Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length

Solution:

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

Here, the denominator of $\frac{y^{2}}{4^{2}}$ 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 $(0, \pm 4)$.

Length of major axis = 2a = 8

      Length of minor axis = 2b = 2

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

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