check the picture below. So the hyperbola looks more or less like so.
notice that the traverse axis is horizontal, thus the positive fraction will be the one with the "x" variable, and the major axis "a" is 4 units, whilst the "c" distance from the center to a focus is 5.
[tex] \bf \textit{hyperbolas, horizontal traverse axis }
\\\\
\cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1
\qquad
\begin{cases}
center\ ( h, k)\\
vertices\ ( h\pm a, k)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{ a ^2 + b ^2}
\end{cases}\\\\
------------------------------- [/tex]
[tex] \bf \begin{cases}
h=0\\
k=0\\
a=4\\
c=5
\end{cases}\implies
\cfrac{(x- 0)^2}{ 4^2}-\cfrac{(y- 0)^2}{ b^2}=1
\\\\\\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b\implies \sqrt{5^2-4^2}=b
\\\\\\
\sqrt{25-16}=b\implies 3=b
\\\\\\
\cfrac{(x- 0)^2}{ 4^2}-\cfrac{(y- 0)^2}{ 3^2}=1\implies \cfrac{x^2}{16}-\cfrac{y^2}{9}=1 [/tex]
