The ellipse E1:9x2+4y2=1 is inscribed in a rectangle R whose sides are parallel to the coordinate axes. Another ellipse E2 passing through the point (0, 4) circumscribes the rectangle R. The eccentricity of the ellipse E2 is
(A)22
(B)23
(C)21✓ Correct
(D)43
Answer: (C)
Solution
The rectangle R has vertices at (±3,±2). The ellipse E2 passes through (0,4) and (3,2). Let E2 be a2x2+b2y2=1. From (0,4): b2=16. From (3,2): a29+164=1, giving a29=43, so a2=12. The eccentricity is e=1−b2a2=1−1612=41=21.
JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Single correct option