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Questions  >>  CBSE XI  >>  Math  >>  Conic Sections
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Equation of the hyperbola with eccentricity $\large\frac{3}{2}$ and foci at $(\pm 2,0)$ is

$\begin{array}{1 1}\large\frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}\\\large\frac{x^2}{9}-\frac{y^2}{9}=\frac{4}{9}\\\large\frac{x^2}{4}-\frac{y^2}{9}\normalsize =1\\\large\frac{x^2}{2}-\frac{y^2}{3}\normalsize =1\end{array} $

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