Chapter 10 - Parametric Equations and Polar Coordinates - 10.5 Conic Sections - 10.5 Exercises - Page 720: 12

Foci: $( \pm \sqrt {28},0)=( \pm 2\sqrt {7},0)$ Vertices : $(\pm 6,0)$

$\frac{x^2}{36}+\frac{y^2}{8}=1$ 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 : $(\pm 6,0)$