Question

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

• A. 9
• B. \( 6\sqrt{5} \)
• C. 18
• D. \( 3\sqrt{5} \)

Hint:

Use \( e = \sqrt{1 - \frac{b^2}{a^2}} \) for ellipse eccentricity. Directrix distance = \( \frac{2a}{e} \).

Answer 1

The Correct Option is C

Given: \[ \begin{align} \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,\quad a^2 = 36,\ b^2 = 20 \Rightarrow a = 6 \] For ellipse, distance between directrices = \( \frac{2a^2}{\sqrt{a^2 - b^2}} \)
Alternatively, for standard horizontal ellipse, directrices are at: \[ \begin{align} x = \pm \frac{a}{e},\quad \text{where } e = \frac{\sqrt{a^2 - b^2}}{a} \Rightarrow e = \frac{\sqrt{36 - 20}}{6} = \frac{\sqrt{16}}{6} = \frac{4}{6} = \frac{2}{3} \] So, \[ \begin{align} \text{Distance between directrices} = \frac{2a}{e} = \frac{2 \cdot 6}{\frac{2}{3}} = \frac{12 \cdot 3}{2} = 18 \]