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

The equation of given ellipse is,
$36x^2+4y^2=144$
$i.e.\frac{36x^2}{14}+\frac{4y^2}{144}=1\Rightarrow \frac{x^2}{4}+\frac{y^2}{36}=1.\text{Now}36>4\Rightarrow a^2=36and{b}^2=4\text{So the equation of ellipse in standard form is}\frac{y^2}{a^2}+\frac{x^2}{b^2}=1$
$\therefore a^2=36\Rightarrow a=6and{b}^2=4\Rightarrow b=2\text{We know that}c=\sqrt{a^2-b^2}\therefore c=\sqrt{36-4}=\sqrt{32}=4\sqrt{2}\therefore \text{Coordinates of foci are}(0,+c)i.e.(0,\pm \sqrt{2})\text{Coordinates of vertices are}(0,\pm a)i.e.(0,\pm 6)\text{Length of major axis}=2a=2\times 6=12\text{Length of minor axis}=2b=2\times 2=4\text{Eccentricity}\left(e\right)=\frac{c}{a}=\frac{4\sqrt{2}}{6}=\frac{2\sqrt{2}}{3}\text{Length of latus rectum}=\frac{2b^2}{a}=\frac{2\times 4}{6}=\frac{4}{3}$