The equation of the hyperbola in the standard | vedclass

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(C) The standard equation of a hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Given length of latus rectum = $\frac{2b^2}{a} = 9$,so $2b^2 = 9a

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$\frac{x^2}{64} - \frac{y^2}{36} = 1$

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(C) The standard equation of a hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Given length of latus rectum = $\frac{2b^2}{a} = 9$,so $2b^2 = 9a$ ... $(i)$. Given eccentricity $e = \frac{5}{4}$. We know $b^2 = a^2(e^2 - 1)$. Substituting $e = \frac{5}{4}$,we get $b^2 = a^2(\frac{25}{16} - 1) = a^2(\frac{9}{16}) = \frac{9a^2}{16}$ ... $(ii)$. Substitute $(ii)$ into $(i)$: $2(\frac{9a^2}{16}) = 9a$. $\frac{9a^2}{8} = 9a \implies a = 8$. Now,$b^2 = \frac{9(8^2)}{16} = \frac{9 	imes 64}{16} = 9 	imes 4 = 36$. Thus,the equation is $\frac{x^2}{8^2} - \frac{y^2}{6^2} = 1$,which is $\frac{x^2}{64} - \frac{y^2}{36} = 1$.

The equation of the hyperbola in the standard form (with transverse axis along the $x$-axis) having the length of the latus rectum = $9$ units and eccentricity = $5/4$ is

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