14. Connecting the intersection points of the circle $x^{2}+y^{2}=9$ and the hyperbola $x y=3$ in sequence, a convex quadrilateral is formed. The area of this convex quadrilateral is
Answer $6 \sqrt{5}$.
Analysis Let $A(m, n)(m>0, n>0)$ be an intersection point of the two curves in the first quadrant. Since the two curves are symmetric about the origin and also about the line $y=x$, the coordinates of the other three intersection points are $B(n, m), C(-m,-n), D(-n,-m)$, so the quadrilateral $A B C D$ is a rectangle, and its area
$$
\begin{aligned}
S & =|A B| \cdot|C D|=\sqrt{2(m-n)^{2}} \cdot \sqrt{2(m+n)^{2}} \\
& =2 \sqrt{\left(m^{2}+n^{2}-2 m n\right)\left(m^{2}+n^{2}+2 m n\right)} \\
& =2 \sqrt{(9-6)(9+6)}=6 \sqrt{5}
\end{aligned}
$$