Question

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity, and the latus rectum of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$.

Answer 1

Given: Equation of ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$
Here, $25>9$

So, the major axis is along the $x-\text{axis}$, while the minor axis is along the $y-\text{axis}$.

On comparing the given equation with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ (general form), we get

$a^2=25\Rightarrow a=5$
and $b^2=9\Rightarrow b=3$

Now, $c=\sqrt{(a^2-b^2)}$
$\Rightarrow c=\sqrt{(25-9)}=\sqrt{16}$
$\Rightarrow c=4$

$a=5,b=3\text{and}c=4$

$\because$ Major axis is along $x-\text{axis},$ while the minor axis is along $y-\text{axis}$.

Then,The coordinates of the foci $=(\pm c,0)$
$=(\pm 4,0)$

Coordinates of the vertices $=(\pm a,0)$
$=(\pm 5,0)$

Length of major axis $=2a=2\left(5\right)=10$
Length of minor axis $=2b=2\left(3\right)=6$

Eccentricity, $e=\frac{c}{a}=\frac{4}{5}=0.8$

Length of Latus Rectum $=\frac{2b^2}{a}=\frac{2\times 9}{5}=3.6$