A hyperbola having its centre at the origin p | vedclass

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(C) The equation of the hyperbola with centre at the origin and transverse axis along the $X$-axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Given t

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$\frac{\sqrt{13}}{2}$

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(C) The equation of the hyperbola with centre at the origin and transverse axis along the $X$-axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Given the length of the transverse axis $2a = 8$,we have $a = 4$,so $a^2 = 16$. The hyperbola passes through $(5, 2)$,so $\frac{25}{16} - \frac{4}{b^2} = 1$. $\frac{4}{b^2} = \frac{25}{16} - 1 = \frac{9}{16}$,which gives $b^2 = \frac{64}{9}$. The conjugate hyperbola is $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$. The eccentricity $e'$ of the conjugate hyperbola is given by $e' = \sqrt{1 + \frac{a^2}{b^2}}$. $e' = \sqrt{1 + \frac{16}{64/9}} = \sqrt{1 + \frac{16 	imes 9}{64}} = \sqrt{1 + \frac{9}{4}} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2}$.

$A$ hyperbola having its centre at the origin passes through the point $(5, 2)$ and has a transverse axis of length $8$ along the $X$-axis. What is the eccentricity of its conjugate hyperbola?

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