22. Let $F$ be the focus of the parabola $y^{2}=2 p x(p>0)$, and $A, B$ be two different points on the parabola. Also, let the midpoint of $A B$ be $(a, t)$.
(1) When $a$ is a non-zero constant, prove that the perpendicular bisector $l$ of $A B$ passes through a fixed point $N$;
(2) When $|F N|=2 a(a>1)$, find the maximum value of the area of $\triangle A N B$.


22. (1) Let the coordinates of $A, B$ be $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$, respectively. Since $M$ is the midpoint of $A B$, then
$2 a=x_{1}+x_{2}, 2 t=y_{1}+y_{2}$.
$\because A, B$ lie on the parabola, $\quad \therefore y_{1}^{2}=2 p x_{1}, y_{2}^{2}=2 p x_{2}$.
$1^{\circ}$ When $t \neq 0$, let the slope of line $A B$ be $k(k \neq 0)$, then
$k=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{2 p\left(y_{1}-y_{2}\right)}{y_{1}^{2}-y_{2}^{2}}=\frac{2 p}{y_{1}+y_{2}}=\frac{p}{t}$.
Therefore, the equation of the perpendicular bisector of $A B$ is $l: y-t=-\frac{1}{k}(x-a)=-\frac{t}{p}(x-a)$.
That is, $p \cdot y=t(p+a-x)$.
From (4), when $x=p+a$, $y=0$ (independent of $t$), hence line $l$ passes through the fixed point $N(p+a, 0)$. $2^{\circ}$ When $t=0$, at this moment $A, B$ are symmetric about the $x$-axis, $l$ is the $x$-axis, also passing through point $N(p+a, 0)$.
Combining $1^{\circ}$ and $2^{\circ}$, the conclusion holds.
(2) Since $F$ is the focus of the parabola $y^{2}=2 p x$, then $F\left(\frac{p}{2}, 0\right)$. And $|F N|=2 a$, i.e., $\left|p+a-\frac{p}{2}\right|=2 a$
$\therefore p=2 a$, thus the equation of the parabola is $y^{2}=4 a$.
$1^{\circ}$ When $t \neq 0$, the equation of $A B$ can be found as $2 a x-t y+t^{2}-2 a^{2}=0$.
Combining (5) and (6), eliminating $x$ and rearranging gives $y^{2}-2 t y+2\left(t^{2}-2 a^{2}\right)=0$.
Since $y_{1}, y_{2}$ are the two real roots of equation (7), then $y_{1}+y_{2}=2 t, y_{1} \cdot y_{2}=2\left(t^{2}-2 a^{2}\right)$,
and $\Delta=4 t^{2}-4 \cdot 2\left(t^{2}-2 a^{2}\right)=4\left(4 a^{2}-t^{2}\right)>0$, thus $-2 a < t < 2 a$.
$2^{\circ}$ When $t=0$, $A B$ is the line $x=a$, and $|A B|=4 a$.
Therefore, $S_{\triangle A B N}=\frac{1}{2} \cdot 4 a \cdot a=2 a^{2}$.
$3^{\circ}$ When $t \neq 0$, $|A B|=\sqrt{1+\frac{4 a^{2}}{t^{2}}}|y_{1}-y_{2}|=\sqrt{1+\frac{4 a^{2}}{t^{2}}} \cdot \sqrt{4 t^{2}-8\left(t^{2}-2 a^{2}\right)}=4 \sqrt{2 a^{2}-t^{2}}$.
The distance from $N$ to $A B$ is $d=\frac{\left|2 a(p+a)-t^{2}+2 a^{2}\right|}{\sqrt{4 a^{2}+t^{2}}}=\frac{4 a^{2}}{\sqrt{4 a^{2}+t^{2}}}$.
Thus, $S_{\triangle A B N}=\frac{1}{2} \cdot 4 \sqrt{2 a^{2}-t^{2}} \cdot \frac{4 a^{2}}{\sqrt{4 a^{2}+t^{2}}}=8 a^{2} \cdot \frac{\sqrt{2 a^{2}-t^{2}}}{\sqrt{4 a^{2}+t^{2}}}$.
Let $u=\frac{t}{a}$, then $-2<u<2$, and $S_{\triangle A B N}=8 a^{2} \cdot \frac{\sqrt{2-u^{2}}}{\sqrt{4+u^{2}}}$.
Let $f(u)=\frac{\sqrt{2-u^{2}}}{\sqrt{4+u^{2}}}$, then $f(u)$ is maximized when $u=0$, and the maximum value is $\frac{\sqrt{2}}{2}$.
Therefore, the maximum value of $S_{\triangle A B N}$ is $8 a^{2} \cdot \frac{\sqrt{2}}{2} = 4 a^{2} \sqrt{2}$.
Combining $2^{\circ}$ and $3^{\circ}$, the maximum value of $S_{\triangle A B N}$ is $(2+a)^{\frac{3}{2}}$.