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