Precalculus (10th Edition)

$(D)$
The given hyperbola has the $y$-axis as the transverse axis; therefore we eliminate $(A)$ and $(B)$. We have to decide between $(C)$ and $(D)$. The vertices for the given hyperbola are: $(0,-1),(0,1)$ Check $(0,1)$ in equation $(C)$: $\dfrac{1^2}{4}-0^2\stackrel{?}{=}1$ $\dfrac{1}{4}\not=1$ So equation $(C)$ doesn't fit. Check $(0,1)$ in equation $(D)$: $1^2-\dfrac{0^2}{4}\stackrel{?}{=}1$ $1=1\checkmark$ So equation $(D)$ fits. Therefore, the correct answer is equation $(D)$.