Question

If the distance between the foci of an ellipse is $6$ and the distance between its directrices is $12$, then the length of its latus rectum is

• A. $2\sqrt{3}$
• B. $\sqrt{3}$
• C. $\frac{3}{\sqrt{2}}$
• D. $3\sqrt{2}$

Answer 1

The correct option is D

$3\sqrt{2}$

Explanation for the correct option:

Step 1: Solve for the major axis and eccentricity of the ellipse

Given that the distance between the foci of an ellipse is $6$ and the distance between its directrices is $12$

Let $a$ be the semi major axis of the ellipse

Let $b$ be the semi minor axis of the ellipse

Let $e$ be the eccentricity of the ellipse

The distance between two foci of the ellipse is $2ae$

$\Rightarrow$$2ae=6$

$\Rightarrow$ $ae=3$ $...\left(i\right)$

The distance between the directrices is $\frac{2a}{e}$

$\Rightarrow$$\frac{2a}{e}=12$

$\Rightarrow$ $\frac{a}{e}=6$ $...\left(ii\right)$

From $\left(i\right),\left(ii\right)$ we get

$\Rightarrow$$\frac{a}{e}\times ae=6\times 3$

$\Rightarrow$ $a^2=9\times 2$

$\Rightarrow$ $a=3\sqrt{2}$

Substituting the value of $a$ in $\left(i\right)$ we get

$\Rightarrow$$2\times 3\sqrt{2}\times e=6$

$\Rightarrow$ $e=\frac{1}{\sqrt{2}}$

Step 2: Solve for length of the latus rectum

The eccentricity of the ellipse is given as

$e^2=1-\frac{b^2}{a^2}$

$\Rightarrow$$\frac{1}{2}=1-\frac{b^2}{18}$

$\Rightarrow$$\frac{b^2}{18}=\frac{1}{2}$

$\Rightarrow$ $b^2=9$

$\Rightarrow$ $b=3$

The length of the latus rectum of the ellipse is given as

$l=\frac{2b^2}{a}$

$\Rightarrow$$l=\frac{2\times 9}{3\sqrt{2}}$

$\Rightarrow$$l=3\sqrt{2}$

Thus the length of the latus rectum is $3\sqrt{2}units$.

Hence option(D) is the correct answer.