Find the coordinates of the foci, the rertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $16 x^{2}+y^{2}=16$
Find the coordinates of the foci, the rertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $16 x^{2}+y^{2}=16$
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 $=\frac{2 b^{2}}{a}=\frac{2 \times 1}{4}=\frac{1}{2}$
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