Question

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

• A. $6x^{2}-y^{2}=3a^{2}$
• B. $x^{2}-3y^{2}=3a^{2}$
• C. $x^{2}-6y^{2} =3a^{2}$
• D. $3x^{2}-y^{2}=3a^{2}$

Answer 1

Answer: (D)

Let $e$ be the eccentricity of hyperbola and length of conjugate axis be $2 b$.
Since, vertex $(a, 0)$ bisects the join of centre $(0,0)$ and focus $(a e, 0)$.
$a =\frac{a e+0}{2} \Rightarrow e=2$
$b^{2} =a^{2}\left(e^{2}-1\right)=3 a^{2}$
$\therefore $ Required hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
$\Rightarrow \frac{x^{2}}{a^{2}}-\frac{y^{2}}{3 a^{2}}=1$
$\Rightarrow 3 x^{2}-y^{2}=3 a^{2}$