## Algebra and Trigonometry 10th Edition

$\frac{y^2}{4}-\frac{x^2}{32}=1$
Transverse axis is vertical. $(0,±a)=(0,±2)$ $a=2$ $(0,±c)=(0,±6)$ $c=6$ $b^2=c^2-a^2=36-4=32$ $b=4\sqrt 2$ Standard form when transverse axis is vertical: $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ $\frac{y^2}{2^2}-\frac{x^2}{(4\sqrt 2)^2}=1$ $\frac{y^2}{4}-\frac{x^2}{32}=1$