Question

Equation of the hyperbola with eccentricity $\frac{3}{2}$ and foci at $(\pm 2,0)$ is

• A. $\frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}$
• B. $\frac{x^2}{9}-\frac{y^2}{9}=\frac{4}{9}$
• C. $\frac{x^2}{4}-\frac{y^2}{9}=1$
• D. None of these

Answer 1

Answer: (A)

Let the equation of hyperbola be
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\dots \left(i\right)$
Given, $e=\frac{3}{2}$ and foci $=\left(\pm ae,0\right)=\left(\pm 2,0\right)$
So, $e=\frac{3}{2}$ and $ae=2$
$\Rightarrow a\times \frac{3}{2}=2$
$\Rightarrow a^2=\frac{16}{9}$
Also, $b^2=a^2\left(e^2-1\right)$
$\Rightarrow b^2=\frac{16}{9}\left(\frac{9}{4}-1\right)=\frac{20}{9}$
On putting the values of $a^2$ and $b^2$ in $\left(i\right)$, we get
$\frac{x^2}{\left(16/9\right)}-\frac{y^2}{\left(20/9\right)}=1$
$\Rightarrow \frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}$