Hyperbola

Concept:For an equilateral triangle with vertex at the origin and base perpendicular to the x-axis, the slope of OP is $\tan 30^\circ$.Explanation:Eccentricity $e = \sqrt{3}$ for $\frac{x^2}{4} - \frac{y^2}{b^2} = 1$.$e^2 = 1 + \frac{b^2}{4} = 3$ gives $b^2 = 8$.Hyperbola: $\frac{x^2}{4} - \frac{y^2}{8} = 1$.Let $P(x_1, y_1)$ and chord PQ perpendicular to x-axis, so $Q(x_1, -y_1)$.$\triangle OPQ$ equilateral implies $\angle POQ = 60^\circ$, so $\angle POX = 30^\circ$.Slope of OP: $\frac{y_1}{x_1} = \tan 30^\circ = \frac{1}{\sqrt{3}} \Rightarrow y_1 = \frac{x_1}{\sqrt{3}}$.Substitute $P$ in hyperbola: $\frac{x_1^2}{4} - \frac{1}{8}\cdot\frac{x_1^2}{3} = 1$.Multiply by 24: $6x_1^2 - x_1^2 = 24 \Rightarrow x_1^2 = \frac{24}{5}$.Area of $\triangle OPQ = \frac{1}{2} \times PQ \times$ height $= \frac{1}{2} \times (2y_1) \times x_1 = x_1 y_1$.Using $y_1 = \frac{x_1}{\sqrt{3}}$, area $= \frac{x_1^2}{\sqrt{3}} = \frac{24}{5\sqrt{3}} = \frac{8\sqrt{3}}{5}$.Answer:Area $= \frac{8\sqrt{3}}{5}$, which matches option C.