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. $4\sqrt{3}$
• B. $\frac{4}{\sqrt{3}}$
• C. $\frac{\sqrt{3}}{4}$
• D. $3\sqrt{2}$

Answer 1

The correct option is D

$3\sqrt{2}$

Explanation for correct option

Step 1: Given data

1. The distance between the foci of an ellipse is $6$
2. The distance between its directrices is $12$,

Step 2: Formulae used

For the general equation of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a,b$ are lengths of semi-major and semi-minor axis respectively

1. Distance between foci is $=2ae$
2. Distance between the directrices$=\frac{2a}{e}$
3. $e$ is the eccentricity of the ellipse and $e=\sqrt{1-\frac{b^2}{a^2}}$
4. Length of Latus rectum is $=\frac{2b^2}{a}$

Step 3: Solve for the value of $a$

Given that the distance between the foci is $6$

$$\begin{gathered} \therefore 2ae=6 \\ \Rightarrow ae=3....\left(i\right) \end{gathered}$$

The distance between its directrices is $12$,

$$\begin{gathered} \therefore \frac{2a}{e}=12 \\ \Rightarrow \frac{a}{e}=6.........\left(ii\right) \end{gathered}$$

Multiplying $\left(i\right)\&\left(ii\right)$ we get

$$\begin{gathered} \Rightarrow ae\times \frac{a}{e}=3\times 6 \\ \Rightarrow a^2=18 \\ \Rightarrow a=\sqrt{18} \\ \Rightarrow a=3\sqrt{2} \end{gathered}$$

Step 4: Solve for the value of $b^2$

From the equation of eccentricity of ellipse we have

$$\begin{gathered} \Rightarrow e=\sqrt{1-\frac{b^2}{a^2}} \\ \Rightarrow a^2-{\left(ae\right)}^2=b^2 \\ \Rightarrow b^2=18-9\left[\because ae=3\mathrm{and}{\mathrm{a}}^2=18\right] \\ \Rightarrow b^2=9 \end{gathered}$$

Step 5: Solve for the length of latus rectum

Length of Latus rectum is $=\frac{2b^2}{a}$

$$\begin{gathered} =\frac{2\times 9}{3\sqrt{2}} \\ =3\sqrt{2} \end{gathered}$$

Hence option (D) is correct i.e. $3\sqrt{2}$