# Find the distance between the directrices

Question:

Find the distance between the directrices of the ellipse x2/36 + y2/20 = 1

Solution:

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.