College Algebra, Chapter 8, 8.3, Section 8.3, Problem 42

Find an equation for the hyperbola with foci $(0,\pm 1)$ and length of the transverse axis of 1.
The hyperbola $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ has length of transverse axis of $2a$ and foci $(0,\pm c)$ where $c^2 = a^2 + b^2$.
Thus, gives us $2a = 1$ and $c = 1$. So, $\displaystyle a=\frac{1}{2}$. Then, by substituting the values, we obtain

$
\begin{equation}
\begin{aligned}
1^2 &= \left( \frac{1}{2} \right)^2 + b^2\\
\\
1 &= \frac{1}{4} + b^2 \\
\\
b^2 &= \frac{3}{4}\\
\\
b &= \frac{\sqrt{3}}{2}
\end{aligned}
\end{equation}
$


Therefore, the equation is

$
\begin{equation}
\begin{aligned}
\frac{y^2}{\left( \frac{1}{2} \right)^2} - \frac{x^2}{\left( \frac{\sqrt{3}}{2} \right)^2} &= 1\\
\\
\frac{y^2}{\frac{1}{4}} - \frac{x^2}{\frac{3}{4}} &= 1\\
\\
4y^2 - \frac{4x^2}{3} &= 1
\end{aligned}
\end{equation}
$