Let PQ be a chord of the hyperbola frac{x^2}{ | vedclass

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(B) Given the hyperbola $\frac{x^2}{4} - \frac{y^2}{b^2} = 1$. The eccentricity $e = \sqrt{1 + \frac{b^2}{4}} = \sqrt{3}$.Squaring both sides, $1 + \f

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$\frac{8\sqrt{3}}{5}$

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(B) Given the hyperbola $\frac{x^2}{4} - \frac{y^2}{b^2} = 1$. The eccentricity $e = \sqrt{1 + \frac{b^2}{4}} = \sqrt{3}$.Squaring both sides, $1 + \frac{b^2}{4} = 3$ $\Rightarrow \frac{b^2}{4} = 2$ $\Rightarrow b^2 = 8$.So, the hyperbola is $\frac{x^2}{4} - \frac{y^2}{8} = 1$.Let $P = (x_0, y_0)$. Since $PQ$ is perpendicular to the $x$-axis and $OPQ$ is an equilateral triangle, the angle $\angle POM = 30^{\circ}$, where $M$ is the foot of the perpendicular from $P$ to the $x$-axis.In $	riangle OMP$, $	an 30^{\circ} = \frac{PM}{OM} = \frac{y_0}{x_0} = \frac{1}{\sqrt{3}} \Rightarrow x_0 = \sqrt{3}y_0$.Since $P$ lies on the hyperbola, $\frac{(\sqrt{3}y_0)^2}{4} - \frac{y_0^2}{8} = 1 \Rightarrow \frac{3y_0^2}{4} - \frac{y_0^2}{8} = 1$.$\frac{6y_0^2 - y_0^2}{8} = 1$ $\Rightarrow \frac{5y_0^2}{8} = 1$ $\Rightarrow y_0^2 = \frac{8}{5}$ $\Rightarrow y_0 = \frac{2\sqrt{2}}{\sqrt{5}}$.The height of the triangle $OPQ$ is $OM = x_0 = \sqrt{3}y_0 = \sqrt{3} 	imes \frac{2\sqrt{2}}{\sqrt{5}} = \frac{2\sqrt{6}}{\sqrt{5}}$.The base of the triangle $PQ = 2y_0 = \frac{4\sqrt{2}}{\sqrt{5}}$.Area of $	riangle OPQ = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{4\sqrt{2}}{\sqrt{5}} 	imes \frac{2\sqrt{6}}{\sqrt{5}} = \frac{4\sqrt{12}}{5} = \frac{8\sqrt{3}}{5}$.

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, where $O$ is the centre of the hyperbola. If the eccentricity of the hyperbola is $\sqrt{3}$, then the area of the triangle $OPQ$ is:

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