4. As shown in the figure, in the Cartesian coordinate system $x-O-y$, $A, B$ are two points on the graph of the function $y=\frac{1}{x}$ in the first quadrant, satisfying $\angle A O B=60^{\circ}$ and $O A=O B$. Then the area of $\triangle O A B$ is ( ).
A: $\quad 2 \sqrt{3}$
B: $\frac{3}{2} \sqrt{3}$
C: $\sqrt{3}$
D: $\frac{\sqrt{3}}{2}$
Answer $\mathrm{C}$.
Analysis According to the problem, $\triangle O A B$ is an equilateral triangle.
Also, $A, B$ are two points on the function $y=\frac{1}{x}$ that satisfy the given conditions, so $A, B$ are symmetric about the line $y=x$. Let point $A\left(a, \frac{1}{a}\right)$, then point $B\left(\frac{1}{a}, a\right), O A^{2}=a^{2}+\frac{1}{a^{2}}, A B^{2}=\left(a-\frac{1}{a}\right)^{2}+\left(\frac{1}{a}-a\right)^{2}=2\left(a-\frac{1}{a}\right)^{2}$.

From $O A=A B$ we get $a^{2}+\frac{1}{a^{2}}=2\left(a-\frac{1}{a}\right)^{2}$, simplifying to $a^{2}+\frac{1}{a^{2}}-4=0$, which is $a^{2}+\frac{1}{a^{2}}=4$.
Therefore, $O A^{2}=4$, so the area of the equilateral $\triangle O A B$ $=\frac{\sqrt{3}}{4} \times O A^{2}=\frac{\sqrt{3}}{4} \times 4=\sqrt{3}$.