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 \( \frac{1}{\sqrt{3}} \). Let the product of their lengths of latus rectums be \( \frac{32}{\sqrt{3}} \) 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. \( \frac{18 \sqrt{6}}{5} \)
• B. \( 6 \sqrt{6} \)
• C. \( \frac{12 \sqrt{6}}{5} \)
• D. \( \frac{24 \sqrt{6}}{5} \)

Hint:

When solving problems with ellipses, use their geometric properties such as eccentricity, lengths of latus rectum, and foci distance to find relationships between the parameters and solve for unknowns.

Answer 1

The Correct Option is C

Step 1: From the equation of the ellipses, we know that the eccentricity \( e \) of both ellipses is given as \( e = \frac{1}{\sqrt{3}} \), which means \( e = \sqrt{1 - \frac{b^2}{a^2}} \) for the first ellipse. Using this, we can solve for \( a \) and \( b \). 
Step 2: Similarly, use the given condition for the lengths of the latus rectum and the distance between the foci to calculate the parameters \( A \) and \( B \) for the second ellipse. 
Step 3: Use the geometric properties of the two ellipses, including the points where they meet, to compute the area of the quadrilateral formed by the intersections, and the result will be \( \frac{12 \sqrt{6}}{5} \). Thus, the correct answer is (3).