Answer
Vertices: $(±12,0)$
Asymptotes:
$y=±\frac{13}{12}x~~$ (Black lines)
Work Step by Step
Standard form when transverse axis is horizontal:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$a^2=144$
$a=12$
$b^2=169$
$b=13$
Vertices when transverse axis is horizontal: $(±a,0)=(±12,0)$
Asymptotes when transverse axis is horizontal:
$y=±\frac{b}{a}x$
$y=±\frac{13}{12}x$
