Answer
Asymptotes: $y=\displaystyle \pm\frac{3}{4}x .$
Foci: $(15, 0)$ and $(-15,0)$.
Work Step by Step
The hyperbola
$\displaystyle \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
has a horizontal transverse axis and two asymptotes
$y=\displaystyle \frac{b}{a}x$ and $y=-\displaystyle \frac{b}{a}x$.
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$a=12, \quad b=9$
For the foci,
$c^{2}=a^{2}+b^{2}=144+81=225$
$c=15 $
Vertices: $(12,\ 0)$ and $(-12,0)$.
Asymptotes: $y=\displaystyle \pm\frac{9}{12}x =\displaystyle \pm\frac{3}{4}x .$
Foci: $($15, $0)$ and $(-15,0)$.
