Chapter 10 - Section 10.5 - Conic Sections - 10.5 Exercises - Page 680: 11

Foci: $(0, \pm \sqrt 2)$ Vertices : $(0, \pm 2)$

$\frac{x^2}{2}+\frac{y^2}{4}=1$ Rewrite as $\frac{x^2}{(\sqrt 2)^2}+\frac{y^2}{(2)^2}=1$ $c^2=a^2-b^2=(2)^2-(\sqrt 2)^2=2$ $c=\sqrt 2$ Foci: $(0, \pm \sqrt 2)$ Vertices : $(0, \pm 2)$