Question

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

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