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