Question

Find the distance between the directrices the ellipse $\frac{x^2}{36}+\frac{y^2}{20}=1$.

Answer 1

Comparing the given equation with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we obtain $a^2=36$ and $b^2=20$.

Let e be the eccentricity of the ellipse. Then,

$b^2=a^2(1-e^2)$

$20=36(1-e^2)$

$36e^2=16$

$e=\frac{2}{3}$

Therefore, distance between the directrices = $\frac{2a}{e}=\frac{2\times 6}{2/3}=18$.