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# Find the equation of the hyperbola with eccentricity $\large\frac{3}{2}$ and foci at $(\pm 2,0)$.

$\begin{array}{1 1}\large\frac{x^2}{5}-\frac{y^2}{4}\normalsize =\large\frac{4}{9}\\\large\frac{y^2}{5}-\frac{x^2}{4}\normalsize =\large\frac{4}{9}\\\large\frac{x^2}{4}-\frac{y^2}{5}\normalsize =\large\frac{4}{9}\\\large\frac{y^2}{4}-\frac{x^2}{5}\normalsize =\large\frac{4}{9}\end{array}$

• General equation of a hyperbola is $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1 • Eccentricity e=\large\frac{\sqrt{a^2-b^2}}{a} Answer : \large\frac{x^2}{4}-\frac{y^2}{5}\normalsize =\large\frac{4}{9} Ecentricity e=\large\frac{3}{2} Given foci ae=2 \therefore a=\large\frac{2}{e} \Rightarrow a=\large\frac{2}{3}$$\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.