Question

Let the ellipse \( E_1 : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), \( a > b \) and \( E_2 : \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 \), \( A < B \), have the same eccentricity \( \sqrt{\frac{1}{3}} \). Let the product of their lengths of latus rectums be \( \sqrt{\frac{32}{3}} \) and the distance between the foci of \( E_1 \) be 4. If \( E_1 \) and \( E_2 \) meet at \( A \), \( B \), \( C \), and \( D \), then the area of the quadrilateral ABCD equals:

• A. \( 18 \sqrt{6} \)
• B. \( 6 \sqrt{6} \)
• C. \( 12 \sqrt{6} \)
• D. \( 24 \sqrt{6} \)

Hint:

To find the area of an intersection of ellipses, use properties of the eccentricity, latus rectum, and geometry of the ellipses.

Answer 1

The Correct Option is C

- First, we use the given data about the eccentricities and the lengths of the latus rectums to find the area of the quadrilateral formed by the intersection points \( A \), \( B \), \( C \), and \( D \).
- Using geometry of the ellipses and their intersection, we find that the area of the quadrilateral is \( 12 \sqrt{6} \).