Question

Find the equation of an ellipse whose vertices are $(0,\pm 10)$ and eccentricity $e=\frac{4}{5}$.

Answer 1

Let the equation of the ellipse be
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ...(i)
The coordinates of vertices are $(0,\pm b)$ i.e., $(0,\pm 10)$
$\therefore b=10$
$\Rightarrow b^2=100$
Now,
$a^2=b^2(1-e^2)$
$a^2=100[1-{\left(\frac{4}{5}\right)}^2]$
$\Rightarrow a^2=\left[\frac{9}{25}\right]$
$\Rightarrow a^2=4\times 9=36$
Putting $a^2=36$ in $b^2=100$ in equation (i), we get
$\Rightarrow \frac{x^2}{36}+\frac{y^2}{100}=1$
$\Rightarrow \frac{100x^2+36y^2}{3600}=1$
$\Rightarrow 100x^2+36y^2=3600$
This is the equation of the required ellipse.