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 $36x^2+4y^2=144$

Answer 1

The given equation is $36x^2 + 4y^2 = 144$
It can be written as
$36x^2 + 4y^2 = 144$
$⇒\dfrac{x^2}{4} + \dfrac{y^2}{36} = 1$
$⇒\dfrac{x^2}{2^2} + \dfrac{y^2}{6^2} = 1$ .......(1)

Here, the denominator of $\dfrac{y^2}{6^2}$ is greater than the denominator of $\dfrac{x^2}{2^2}$
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.
On comparing the given equation (1) with $\dfrac{x^2}{b^2} + \dfrac{y^2}{6^2} = 1$, we obtain $b = 2$ and $a =6.$
∴

$c = √(a^2 – b^2)$

$= √(36-4)$

$= √32$

$= 4√2$

Therefore,
The coordinates of the foci are $(0, ±4√2).$
The coordinates of the vertices are$(0, ±6).$
Length of major axis = $2a= 12$
Length of the minor axis = $2b = 4$
Eccentricity,$e = \dfrac{c}{a} = \dfrac{4√2}{6} = \dfrac{2√2}{3}$
Length of latus rectum = $\dfrac{2b^2}{a}$

$= \dfrac{(2×2^2)}{6}$

$= \dfrac{(2×4)}{6}$

$= \dfrac{4}{3}$ (Ans)