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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 $\frac{x^{2}}{25}+\frac{y^{2}}{100}=1$
Solution
The given equation is $\frac{x^{2}}{25}+\frac{y^{2}}{100}=1$ or $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{10^{2}}=1$
Here, the denominator of $\frac{y^{2}}{100}$ is greater than the denominator of $\frac{x^{2}}{25}$
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 $b=5$ and $a=10$
$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{100-25}=\sqrt{75}=5 \sqrt{3}$
Therefore,
The coordinates of the foci are $(0, \,\pm 5 \sqrt{3})$
The coordinates of the vertices are $(0,\,±10)$
Length of major axis $=2 a=20$
Length of minor axis $=2 b=10$
Eccentricity, $e=\frac{c}{a}=\frac{5 \sqrt{3}}{10}=\frac{\sqrt{3}}{2}$
Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 25}{10}=5$
