# Equation of the hyperbola with eccentricity $\large\frac{3}{2}$ and foci at $(\pm2,0)$ is

$\begin{array}{1 1}(A)\;\frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9} \\(B)\;\frac{x^2}{9}-\frac{y^2}{9}=\frac{4}{9} \\(C)\;\frac{x^2}{4}-\frac{y^2}{9}=1 \\(D)\;none\;of\;these \end{array}$

$e=\large\frac{3}{2}$
$2ae=4$
$\therefore a=\large\frac{4}{3}$
$b^2=a^2(e^2-1)$
$\large\frac{16}{9} . \frac{5}{4}=\frac{20}{9}$
$9 \bigg(\frac{x^2}{16} -\frac{y^2}{20} \bigg)=1$
$\large\frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}$
Hence A is the correct answer.