$\text{If the distances of one focus of hyperbola from its directrices are 5 and 3, then eccentricity of hyperbola is}$

$\text{According to the question,}$

$\begin{array}{r}ae-\frac{a}{e}=3----\left(1\right)\\ ae+\frac{a}{e}=5----\left(2\right)\end{array}$

$\begin{array}{r}\text{ Solving (1) and (2), we get }\\ \begin{array}{l}ae=4\text{ and }\frac{a}{e}=1\\ \text{ So, }{a}^2=4\Rightarrow a=2\Rightarrow e=2\end{array}\end{array}$