Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the length of the latus rectum of the ellipse $16x^{2} + y^{2} = 16$.

(N/A) The given equation is $16x^{2} + y^{2} = 16$.
Dividing both sides by $16$,we get:
$\frac{x^{2}}{1} + \frac{y^{2}}{16} = 1$
Or,$\frac{x^{2}}{1^{2}} + \frac{y^{2}}{4^{2}} = 1$ $(1)$
Comparing equation $(1)$ with the standard form $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1$,we observe that $a^{2} = 16$ and $b^{2} = 1$,so $a = 4$ and $b = 1$.
Since $a > b$,the major axis is along the $y$-axis.
We calculate $c$ using $c = \sqrt{a^{2} - b^{2}} = \sqrt{16 - 1} = \sqrt{15}$.
$1$. Coordinates of the foci: $(0, \pm c) = (0, \pm \sqrt{15})$.
$2$. Coordinates of the vertices: $(0, \pm a) = (0, \pm 4)$.
$3$. Length of the major axis: $2a = 2(4) = 8$.
$4$. Length of the minor axis: $2b = 2(1) = 2$.
$5$. Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{15}}{4}$.
$6$. Length of the latus rectum: $\frac{2b^{2}}{a} = \frac{2(1)}{4} = \frac{1}{2}$.