Find the distance between the directrices of the ellipse x2/36 + y2/20 = 1
We know that equation of an ellipse
$=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Also we have Length of latus rectum $=\frac{2 b^{2}}{a}$
Length of minor axis $=2 \mathrm{~b}$
$\mathrm{b}^{2}=\mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right) \ldots \ldots \ldots \ldots 1$
Now by substituting we get
$\frac{x^{2}}{(\sqrt{36})^{2}}+\frac{y^{2}}{(\sqrt{20})^{2}}=1$
$\frac{x^{2}}{(6)^{2}}+\frac{y^{2}}{(2 \sqrt{5})^{2}}=1$
From equation 1, we have
$20=36\left(1-e^{2}\right)$
On rearranging we get
$\frac{20}{36}=1-\mathrm{e}^{2}$
$\frac{20}{36}=1-\mathrm{e}^{2}$
$e^{2}=1-\frac{20}{36}$
$\mathrm{e}^{2}=\frac{16}{36}=\frac{4}{9}$
$e=\frac{2}{3}$
Directrices $=\pm \frac{\mathrm{a}}{\mathrm{e}}$
Distance between directrices $=$ $\frac{2 \mathrm{a}}{\mathrm{e}}=\frac{2 \times 6}{\frac{2}{2}}=18$
Hence the distance between directrices is 18 units.