Question

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 $\frac{x^2}{25}+\frac{y^2}{100}=1$

Answer 1

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 $a=10$
$\therefore ae=c=\sqrt{a^2-b^2}=\sqrt{100-25}=\sqrt{75}=5\sqrt{3}$
The coordinates of the foci are $(0,\pm 5\sqrt{3})$
The coordinates of the vertices are $(0,\pm 10)$
Length of major axis $=2a=20$
Length of minor axis $=2b=10$
Eccentricity $e=$ $\frac{c}{a}=\frac{5\sqrt{3}}{10}=\frac{\sqrt{3}}{2}$
Length of latus rectum $=$ $\frac{2b^2}{a}=\frac{2\times 25}{10}=5$