Answer
Foci: $( \pm \sqrt {28},0)=( \pm 2\sqrt {7},0)$
Vertices : $(\pm 6,0)$
Work Step by Step
$\frac{x^2}{36}+\frac{y^2}{8}=1$
Rewrite as $\frac{x^2}{(6)^2}+\frac{y^2}{(2\sqrt 2)^2}=1$
$c^2=a^2-b^2=36-8=28$
$c=\sqrt {28}$
Foci: $( \pm \sqrt {28},0)=( \pm 2\sqrt {7},0)$
Vertices : $(\pm 6,0)$
