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AIEEE 2012

JEE Main 2012 Mathematics question on Conic Sections

The ellipse E1:x29+y24=1E_1: \frac{x^2}{9} + \frac{y^2}{4} = 1 is inscribed in a rectangle RR whose sides are parallel to the coordinate axes. Another ellipse E2E_2 passing through the point (0,4)(0, 4) circumscribes the rectangle RR. The eccentricity of the ellipse E2E_2 is
  1. A.
    22\frac{\sqrt{2}}{2}
  2. B.
    32\frac{\sqrt{3}}{2}
  3. C.
    12\frac{1}{2}
  4. D.
    34\frac{3}{4}
Show answer and solution

Correct option: C

Solution

The rectangle R has sides parallel to the axes and is inscribed around E₁, so its vertices are at (±3, ±2). E₂ circumscribes this rectangle, so it passes through (3, 2) and (0, 4).

Let E₂: x²/a² + y²/b² = 1. Since it passes through (0, 4), b = 4. Since it passes through (3, 2): 9/a² + 4/16 = 1, giving 9/a² = 3/4, so a² = 12.

Since b² = 16 > a² = 12, the major axis is along y. Then c² = b² - a² = 16 - 12 = 4, so e = c/b = 2/4 = 1/2.

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