Question:medium

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:

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Remember that the product of the focal distance ($2ae$) and the directrix distance ($2a/e$) is $4a^2$. This is a fast way to find $a^2$ directly!
Updated On: Jun 3, 2026
  • $3\sqrt{2}$
  • $\sqrt{2}$
  • $2\sqrt{2}$
  • $4\sqrt{2}$
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The Correct Option is A

Solution and Explanation

Step 1: Recall the three ellipse facts.
For an ellipse with the bigger axis along x, three things matter. The gap between the two foci is $2ae$. The gap between the two directrix lines is $\frac{2a}{e}$. The latus rectum length is $\frac{2b^2}{a}$. Here $a$ is the big radius, $b$ is the small radius, and $e$ is the eccentricity. We want the latus rectum, so we must find $a$ and $b$ first.

Step 2: Write down what is given.
We are told the foci are $6$ apart and the directrices are $12$ apart. So $2ae = 6$, which means $ae = 3$. Also $\frac{2a}{e} = 12$, which means $\frac{a}{e} = 6$. Now we have two simple facts about $a$ and $e$.

Step 3: Multiply the two facts to get $a$.
Multiply $ae = 3$ by $\frac{a}{e} = 6$. The $e$ cancels out neatly.
\[ ae \times \frac{a}{e} = 3 \times 6 \] So $a^2 = 18$, which gives $a = 3\sqrt{2}$.

Step 4: Divide the two facts to get $e$.
Now divide $ae = 3$ by $\frac{a}{e} = 6$. This time the $a$ cancels out.
\[ \frac{ae}{a/e} = \frac{3}{6} \] So $e^2 = \frac{1}{2}$. We do not even need $e$ by itself, just $e^2$.

Step 5: Find $b^2$ from $a$ and $e$.
The link between the radii and eccentricity is $b^2 = a^2(1 - e^2)$. Put in the numbers.
\[ b^2 = 18\left(1 - \tfrac{1}{2}\right) = 18 \times \tfrac{1}{2} = 9 \] So the small radius squared is $9$.

Step 6: Plug into the latus rectum formula.
Now use $\frac{2b^2}{a}$ with $b^2 = 9$ and $a = 3\sqrt{2}$.
\[ \frac{2 \times 9}{3\sqrt{2}} = \frac{18}{3\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2} \] So the latus rectum length is $3\sqrt{2}$.
\[ \boxed{3\sqrt{2}} \]
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