For a hyperbola frac{x^2}{a^2} - frac{y^2}{b^ | vedclass

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(A) Given,the length of the transverse axis is $2a = 8$,which implies $a = 4$.The distance between the foci is $2ae = 2\sqrt{41}$,which implies $ae =

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$\frac{25}{2}$

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(A) Given,the length of the transverse axis is $2a = 8$,which implies $a = 4$.The distance between the foci is $2ae = 2\sqrt{41}$,which implies $ae = \sqrt{41}$.Using the relation $a^2e^2 = a^2 + b^2$,we substitute the values:$(\sqrt{41})^2 = 4^2 + b^2$$41 = 16 + b^2$$b^2 = 41 - 16 = 25$.The length of the latus rectum is given by $\frac{2b^2}{a}$.Substituting the values,we get $\frac{2 	imes 25}{4} = \frac{50}{4} = \frac{25}{2}$.

For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,if the length of the transverse axis is $8$ and the distance between the foci is $2\sqrt{41}$,then the length of its latus rectum is

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