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:
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: Concept
For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ($a > b$), the distance between the foci is $2ae$, the distance between the directrices is $\frac{2a}{e}$, and the length of the latus rectum is $\frac{2b^2}{a}$.
Step 2: Meaning
We are given $2ae = 6 \implies ae = 3$ and $\frac{2a}{e} = 12 \implies \frac{a}{e} = 6$. We need to find $a$, $e$, and then $b^2$ to compute $\frac{2b^2}{a}$.
Step 3: Analysis
Multiplying $ae = 3$ and $\frac{a}{e} = 6$: \[ (ae) \cdot \left(\frac{a}{e}\right) = 3 \times 6 \implies a^2 = 18 \implies a = 3\sqrt{2} \] Dividing $ae = 3$ by $\frac{a}{e} = 6$: \[ e^2 = \frac{3}{6} = \frac{1}{2} \implies e = \frac{1}{\sqrt{2}} \] Now, find $b^2$ using $b^2 = a^2(1 - e^2)$: \[ b^2 = 18\left(1 - \frac{1}{2}\right) = 9 \] The length of the latus rectum is: \[ \text{L.R.} = \frac{2b^2}{a} = \frac{2(9)}{3\sqrt{2}} = \frac{18}{3\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2} \]
Step 4: Conclusion
The length of the latus rectum is $3\sqrt{2}$.
Final Answer: (A)
For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ($a > b$), the distance between the foci is $2ae$, the distance between the directrices is $\frac{2a}{e}$, and the length of the latus rectum is $\frac{2b^2}{a}$.
Step 2: Meaning
We are given $2ae = 6 \implies ae = 3$ and $\frac{2a}{e} = 12 \implies \frac{a}{e} = 6$. We need to find $a$, $e$, and then $b^2$ to compute $\frac{2b^2}{a}$.
Step 3: Analysis
Multiplying $ae = 3$ and $\frac{a}{e} = 6$: \[ (ae) \cdot \left(\frac{a}{e}\right) = 3 \times 6 \implies a^2 = 18 \implies a = 3\sqrt{2} \] Dividing $ae = 3$ by $\frac{a}{e} = 6$: \[ e^2 = \frac{3}{6} = \frac{1}{2} \implies e = \frac{1}{\sqrt{2}} \] Now, find $b^2$ using $b^2 = a^2(1 - e^2)$: \[ b^2 = 18\left(1 - \frac{1}{2}\right) = 9 \] The length of the latus rectum is: \[ \text{L.R.} = \frac{2b^2}{a} = \frac{2(9)}{3\sqrt{2}} = \frac{18}{3\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2} \]
Step 4: Conclusion
The length of the latus rectum is $3\sqrt{2}$.
Final Answer: (A)
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