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.
$\frac{x^2}{16}+\frac{y^2}{9}=1$

Answer 1

The equation of given ellipse is
$\frac{x^2}{16}+\frac{y^2}{9}=1$
Now $16>9\Rightarrow a^2=16and{b}^2=9$
So the equationof ellipse in standard form is
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$a^2=16\Rightarrow a=4and{b}^2=9\Rightarrow b=3$
We know that $c=\sqrt{a^2-b^2}\therefore c=\sqrt{16-9}=\sqrt{7}$
\thereofre Coordinates fo foci are $(\pm c,0)i.e.(\pm \sqrt{7},0)$
coordinates fo vertices are $(\pm a,0)i.e.(\pm 4,0)$
Length of major axis $=2a=2\times 4=8$
Length of minor axis $=2b=2\times 3=6$
Eccentricity (e) $=\frac{c}{a}=frac\sqrt{7}4$
Length of latus rectum $=\frac{2b^2}{a}=\frac{2\times 9}{4}=\frac{9}{2}.$