Answer
$\frac{17y^2}{1024}-\frac{17x^2}{64}=1$
Work Step by Step
Transverse axis is vertical.
$(0,±c)=(0,±8)$
$c=8$
The equation of the asymptotes when transverse axis is vertical:
$y=±\frac{a}{b}x$
The equation of the asymptotes:
$y=±4x$
So:
$\frac{a}{b}=4$
$a=4b$
$a^2=16b^2$
$c^2=a^2+b^2=16b^2+b^2=17b^2$
$64=17b^2$
$b^2=\frac{64}{17}$
$a^2=16(\frac{64}{17})=\frac{1024}{17}$
Standard form when transverse axis is vertical:
$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$\frac{y^2}{\frac{1024}{17}}-\frac{x^2}{\frac{64}{17}}=1$
$\frac{17y^2}{1024}-\frac{17x^2}{64}=1$
