again, this one, notice, the positive fraction is the one with the "x" variable in it, thus is also a horizontal hyperbola,
[tex]\bf \cfrac{x^2}{4}-\cfrac{y^2}{60}=1\implies \cfrac{(x-0)^2}{2^2}-\cfrac{(y-0)^2}{(\sqrt{60})^2}=1\qquad
\begin{cases}
a=2\\
b=\sqrt{60}
\end{cases}
\\\\\\
c=\sqrt{2^2+(\sqrt{60})^2}\implies c=\sqrt{4+60}\implies c=8\\\\
-------------------------------\\\\
vertices~~(0\pm 2,0)\implies
\begin{cases}
(2,0)\\
(-2,0)
\end{cases}\qquad foci~~(0\pm 8,0)\implies
\begin{cases}
(8,0)\\
(-8,0)
\end{cases}[/tex]
