If $\sqrt{5} y-\sqrt{8}=0$ is the equation of the directrix of a hyperbola…
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 $\frac1a=$
$=\frac{1}{\sqrt{2}}$
$\sqrt{3}$
$\sqrt{5}$
$\sqrt{6}$
Solution
$\frac{x^2}{a^2}-\frac{y^2}{b^2}+1=0$
i.e. $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$
It is conjugate hyperbola
Equation of directrix must be $y=\frac{b}{e}$
Comparing with $\sqrt{5} y-\sqrt{8}=0$
$\begin{aligned} & y=\frac{\sqrt{8}}{\sqrt{5}}=\frac{b}{e} \\ & \because e=\sqrt{\frac{5}{2}}\end{aligned}$
$\begin{aligned} & b=\frac{\sqrt{8}}{\sqrt{5}} e=\frac{\sqrt{8}}{\sqrt{5}} \times \frac{\sqrt{5}}{2}=\frac{\sqrt{8}}{2} \\ & b=\sqrt{2}\end{aligned}$
$\because e=\sqrt{1+\frac{a^2}{b^2}}$
$\begin{aligned} & \frac{\sqrt{5}}{2}=\sqrt{1+\frac{a^2}{2}} \Rightarrow \frac{5}{4}=1+\frac{a^2}{2} \\ & \frac{a^2}{2}=\frac{1}{4} \Rightarrow a=\frac{1}{\sqrt{2}}\end{aligned}$