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(D) Let $e$ be the eccentricity of the hyperbola. The centre is $(0, 0)$, the vertex is $(a, 0)$, and the focus is $(ae, 0)$.Since the vertex $(a, 0)$

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$3x^{2}-y^{2}=3a^{2}$

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(D) Let $e$ be the eccentricity of the hyperbola. The centre is $(0, 0)$, the vertex is $(a, 0)$, and the focus is $(ae, 0)$.Since the vertex $(a, 0)$ bisects the line segment joining the centre $(0, 0)$ and the focus $(ae, 0)$, we have:$a = \frac{ae + 0}{2}$ $\Rightarrow 2a = ae$ $\Rightarrow e = 2$.For a hyperbola, $b^{2} = a^{2}(e^{2} - 1)$.Substituting $e = 2$, we get $b^{2} = a^{2}(2^{2} - 1) = 3a^{2}$.The equation of the hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.Substituting $b^{2} = 3a^{2}$, we get $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{3a^{2}} = 1$.Multiplying by $3a^{2}$, we obtain $3x^{2} - y^{2} = 3a^{2}$.

The transverse axis of a hyperbola is along the $x$-axis and its length is $2a$. The vertex of the hyperbola bisects the line segment joining the centre and the focus. The equation of the hyperbola is

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