Chapter 10 - Section 10.5 - Conic Sections - 10.5 Exercises - Page 680: 13

Foci: $( \pm \sqrt 8,0)=( \pm 2 \sqrt 2,0)$ Vertices : $( \pm 3,0)$

$x^2+9y^2=0$ Rewrite as $\frac{x^2}{9}+\frac{y^2}{1}=1$ $\frac{x^2}{(3)^2}+\frac{y^2}{(1)^2}=1$ $c^2=a^2-b^2=9-1=8$ $c=\sqrt 8$ Foci: $( \pm \sqrt 8,0)=( \pm 2 \sqrt 2,0)$ Vertices : $( \pm 3,0)$