Question

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

Answer 1

Since the vertices of the ellipse lie on the y-axis, it is a vertical ellipse.

Let the required equation be $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1,$ and $a^2>b^2.$

Its vertices are $(0,\pm a)$ and therefore, $a=10.$

Let $c^2=(a^2-b^2).$

Then, $e=\frac{c}{a}\Rightarrow c=ae=(10\times \frac{4}{5})=8.$

Now, $c^2=(a^2-b^2)\iff b^2=(a^2-c^2)=(100-64)=36.$

$\therefore a^2=(10{)}^2=100$ and $b^2=36.$

Hence, the required equation is $\frac{x^2}{36}+\frac{y^2}{100}=1.$