Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $36 x^{2}+4 y^{2}=144$
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $36 x^{2}+4 y^{2}=144$
The given equation is $36 x^{2}+4 y^{2}=144$
It can be written as
$36 x^{2}+4 y^{2}=114$
Or , $\frac{ x ^{2}}{4}+\frac{y^{2}}{36}=1$
Or, $\frac{x^{2}}{2^{2}}+\frac{y^{2}}{6^{2}}=1$ ........ $(1)$
Here, the denominator of $\frac{y^{2}}{6^{2}}$ is greater than the denominator of $\frac{x^{2}}{2^{2}}$
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 $b =2$ and $a =6$
$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{36-4}=\sqrt{32}=4 \sqrt{2}$
Therefore,
The coordinates of the foci are $(0, \,\pm 4 \sqrt{2})$
The coordinates of the vertices are $(0,\,±6)$
Length of major axis $=2 a=12$
Length of minor axis $=2 b=4$
Eccentricity, $e=\frac{c}{a}=\frac{4 \sqrt{2}}{6}=\frac{2 \sqrt{2}}{3}$
Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 4}{6}=\frac{4}{3}$
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