Question

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

$\frac{x^{2}}{4}-\frac{y^{2}}{5}=\frac{4}{9}$

Answer 2

Let the equation of hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\, \ldots\left(i\right)$ Given, $e=\frac{3}{2}$ and foci $=\left(\pm ae, 0\right)=\left(\pm2, 0\right)$ 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 $\frac{x^{2}}{\left(16 / 9\right)}-\frac{y^{2}}{\left(20 / 9\right)}=1$ $\Rightarrow \frac{x^{2}}{4}-\frac{y^{2}}{5}=\frac{4}{9}$