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$\dfrac{x^2}{36} + \dfrac{y^2}{16} = 1$

Answer 1

The given equation is $\dfrac{x^2}{36} + \dfrac{y^2}{16} = 1$
Here, the denominator of $\dfrac{x^2}{36}$ is greater than the denominator of $\dfrac{y^2}{16}$. Therefore, the major axis is along the $x-axis$, while the minor axis is along the $y-axis.$

On comparing the given equation with $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ we obtain $a = 6$ and $b = 4.$

$∴ c = √(a^2 – b^2)$

$= √(36-16)$

$= √20$

$= 2√5$
Therefore, the coordinates of the foci are $(2√5, 0)$ and $(-2√5, 0).$
The coordinates of the vertices are $(6, 0)$ and $(-6, 0).$
Length of major axis =$2a= 12$
Length of minor axis = $2b= 8$

Eccentricity, $e = \dfrac{c}{a} = \dfrac{2√5}{6} = \dfrac{√5}{3}$

Length of latus rectum = $\dfrac{2b^2}{a} = \dfrac{(2×16)}{6} = \dfrac{16}{3}.$ (Ans)