Question

Let the distance between the foci of an ellipse \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (a>b) \] be \(4\) and the distance between its directrices be \(10\). Then the length of its latus rectum is:

• A. \( \dfrac{12}{\sqrt{10}} \)
• B. \( \dfrac{6}{\sqrt{10}} \)
• C. \( \dfrac{8}{\sqrt{5}} \)
• D. \( \sqrt{10} \)

Hint:

For ellipse problems, memorize standard results for foci, directrices, eccentricity, and latus rectum to quickly link given data.

Answer 1

The Correct Option is A

Concept: For an ellipse:
Distance between the foci \(=2c\), where \(c^2=a^2-b^2\)
Eccentricity \( e=\dfrac{c}{a} \)
Distance between the directrices \(= \dfrac{2a}{e} \)
Length of latus rectum \(= \dfrac{2b^2}{a} \)
Step 1: Use the distance between the foci \[ 2c = 4 \Rightarrow c = 2 \]
Step 2: Use the distance between the directrices \[ \frac{2a}{e} = 10 \] Since \( e = \dfrac{c}{a} \), \[ \frac{2a}{c/a} = 10 \Rightarrow \frac{2a^2}{c} = 10 \] Substitute \( c=2 \): \[ a^2 = 10 \Rightarrow a = \sqrt{10} \]
Step 3: Find \(b^2\) \[ b^2 = a^2 - c^2 = 10 - 4 = 6 \]
Step 4: Find the length of the latus rectum \[ \text{Latus rectum} = \frac{2b^2}{a} = \frac{2\times 6}{\sqrt{10}} = \frac{12}{\sqrt{10}} \]