3. Given point $A$ is the left vertex of the hyperbola $x^{2}-y^{2}=1$, points $B$ and $C$ are on the right branch of the hyperbola, and $\triangle A B C$ is an equilateral triangle, then the area of $\triangle A B C$ is
(A) $\frac{\sqrt{3}}{3}$
(B) $\frac{3 \sqrt{3}}{2}$
(C) $3 \sqrt{3}$
(D) $6 \sqrt{3}$
(C)
3.【Analysis and Solution】Assume point $B$ is above the $x$-axis. Since $\triangle ABC$ is an equilateral triangle, we know the slope of line $AB$ is $k=\frac{\sqrt{3}}{3}$. Since the line passes through point $A(-1,0)$, the equation is $y=\frac{\sqrt{3}}{3} x+\frac{\sqrt{3}}{3}$. Substituting this into the hyperbola equation $x^{2}-y^{2}=1$, we find the coordinates of point $B$ to be $(2, \sqrt{3})$. Similarly, the coordinates of point $C$ are $(2,-\sqrt{3})$. Therefore, the area of $\triangle ABC$ is $[2-(-1)] \sqrt{3}=$ $3 \sqrt{3}$.