$$
\begin{array}{l}
\text { 9. Let point } A(\sec \theta, \tan \theta)\left(0<\theta<\frac{\pi}{2}\right) \text { be a vertex of square } A B C D \text {, and point } D \text { lies on the hyperbola } \\
\frac{x^{2}}{4}-\frac{y^{2}}{9}=1 \\
\text { Thus, } x_{D}=\frac{2}{3} \sqrt{9+\tan ^{2} \theta}, y_{D}=\tan \theta. \\
\text { Then }|A D|=\left|x_{D}-x_{A}\right| \\
=\frac{2}{3} \sqrt{9+\tan ^{2} \theta}-\sec \theta . \\
\text { Since }|A B|=|A D| \\
\Rightarrow 2 \tan \theta=\frac{2}{3} \sqrt{9+\tan ^{2} \theta}-\sec \theta \\
\Rightarrow 4\left(9+\tan ^{2} \theta\right)=9(2 \tan \theta+\sec \theta)^{2} \\
\Rightarrow 68 \sin ^{2} \theta+36 \sin \theta-27=0 \\
\Rightarrow \sin \theta=\frac{-9+\sqrt{540}}{34} .
\end{array}
$$

Therefore, the area of square $A B C D$
$$
S=4 \tan ^{2} \theta=\frac{4 \sin ^{2} \theta}{1-\sin ^{2} \theta}=0.8506 .
$$
9. As shown in Figure 2, on the right branches of the hyperbolas $x^{2}-y^{2}=1$ and $\frac{x^{2}}{4}-\frac{y^{2}}{9}=1$, there are points $A, B$ and $C, D$ respectively, such that quadrilateral $A B C D$ is a square with sides parallel to the coordinate axes. Find the area of square $A B C D$ (accurate to 0.0001).