first off, let's take a peek at the picture above
hmmm the hyperbola is opening sideways, that means it has a horizontal traverse axis, it also means that the positive fraction will be the one with the "x" variable in it.
now, the length of the horizontal traverse axis is 4 units, from vertex to vertex, that means the "a" component of the hyperbola is half that or 2 units, and 2² = 4, with a center at the origin.
![\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 √( a ^2 + b ^2) \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(x- 0)^2}{ 2^2}-\cfrac{(y- 0)^2}{ (√(3))^2}=1\implies {\Large \begin{array}{llll} \cfrac{x^2}{4}-\cfrac{y^2}{3}=1 \end{array}}](https://img.qammunity.org/2024/formulas/mathematics/college/pfv9wv79tvo7tiftv61425mlu6qirpwqy6.png)