Question

Let the length of the latus rectum of an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) be equal to the length of its semi-major axis. If the radius of its director circle is \( \sqrt{3} \) and \( e \) is its eccentricity, then the length of its latus rectum is:

• A. \( \frac{1}{a} \)
• B. \( \frac{1}{b} \)
• C. \( \frac{1}{e} \)
• D. \( \frac{1}{ab} \)

Hint:

To find the length of the latus rectum, use the formula \( \frac{2b^2}{a} \) and relate it to the eccentricity \( e \).

Answer 1

The Correct Option is C

We are given that the length of the latus rectum of an ellipse is equal to the length of its semi-major axis. The formula for the length of the latus rectum of an ellipse is: \[ \text{Latus rectum} = \frac{2b^2}{a} \] We are also given that the radius of the director circle is \( \sqrt{3} \) and that the eccentricity \( e \) is related to \( a \) and \( b \) by: \[ e = \sqrt{1 - \frac{b^2}{a^2}} \] From the given conditions, we can find that the length of the latus rectum is \( \frac{1}{e} \). Thus, the correct answer is option (3), \( \frac{1}{e} \).