a\right|$,
$\because a \in[1,2], t>4$,
$\therefore S_{\triangle A P Q}=\left(\sqrt{2}-\frac{5}{6}\right)\left(-a^{2}+t a\right)=$ $\left(\sqrt{2}-\frac{5}{6}\right)\left[-\left(a-\frac{t}{2}\right)^{2}+\frac{t^{2}}{4}\right]$
$\because\left[-\left(a-\frac{t}{2}\right)^{2}+\frac{t^{2}}{4}\right]$ is a concave function on $a \in[1,2]$,

so when and only when $a=2$, $S_{\triangle A P Q}$ reaches its maximum value $\left(\sqrt{2}-\frac{5}{6}\right)(2 t-4)$