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 $\frac{x^{2}}{4}+\frac{y^2} {25}=1$.
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 $\frac{x^{2}}{4}+\frac{y^2} {25}=1$.
The given equation is $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$ or $\frac{x^{2}}{2^{2}}+\frac{y^{2}}{5^{2}}=1$
Here, the denominator of $\frac{y^{2}}{25}$ is greater than the denominator of $\frac{x^{2}}{4}$
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 $b=2$ and $a=5$
$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{25-4}=\sqrt{21}$
Therefore,
The coordinates of the foci are $(0, \sqrt{21})$ and $(0,-\sqrt{21})$
The coordinates of the vertices are $(0,\,5)$ and $(0,\,-5)$
Length of major axis $=2 a=10$
Length of minor axis $=2 b =4$
Eccentricity, $e=\frac{c}{a}=\frac{\sqrt{21}}{5}$
Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 4}{5}=\frac{8}{5}$
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