Explanation

The given hyperbola is

$$ \frac{x^2}{4}-\frac{y^2}{b^2}=1 $$

Here,

$$ a^2=4 \Rightarrow a=2 $$

For a hyperbola,

$$ e=\sqrt{1+\frac{b^2}{a^2}} $$

Given $e=\sqrt{3}$, so

$$ \sqrt{3}=\sqrt{1+\frac{b^2}{4}} $$

Squaring both sides,

$$ 3=1+\frac{b^2}{4} $$

$$ \frac{b^2}{4}=2 $$

$$ b^2=8 $$

Let the chord $PQ$ be perpendicular to the x-axis at $x=h$.

Coordinates of $P$ and $Q$ are $(h,y)$ and $(h,-y)$.

Substitute $x=h$ in the hyperbola equation:

$$ \frac{h^2}{4}-\frac{y^2}{8}=1 $$

$$ \frac{y^2}{8}=\frac{h^2}{4}-1 $$

$$ y^2=2(h^2-4) $$

So,

$$ PQ=2y=2\sqrt{2(h^2-4)} $$

Distance $OP$ is:

$$ OP=\sqrt{h^2+y^2} $$

$$ =\sqrt{h^2+2(h^2-4)} $$

$$ =\sqrt{3h^2-8} $$

Since $OPQ$ is an equilateral triangle,

$$ OP=PQ $$

$$ \sqrt{3h^2-8}=2\sqrt{2(h^2-4)} $$

Squaring both sides,

$$ 3h^2-8=8(h^2-4) $$

$$ 3h^2-8=8h^2-32 $$

$$ 5h^2=24 $$

$$ h^2=\frac{24}{5} $$

Side of equilateral triangle:

$$ OP=\sqrt{3h^2-8}=\sqrt{\frac{72}{5}-8} $$

$$ =\sqrt{\frac{32}{5}} $$

Area of equilateral triangle:

$$ \text{Area}=\frac{\sqrt{3}}{4}\times OP^2 $$

$$ =\frac{\sqrt{3}}{4}\times\frac{32}{5} $$

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

Hence, the area of triangle $OPQ$ is $\dfrac{8\sqrt{3}}{5}$.

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