Let the ellipse E1 : x2a2 + y2b2 = 1, a > b and E2 : x2A2 + y2B2 = 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 :
- 24 65
- 18 65
- 6 6
- 12 65
Answer
(1) 24 65
Solution
For E1, 2ae = 4 gives a = 2 3, a2 = 12, and b2 = a2(1 - e2) = 12 23 = 8.
So the latus rectum of E1 is 2b2a = 162 3 = 8 3, and that of E2 is 32 3 8 3 = 4.
For E2 (major axis along y-axis): A2 = B2(1 - e2) = 23B2
2A2B = 4 4B3 = 4 B2 = 9, A2 = 6
The intersection points satisfy x212 + y28 = 1 and x26 + y29 = 1, i.e.
2x2 + 3y2 = 24 and 3x2 + 2y2 = 18
x2 = 65, y2 = 365
ABCD is a rectangle with sides 2|x| and 2|y|.
Area = 4|x||y| = 4 65 365 = 24 65
Answer: (1)