If a hyperbola has length of its conjugate ax | vedclass

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(A) The length of the conjugate axis is $2b = 5$,so $b = \frac{5}{2}$.The distance between the foci is $2ae = 13$,so $ae = \frac{13}{2}$.For a hyperbo

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$\frac{13}{12}$

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(A) The length of the conjugate axis is $2b = 5$,so $b = \frac{5}{2}$.The distance between the foci is $2ae = 13$,so $ae = \frac{13}{2}$.For a hyperbola,the relationship between $a$,$b$,and $e$ is given by $a^2e^2 = a^2 + b^2$.Substituting the values,we get $(ae)^2 = a^2 + b^2$.$\left(\frac{13}{2}\right)^2 = a^2 + \left(\frac{5}{2}\right)^2$.$\frac{169}{4} = a^2 + \frac{25}{4}$.$a^2 = \frac{169 - 25}{4} = \frac{144}{4} = 36$.Thus,$a = 6$.The eccentricity $e$ is given by $e = \frac{ae}{a} = \frac{13/2}{6} = \frac{13}{12}$.

If a hyperbola has length of its conjugate axis equal to $5$ and the distance between its foci is $13$,then the eccentricity of the hyperbola is

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