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 $\frac{x^2}{100}+\frac{y^2}{400}=1.$

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