Question

The equation of hyperbola whose axes are the coordinates axes, one vertex is $(-3,0)$ and eccentricity is $\sqrt{\frac{5}{2}}$ is

• A. $\frac{x^2}{9}-\frac{2y^2}{27}=1$
• B. $\frac{x^2}{9}-\frac{y^2}{27}=1$
• C. $\frac{x^2}{15}-\frac{y^2}{26}=1$
• D. $\frac{x^2}{9}-\frac{3y^2}{27}=1$

Answer 1

The correct option is A

$\frac{x^2}{9}-\frac{2y^2}{27}=1$

$e^2=\frac{5}{2}$
$\Rightarrow 1+\frac{b^2}{a^2}=\frac{5}{2}\Rightarrow b^2=\frac{3}{2}{a}^2\cdots \left(1\right)$
Distance between center and vertex $=a=3$
Putting in $\left(1\right)$ we get,
$b^2=\frac{3}{2}\left(9\right)\Rightarrow b^2=\frac{27}{2}$
$\frac{x^2}{9}-\frac{y^2}{\frac{27}{2}}=1$
$\Rightarrow \frac{x^2}{9}-\frac{2y^2}{27}=1$