Answer
$e=\sqrt 2$
and asymptotes are : $y=\pm x$ and focus: $F: (\pm \sqrt 2,0)$;
Directrices are: $x=0 \pm \dfrac{a}{e}=\pm \dfrac{1}{\sqrt 2}$
Work Step by Step
The eccentricity of the ellipse $\dfrac{x^2}{m^2}+\dfrac{y^2}{n^2}=1$ when $m \gt n$ is given by:
$e=\dfrac{\sqrt {m^2-n^2}}{m}$
The foci of the ellipse are: $(\pm me,0)$ and the directrices are given as: $x=\pm \dfrac{m}{e}$
Given: $x^2-y^2=1$
so, $c=\sqrt {m^2+n^2}=\sqrt {1+1}=\sqrt 2$
The eccentricity of the ellipse: $e=\sqrt 2$;
Asymptotes are $y=\pm x$ and focus is: $F: (\pm \sqrt 2,0)$ ;
The directrices are: $x=0 \pm \dfrac{a}{e}=\pm \dfrac{1}{\sqrt 2}$
