Answer : $\large\frac{x^2}{4}-\frac{y^2}{5}\normalsize =\large\frac{4}{9}$

Ecentricity $e=\large\frac{3}{2}$

Given foci $ae=2$

$\therefore a=\large\frac{2}{e}$

$\Rightarrow a=\large\frac{2}{3}$$\times 2=\large\frac{4}{3}$

$\therefore a^2=\large\frac{16}{9}$

$b^2=a^2e^2-a^2$

$\;\;\;\;=4-\large\frac{16}{9}$

$\;\;\;\;=\large\frac{20}{9}$

Hence equation of the hyperbola is $\large\frac{x^2}{\Large\frac{16}{9}}-\frac{y^2}{\Large\frac{20}{9}}$$=1$

$\large\frac{9x^2}{16}-\frac{9y^2}{20}$$=1$

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

$\large\frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}$ is the equation of the hyperbola.