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 $\frac{x^{2}}{100}+\frac{y^{2}}{400}=1$.
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 $\frac{x^{2}}{100}+\frac{y^{2}}{400}=1$.
The given equation is $\frac{x^{2}}{100}+\frac{y^{2}}{400}=1$ or $\frac{x^{2}}{10^{2}}+\frac{y^{2}}{20^{2}}=1$
Here, the denominator of is greater than the denominator of $\frac{x^{2}}{100}$.
Therefore, the major axis is along the $y-$ axis, while the minor axis is along the $x-$ axis.
On comparing the given equation with, $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$ we obtain $b=10$ and $a=20$
$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{400-100}=\sqrt{300}=10 \sqrt{3}$
Therefore,
The coordinates of the foci are $(0,\, \pm 10 \sqrt{3})$
The coordinates of the vertices are $(0,\,±20)$
Length of major axis $=2 a =40$
Length of minor axis $=2 b =20$
Eccentricity, $e=\frac{c}{a}=\frac{10 \sqrt{3}}{20}=\frac{\sqrt{3}}{2}$
Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 100}{20}=10$
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