Example 13 Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, find the hyperbola that intersects this ellipse, such that the quadrilateral formed by their intersection points has the maximum area, and find the coordinates of the vertices of the corresponding quadrilateral.
Let the equation of the hyperbola be $\frac{x^{2}}{m^{2}}-\frac{y^{2}}{n^{2}}=1(m>0, n>0)$. Given the condition $c^{2}=a^{2}-b^{2}=m^{2}+n^{2}$, solving the system of equations 
\[
\left\{\begin{array}{l}
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, \\
\frac{x^{2}}{m^{2}}-\frac{y^{2}}{c^{2}-m^{2}}=1,
\end{array}\right.
\]
we find that the coordinates of the intersection points of the ellipse and the hyperbola satisfy $|x|=\frac{a m}{c},|y|=\frac{b m}{c}$. Since the ellipse and the hyperbola are symmetric about the $x$-axis and $y$-axis, the quadrilateral formed by their intersection points is a rectangle, and thus its area is
\[
S=4|x y|=\frac{4 a b m n}{c^{2}} \leqslant \frac{2 a b\left(m^{2}+n^{2}\right)}{c^{2}}=2 a b,
\]
with equality holding if and only if $m=n=\frac{c}{\sqrt{2}}=\sqrt{\frac{a^{2}-b^{2}}{2}}$. Therefore, the equation of the hyperbola is
\[
\frac{x^{2}}{\frac{a^{2}-b^{2}}{2}}-\frac{y^{2}}{\frac{a^{2}-b^{2}}{2}}=1,
\]
and the coordinates of the four vertices of the corresponding quadrilateral are $\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right),\left(-\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right),\left(-\frac{a}{\sqrt{2}},-\frac{b}{\sqrt{2}}\right),\left(\frac{a}{\sqrt{2}},-\frac{b}{\sqrt{2}}\right)$.