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

The key idea is that for an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$, the directrices are vertical lines at $x = \pm \frac{a}{e}$, where $e$ is the eccentricity.

Step 1: Identify $a^2 = 36$ and $b^2 = 20$, so $a = 6$ and $b = \sqrt{20} = 2\sqrt{5}$. Since $a > b$, the major axis is along the $x$-axis.

Step 2: Compute eccentricity:

$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{20}{36}} = \sqrt{\frac{16}{36}} = \frac{2}{3}$. …

The directrices of an ellipse are two vertical lines (for a horizontal major axis) that lie symmetrically on either side of the centre. They are not part of the curve itself, but they play a key role in the ellipse’s definition: for any point on the ellipse, the ratio of its distance to a focus to its distance to the corresponding directrix is constant — that constant is the eccentricity $e$.

For the standard ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$, the foci are at $(\pm ae, 0)$ and the directrices are the lines $x = \pm a/e$. So the distance between the two directrices is simply the distance between these two vertical lines: $2a/e$.

Let’s apply this to the given ellipse.

1. Identify $a$ and $b$.

   The equation is $\frac{x^2}{36} + \frac{y^2}{20} = 1$.

   Here $a^2 = 36$, so $a = 6$ (the semi-major axis, since $36 > 20$).

   And $b^2 = 20$, so $b = \sqrt{20} = 2\sqrt{5}$.

2. Find the eccentricity $e$.

   For an ellipse, $e = \sqrt{1 - \frac{b^2}{a^2}}$.

3. Compute the distance between the directrices. Each directrix is $x = \pm \frac{a}{e} = \pm \frac{6}{2/3} = \pm 9$. …