Let PQ be a chord of the hyperbola $$\frac{x^{2}}{4}-\frac{y^{2}}{b^{2}}=1$$, perpendicular to the x-axis

such that OPQ is an equilateral triangle, O being the centre of the hyperbola. If the eccentricity of the hyperbola is $$\sqrt{3}.$$ then the area of the triangle OPQ is

$$e^2 = 1 + \frac{b^2}{a^2} \implies 3 = 1 + \frac{b^2}{4} \implies \frac{b^2}{4} = 2 \implies b^2 = 8$$.

Equation: $$\frac{x^2}{4} - \frac{y^2}{8} = 1$$. 
Equilateral Triangle Property: In $$\triangle OPQ$$, if $$P = (x_1, y_1)$$, then because it's equilateral and $$PQ$$ is vertical, the angle $$POX$$ is $$30^\circ$$.

Thus, $$\tan 30^\circ = \frac{y_1}{x_1} \implies \frac{1}{\sqrt{3}} = \frac{y_1}{x_1} \implies y_1 = \frac{x_1}{\sqrt{3}}$$.

 Solve for Point P:

Substitute $$y^2 = \frac{x^2}{3}$$ into the hyperbola:

$$\frac{x^2}{4} - \frac{x^2/3}{8} = 1 \implies \frac{x^2}{4} - \frac{x^2}{24} = 1 \implies \frac{6x^2 - x^2}{24} = 1 \implies 5x^2 = 24 \implies x^2 = \frac{24}{5}$$.
Area of Triangle:

Area $$= \frac{\sqrt{3}}{4} (\text{side})^2$$. Side $$PQ = 2y_1$$.

Side length squared $$s^2 = (2y_1)^2 = 4 \cdot \frac{x^2}{3} = 4 \cdot \frac{24/5}{3} = \frac{32}{5}$$.

Area $$= \frac{\sqrt{3}}{4} \cdot \frac{32}{5} = \frac{8\sqrt{3}}{5}$$.

Correct Answer: B ($$8\sqrt{3}/5$$)