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.

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

Answer 1

The given equation is ${\displaystyle \frac{x^2}{100}+\frac{y^2}{100}=1}$ or ${\displaystyle \frac{x^2}{{10}^2}+\frac{y^2}{{20}^2}=1}$

Here, the denominator of ${\displaystyle \frac{y^2}{400}}$ is greater than the denominator of ${\displaystyle \frac{x^2}{100}}$.

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation with ${\displaystyle \frac{x^2}{b^2}+\frac{y^2}{a^2}=1}$ we obtain b = 10 and a = 20.

∴ ${\displaystyle c=\sqrt{a^2-b^2}=\sqrt{400-100}=\sqrt{300}=10\sqrt{3}}$

Therfore,

Coordinates of foci are (0, ± c) i.e. ${\displaystyle (0,\pm 10\sqrt{3})}$

Coordinates of vertices are (0, ±a) i.e. (0, ± 20).

Length of major axis = 2a = 2 x 20 = 40

Length of minor axis= 2b = 2 x 10 = 20

Eccentricity (e) = ${\displaystyle \frac{c}{a}=\frac{10\sqrt{3}}{20}=\frac{\sqrt{3}}{2}}$

Length of latus rectum = ${\displaystyle \frac{2b^2}{a}=\frac{2\times 100}{20}=10}$