Question

Find the coordinatesof the foce, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
$\frac{x^2}{4}+\frac{y^2}{25}=1.$

Answer 1

The equation of given ellipse is
$\frac{x^2}{4}+\frac{y^2}{25}=1.$
now $25>4\Rightarrow a^2=25and{b}^2=4$
So the equation of ellipse in standard form is
$\frac{y^2}{a^2}+\frac{x^2}{b^2}=1\therefore a^2=25\Rightarrow a=5and{b}^2=4\Rightarrow b=2$
We know that c = \sqrt{a^2 - b^2}\\
$\therefore c=\sqrt{25-4}=\sqrt{21}$
$\therefore$ Coordinates fo foci are $(0,\pm c)i.e.(0,\pm \sqrt{21})$
Coordinates fo vertices are $(0,\pm a)i.e.(0,\pm 5)$
Length of minor axis $=2b=2\times 2=4$
Eccentricity $\left(e\right)=\frac{c}{a}=\frac{21}{5}$
Length of latus rectum $=\frac{2b^2}{a}=\frac{2\times 4}{5}=\frac{8}{5}$