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

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Solution

The correct option is D 18
Comparing the given equation with $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , we obtain $a^{2} = 36, b^{2} = 20$ .

Let e be the eccentricity of the ellipse. Then,
$b^{2} = a^{2} (1 - e^{2})$
$20 = 36 (1 - e^{2})$
$36 e^{2} = 16$
$e = \frac{2}{3}$

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

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