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

Answer 1

The given equation is $\dfrac{x^2}{25}+\dfrac{y^2}{100}=1$ or $\dfrac{x^2}{5^2}+\dfrac{y^2}{10^2}=1$

Here, the denominator of $\dfrac{y^2}{100}$ is greater than the denominator of $\dfrac{x^2}{25}$.
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis. On comparing the given equation with $\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$, we obtain $b = 5$ and $a = 10.$

$∴ c = √(a^2 – b^2)$

$= √(100-25)$

$= √75$

$= 5√3$
Therefore,
The coordinates of the foci are $(0, ±5√3) .$
The coordinates of the vertices are $(0, ±10)$
Length of major axis = $2a= 20$
Length of minor axis = $2b= 10$

Eccentricity,$e = \dfrac{c}{a} = \dfrac{5√3}{10} = \dfrac{√3}{2}$

Length of latus rectum = $\dfrac{2b^2}{a} = \dfrac{(2×5^2)}{10} = \dfrac{(2×25)}{10} = 5.$