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.

$\frac{x^2}{49}+\frac{y^2}{36}=1$

Answer 1

Now $49>36\Rightarrow a^2=49and{b}^2=36$
So the equation of ellipse in standard form is
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\therefore a^2=49\Rightarrow a=7and{b}^2=36\Rightarrow b=6$
We know that, $c=\sqrt{a^2-b^2}\therefore c=\sqrt{49-36}=\sqrt{13}$
$\therefore$ Coordinates of foci are, $(\pm c,0)i.e.(\pm \sqrt{13},0)\text{Coordinates of vertices are}(\pm a,0)i.e.(\pm 7,0)\text{Length of major axis =}2a=2\times 7=14\text{Length of minor axis}=2b=2\times 6=16\text{Eccentricity}\left(e\right)=\frac{c}{a}=\frac{\sqrt{13}}{7}\text{Length of latus rectum}=\frac{2b^2}{a}=\frac{2\times 36}{7}=\frac{72}{7}$