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(A) Given,in a hyperbola,the length of the transverse axis is twice that of the conjugate axis. The equation of the standard hyperbola is $\frac{x^2}{

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$\frac{8b}{\sqrt{5}}$

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(A) Given,in a hyperbola,the length of the transverse axis is twice that of the conjugate axis. The equation of the standard hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Since the length of the transverse axis $= 2 	imes$ length of the conjugate axis,we have $2a = 2(2b)$,which implies $a = 2b$.We know that $b^2 = a^2(e^2 - 1)$.Substituting $b^2 = \frac{a^2}{4}$,we get $\frac{a^2}{4} = a^2(e^2 - 1)$.Dividing by $a^2$,we get $\frac{1}{4} = e^2 - 1$,so $e^2 = 1 + \frac{1}{4} = \frac{5}{4}$.Thus,$e = \frac{\sqrt{5}}{2}$.The distance between the directrices is given by $\frac{2a}{e}$.Substituting $a = 2b$ and $e = \frac{\sqrt{5}}{2}$,the distance is $\frac{2(2b)}{\sqrt{5}/2} = \frac{4b 	imes 2}{\sqrt{5}} = \frac{8b}{\sqrt{5}}$ units.

In a hyperbola,if the length of the transverse axis is twice that of the conjugate axis,then the distance between its directrices is..... units.

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