Let the ellipse E₁: x²/a² + y²/b² = 1, a > b and E₂: x²/A² + y²/B² = 1, A < B have same eccentricity 1/√3. Let the product of their lengths of latus rectums be 32/√3 and the distance between the foci of E₁ be 4. If E₁ and E₂ meet at A, B, C and D, then the area of the quadrilateral ABCD equals :

Solution

Maths sir · 5 steps

Find both ellipses first, then solve for x² and y² as a simple linear pair.

1. Fix E1. Focal distance 2ae = 4 with e = 1/√3 gives a = 2√3, so a² = 12. Then b² = a²(1 - e²) = 12 × 2/3 = 8. Its latus rectum is 2b²/a = 16/2√3 = 8/√3.
2. Fix E2. The product of latus rectums is 32/√3, so the latus rectum of E₂ is 4. Since A < B, the major axis is along y, so 2A²/B = 4 and A² = B²(1 - e²) = 2B²/3. This gives 4B/3 = 4, so B² = 9 and A² = 6.
3. Intersect them. Take X = x² and Y = y². Multiplying out X/12 + Y/8 = 1 and X/6 + Y/9 = 1 gives 2X + 3Y = 24 and 3X + 2Y = 18. Solving gives Y = 36/5 and X = 6/5.
4. Rectangle area. Both curves are symmetric about both axes, so the four points (± x, ± y) form a rectangle with sides 2|x| and 2|y|.

   Area = 4|x||y| = 4√(6/5)√(36/5) = 4 × 6√6/5 = 24√6/5

5. So. The area of ABCD is 24√6/5, which is option (a).

The idea to keep. Two centred ellipses with axes along the coordinate axes always meet in a rectangle. Solve for x² and y² as a linear pair, then the area is 4|x||y|.

Answer from the NTA final answer key. Solution by PrepWiser.