Question

Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity, and the length of the latus rectum of the ellipse $\dfrac{x^2}{100}+\dfrac{y^2}{400}=1$

Answer 1

The given equation is $\dfrac{x^2}{100}+\dfrac{y^2}{400}=1$ or $\dfrac{x^2}{10^2}+\dfrac{y^2}{20^2}=1$

Here, the denominator of $\dfrac{y^2}{400}$ is greater than the denominator of $\dfrac{x^2}{100}.$
Therefore, the major axis is along the $y-axis$, while the minor axis is along the$x-axis.$
On comparing the given equation with $\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$, we obtain $b = 10$ and $a = 20.$
$∴ c = √(a^2 – b^2)$

$= √(400-100)$

$= √300$

$= 10√3$

Therefore,
The coordinates of the foci are $(0, ±10√3).$
The coordinates of the vertices are $(0, ±20)$
Length of major axis =$2a= 40$
Length of the minor axis = $2b = 2 (10) = 20$
Eccentricity, $e = c/a = 10√3/20 = √3/2$
Length of latus rectum = $\dfrac{2b^2}{a} = \dfrac{(2×10^2)}{20} =\dfrac{(2×100)}{20} = 10$