If sqrt{5} y - sqrt{8} = 0 is the equation of | vedclass

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(A) The given equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} + 1 = 0$,which can be rewritten as $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$. This is a con

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$\frac{1}{\sqrt{2}}$

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(A) The given equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} + 1 = 0$,which can be rewritten as $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$. This is a conjugate hyperbola. The equation of the directrix for this hyperbola is $y = \frac{b}{e}$. Comparing this with the given directrix $\sqrt{5} y - \sqrt{8} = 0$,we get $y = \frac{\sqrt{8}}{\sqrt{5}}$. Thus,$\frac{b}{e} = \frac{\sqrt{8}}{\sqrt{5}}$. Given $e = \frac{\sqrt{5}}{2}$,we have $b = \frac{\sqrt{8}}{\sqrt{5}} 	imes \frac{\sqrt{5}}{2} = \frac{\sqrt{8}}{2} = \sqrt{2}$. For a conjugate hyperbola,$e^2 = 1 + \frac{a^2}{b^2}$. Substituting the values,$(\frac{\sqrt{5}}{2})^2 = 1 + \frac{a^2}{(\sqrt{2})^2} \implies \frac{5}{4} = 1 + \frac{a^2}{2}$. $\frac{a^2}{2} = \frac{5}{4} - 1 = \frac{1}{4} \implies a^2 = \frac{1}{2} \implies a = \frac{1}{\sqrt{2}}$. Therefore,$\frac{1}{a} = \sqrt{2}$.

If $\sqrt{5} y - \sqrt{8} = 0$ is the equation of the directrix of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} + 1 = 0$ and $\frac{\sqrt{5}}{2}$ is its eccentricity,then $\frac{1}{a} =$

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