Question

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity, and the length of the latus rectum of the ellipse $36x^2+4y^2=144$

Answer 1

Given: Equation of ellipse $36x^2+4y^2=144$
i.e., $\frac{x^2}{4}+\frac{y^2}{36}=1$
Here, $4<36$
So, the major axis is along the $y$-axis, while the minor axis is along the $x$-axis.
On comparing the given equation with $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ (general form), we get
$a^2=36\Rightarrow a=6$ and $b^2=4\Rightarrow b=2$
Now, $c=\sqrt{(a^2-b^2)}=\sqrt{(36-4)}=\sqrt{32}$
$\therefore c=4\sqrt{2}$
$a=6,b=2,c=4\sqrt{2}$
$\because$ The major axis is along $y$-axis, while minor axis is along $x$-axis.
Then, coordinates of foci $=(0,\pm c)=(0,\pm 4\sqrt{2})$
Corrdinates of vertices $=(0,\pm a)=(0,\pm 6)$
Length of major axis $=2a=2\left(6\right)=12$
Length of minor axis $=2b=2\left(2\right)=4$
Eccentricity, $e=\frac{c}{a}=\frac{4\sqrt{2}}{6}=\frac{2\sqrt{2}}{3}$
Length of latus rectum $=\frac{2b^2}{a}=\frac{2\times 4}{6}=\frac{4}{3}$