Question

Find the coordinatesof the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
$\frac{x^2}{36}+\frac{y^2}{16}=1.$

Answer 1

The equation of given ellipse is
$\frac{x^2}{36}+\frac{y^2}{16}=1.$
Now ($36>16\Rightarrow a^2=36and{b}^2=16$
So the equation of ellipse in standard form is,
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\therefore a^2=36\Rightarrow a=6and{b}^2=16\Rightarrow b=4$
We know that $c=\sqrt{a^2-b^2}\therefore c=\sqrt{36-16}=\sqrt{20}=2\sqrt{5}$
$\therefore$ Coordinates of foci are $(\pm c,0)i.e.(\pm 2\sqrt{5},0)$
Coordinates fo vertices are $(\pm a,0)i.e.(\pm 6,0)$
Length of major axis $=2a=2\times 6=12$
Length of minor axis $=2b=2\times 4=8$
Eccentricity $\left(e\right)=\frac{c}{a}=\frac{2\sqrt{5}}{6}=\frac{\sqrt{5}}{3}$
Length of latus rectum $=\frac{2b^2}{a}=\frac{2\times 16}{6}=\frac{16}{3}.$