The given equation is $16 x^{2}+y^{2}=16$

It can be written as

$16 x^{2}+y^{2}=16$

Or, $\frac{x^{2}}{1}+\frac{y^{2}}{16}=1$

Or,  $\frac{ x ^{2}}{1^{2}}+\frac{y^{2}}{4^{2}}=1$    …….. $(1)$

Here, the denominator of $\frac{ x ^{2}}{4^{2}}$ is greater than the denominator of $\frac{ x ^{2}}{1^{2}}$. 

Therefore, the major axis is along the $y-$ axis, while the minor axis is along the $x-$ axis.

On comparing equation $(1)$ with $\frac{ x ^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1,$ we obtain $b =1$ and $a =4$

$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{16-1}=\sqrt{15}$

Therefore,

The coordinates of the foci are $(0, \,\pm \sqrt{15})$

The coordinates of the vertices are $(0,\,±4)$

Length of major axis $=2 a=8$

Length of minor axis $=2 b =2$

Eccentricity, $e=\frac{c}{a}=\frac{\sqrt{15}}{4}$

Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 1}{4}=\frac{1}{2}$