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 $\dfrac{x^2}{4}+\dfrac{y^2}{25}=1$

Answer 1

The given equation is $\dfrac{x^2}{4}+\dfrac{y^2}{25}=1$ or $\dfrac{x^2}{2^2}+\dfrac{y^2}{5^2}=1$
Here, the denominator of $\dfrac{y^2}{25}$ is greater than the denominator of $\dfrac{x}{4}.$
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis. On comparing the given equation with $\dfrac{x^2}{a^2} \+ \dfrac{y^2}{b^2} = 1$, , we obtain $b = 2$ and $a = 5.$
$∴ c = √(a^2 – b^2)$

$= √(25-4)$

$= √21$
Therefore, The coordinates of the foci are$(0, √21)$ and $(0, -√21).$
The coordinates of the vertices are $(0, 5)$ and $(0, -5)$
Length of major axis = $2a= 10$
Length of minor axis = $2b= 4$

Eccentricity, $e =\dfrac{c}{1} = \dfrac{√21}{5}$

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

                                       $= \dfrac{(2×4)}{5} = \dfrac{8}{5}$  (Ans. )