Question

Let \( E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), \( a > b \) be an ellipse of eccentricity \( \frac{1}{\sqrt{3}} \). If the distance between the foci of \( E \) is 4, then the length of its latus rectum is :

• A. \( \frac{4\sqrt{3}}{3} \)
• B. \( \frac{6\sqrt{3}}{3} \)
• C. \( \frac{8\sqrt{3}}{3} \)
• D. \( \frac{10\sqrt{3}}{3} \)

Hint:

Use \( 2ae = 4 \) to find a, then \( b^2 = a^2(1 - e^2) \) and latus rectum \( = \frac{2b^2}{a} \).

Answer 1

The Correct Option is C

Let c be the focal distance, \( c = ae = 2 \). Since \( c^2 = a^2 - b^2 \) and \( e^2 = \frac{c^2}{a^2} = \frac{1}{3} \), we get \( a^2 = 3c^2 = 12 \) and \( b^2 = a^2 - c^2 = 12 - 4 = 8 \).
The length of the latus rectum is \( \frac{2b^2}{a} = \frac{2 \times 8}{2\sqrt{3}} = \frac{8}{\sqrt{3}} = \frac{8\sqrt{3}}{3} \).
Answer: (3) \( \frac{8\sqrt{3}}{3} \)