Answer
Vertices: $(0,±6)$
Asymptotes: $y=±\frac{3}{5}x$
Work Step by Step
Standard form when transverse axis is vertical:
$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$a^2=36$
$a=6$
$b^2=100$
$b=10$
Vertices when transverse axis is vertical:
$(0,±a)=(0,±6)$
Asymptotes when transverse axis is vertical:
$y=±\frac{a}{b}x$
$y=±\frac{6}{10}x=±\frac{3}{5}x$
