Find the equation of the hyperbola with eccen | vedclass

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(D) The standard form of a hyperbola with foci on the $x$-axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Given foci are $(\pm ae, 0) = (\pm 2, 0)$,so

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None of these

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(D) The standard form of a hyperbola with foci on the $x$-axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Given foci are $(\pm ae, 0) = (\pm 2, 0)$,so $ae = 2$.Given eccentricity $e = 3/2$.Substituting $e$ in $ae = 2$,we get $a(3/2) = 2$,which implies $a = 4/3$.For a hyperbola,$b^2 = a^2(e^2 - 1)$.$b^2 = (4/3)^2 ((3/2)^2 - 1) = (16/9) (9/4 - 1) = (16/9) (5/4) = 20/9$.Substituting $a^2 = 16/9$ and $b^2 = 20/9$ into the standard equation:$\frac{x^2}{16/9} - \frac{y^2}{20/9} = 1 \implies \frac{9x^2}{16} - \frac{9y^2}{20} = 1$.None of the given options match this result.

Find the equation of the hyperbola with eccentricity $e = 3/2$ and foci at $(\pm 2, 0)$.

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