15. In the Cartesian coordinate system, it is known that the parabola $\Gamma: y=a x^{2}$ (where $a$ is a given positive real number). For a given positive real number $b$, if a circle with center $P(0, b)$ and variable radius $r(r>0)$ intersects the parabola $\Gamma$ at four points $A, B, C, D$, find the maximum value of the area $S$ of the quadrilateral with vertices $A, B, C, D$.
15. Let the circle with center $P(0, b)$ and variable radius $r$ be $\Gamma_{r}$. Assume points $A, B, C, D$ are arranged in a clockwise order, and $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$ are in the first quadrant. Then $x_{1}>x_{2}>0, y_{1}>y_{2}>0$.

Since the equation of the circle $\Gamma_{r}$ is $x^{2}+(y-b)^{2}=r^{2}$, and it intersects with $y=a x^{2}$, we get
$$
y^{2}+\left(\frac{1}{a}-2 b\right) y+b^{2}-r^{2}=0 .
$$

By Vieta's formulas, we have
$y_{1}+y_{2}=2 b-\frac{1}{a}=\frac{2 a b-1}{a}>0$,
$y_{1} y_{2}=b^{2}-r^{2}>0$.
Thus, $4 a b-1>2 a b-1>0, r>0$
$\Rightarrow \frac{\sqrt{4 a b-1}}{2 a}<r<b$.
By symmetry, quadrilateral $A B C D$ is an isosceles trapezoid.
Therefore, $S=\frac{1}{2}\left(2 x_{1}+2 x_{2}\right)\left(y_{1}-y_{2}\right)$
$$
=\left(x_{1}+x_{2}\right)\left(y_{1}-y_{2}\right) \text {. }
$$

Next, we find the maximum value of $S$.
Substituting $y_{1}=a x_{1}^{2}, y_{2}=a x_{2}^{2}$ into the above equation, we get
$$
\begin{array}{l}
S=a\left(x_{1}+x_{2}\right)^{2}\left(x_{1}-x_{2}\right) \\
=2 a \sqrt{\frac{\left(x_{1}+x_{2}\right)^{2}}{2} \cdot \frac{\left(x_{1}+x_{2}\right)^{2}}{2}\left(x_{1}-x_{2}\right)^{2}} \\
\leqslant 2 a\left(\frac{2\left(x_{1}^{2}+x_{2}^{2}\right)}{3}\right)^{\frac{3}{2}}=2 a\left(\frac{2\left(y_{1}+y_{2}\right)}{3 a}\right)^{\frac{3}{2}} \\
=2 a\left(\frac{2(2 a b-1)}{3 a^{2}}\right)^{\frac{3}{2}}=\frac{4 \sqrt{6}}{9} \cdot \frac{(2 a b-1)^{\frac{3}{2}}}{a^{2}} .
\end{array}
$$

The equality holds if and only if
$$
\begin{array}{l}
\frac{\left(x_{1}+x_{2}\right)^{2}}{2}=\left(x_{1}-x_{2}\right)^{2} \\
\Leftrightarrow x_{1}^{2}+x_{2}^{2}=6 x_{1} x_{2} \\
\Rightarrow \frac{2 a b-1}{a^{2}}=6 \sqrt{\frac{b^{2}-r^{2}}{a}} \\
\Rightarrow r=\frac{\sqrt{(8 a b-1)(4 a b+1)}}{6 a} \in\left(\frac{\sqrt{4 a b-1}}{2 a}, b\right) .
\end{array}
$$

Therefore, when $r=\frac{\sqrt{(8 a b-1)(4 a b+1)}}{6 a}$, $S$ reaches its maximum value $\frac{4 \sqrt{6}(2 a b-1)^{\frac{3}{2}}}{9 a^{2}}$.