2. Connecting the intersection points of $x^{2}+y^{2}=10$ and $y=\frac{4}{x}$ in sequence, a convex quadrilateral is formed. The area of this quadrilateral is $\qquad$
2. 12 .

Let $A\left(x_{0}, y_{0}\right)\left(x_{0}>0, y_{0}>0\right)$ be the intersection point of the two curves in the first quadrant.

Since the two curves are symmetric with respect to the origin and the line $y=x$, the coordinates of the other three intersection points are
$$
B\left(y_{0}, x_{0}\right), C\left(-x_{0},-y_{0}\right), D\left(-y_{0},-x_{0}\right) \text {. }
$$

Thus, quadrilateral $A B C D$ is a rectangle, and its area is
$$
\begin{array}{l} 
S=|A B||A D| \\
= \sqrt{2\left(x_{0}-y_{0}\right)^{2}} \sqrt{2\left(x_{0}+y_{0}\right)^{2}} \\
= 2 \sqrt{x_{0}^{2}+y_{0}^{2}-2 x_{0} y_{0}} \sqrt{x_{0}^{2}+y_{0}^{2}+2 x_{0} y_{0}} \\
=2 \sqrt{10-8} \sqrt{10+8}=12 .
\end{array}
$$