## College Algebra (10th Edition)

$(D)$
The major axis is along the $y$-axis, center is at (0,0). The denominator under $y^{2}$ is going to be larger. The vertices are $(0,\pm 4),\quad a=4, \quad a^{2}=16$ The minor axis has length $2b=4,$ so $\quad b=2, \quad b^{2}=4$ The equation is$\quad$ $\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{16}=1$, ....choice $(D)$