Answer
The eccentricity of the ellipse is $e=\sqrt 2$
and asymptotes are $y=\pm x$ and $F: (\pm \sqrt 2,0)$ and
The directrices are: $x=\pm \dfrac{1}{\sqrt 2}$
Work Step by Step
The eccentricity of the ellipse $\dfrac{x^2}{p^2}+\dfrac{y^2}{q^2}=1$ when $p \gt q$ is given by:
$e=\dfrac{\sqrt {p^2-q^2}}{p}$
The foci of the ellipse are: $(\pm pe,0)$ and the directrices are given as: $x=\pm \dfrac{p}{e}$
We are given that $x^2-y^2=1$
so, $c=\sqrt {a^2+b^2}=\sqrt {1+1}=\sqrt 2$
The eccentricity of the ellipse is $e=\dfrac{\sqrt 2}{1}=\sqrt 2$
and asymptotes are $y=\pm x$ and $F: (\pm \sqrt 2,0)$ and
The directrices are: $x=0 \pm \dfrac{a}{e}=\pm \dfrac{1}{\sqrt 2}$
