Answer
$\quad \displaystyle \frac{x^{2}}{16}+y^{2}=1$
Work Step by Step
The center is at the midpoint of the vertices, $(0,0)$
The foci, center and vertices lie on the same line, the main axis.
Main axis here: $y=0$ (the $x$-axis).
Table 3: Major axis horizontal (parallel to the x-axis)$\begin{array}{lll}
\text{ Foci}&\text{ Vertices}&\text{ Equation}\\
{(h+c, k)}&{(h+a,k)}&{\displaystyle \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1}\\
{(h-c,k)}&{(h-a,k)}&{a \gt b \gt 0\text{ and }b^{2}=a^{2}-c^{2}}\\ \end{array}$
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$y$ -intercepts are $\pm 1 \quad \Rightarrow \quad b=1$
Vertices at $(\pm 4,0) \quad \Rightarrow \quad a=4$
$c^{2}=a^{2}-b^{2}=16-1=15$
$c=\sqrt{15}\approx 3.873$
Foci at $(\pm 3.873)$
The equation is $\quad \displaystyle \frac{x^{2}}{16}+\frac{y^{2}}{1}=1$
or $\quad \displaystyle \frac{x^{2}}{16}+y^{2}=1$
