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 $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$.

Answer 1

The given equation is $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$ or $\dfrac{x^2}{4^2}+ \dfrac{y^2}{3^2} = 1$
Here, the denominator of $\dfrac{x^2}{16}$ is greater than the denominator of $\dfrac{y^2}{9} .$
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 $a = 4$ and $b = 3.$
$∴ c = √(a^2 – b^2)$

$= √(16-9)$

$= √7$

Therefore,
The coordinates of the foci are$(√7, 0)$ and $(-√7, 0)$.
The coordinates of the vertices are$(4, 0)$ and $(-4, 0).$
Length of major axis =$2a= 8$
Length of minor axis = $2b= 6$

Eccentricity, $e = \dfrac{c}{a} = \dfrac{√7}{4}$

Length of the latus rectum $= \dfrac{2b^2}{a} = \dfrac{(2×3^2)}{4}$

                                       $     = \dfrac{(2×9)}{4} $

                                       $= \dfrac{18}{4} $

                                       $ = \dfrac{9}{2}$  (Ans)