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The transverse axis of a hyperbola is along the $x$ -axis and its length is $2 a$. 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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The correct answer is:
$3 x^{2}-y^{2}=3 a^{2}$
Let e be the eccentricity of hyperbola and length of conjugate axis be $2 h$. since, vertex $(a, 0)$ bisects the join of centre (0,0) and focus $(a e, 0)$
$$
\begin{array}{l}
a=\frac{a e+0}{2} \Rightarrow e=2 \\
b^{2}=a^{2}\left(c^{2}-1\right)=3 a^{2}
\end{array}
$$
$\therefore$ Required hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
$\Rightarrow \quad \frac{x^{2}}{a^{2}}-\frac{y^{2}}{3 a^{2}}=1 \Rightarrow 3 x^{2}-y^{2}=3 a^{2}$
$$
\begin{array}{l}
a=\frac{a e+0}{2} \Rightarrow e=2 \\
b^{2}=a^{2}\left(c^{2}-1\right)=3 a^{2}
\end{array}
$$
$\therefore$ Required hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
$\Rightarrow \quad \frac{x^{2}}{a^{2}}-\frac{y^{2}}{3 a^{2}}=1 \Rightarrow 3 x^{2}-y^{2}=3 a^{2}$
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