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AIEEE 2006

JEE Main 2006 Mathematics question on Conic Sections

If e1e_1 is the eccentricity of the ellipse x216+y225=1\frac{x^2}{16} + \frac{y^2}{25} = 1 and e2e_2 is the eccentricity of the hyperbola passing through the foci of the ellipse and e1e2=1e_1 e_2 = 1, then equation of the hyperbola is
  1. A.
    x29−y216=1\frac{x^2}{9} - \frac{y^2}{16} = 1
  2. B.
    x216−y29=−1\frac{x^2}{16} - \frac{y^2}{9} = -1
  3. C.
    x29−y225=1\frac{x^2}{9} - \frac{y^2}{25} = 1
  4. D.
    None of these
Show answer and solution

Correct option: B

Solution

For the ellipse x²/16 + y²/25 = 1, since 25 > 16, the major axis is along y. So a² = 25, b² = 16, giving c² = 25 - 16 = 9, c = 3. Thus e₁ = 3/5, and since e₁e₂ = 1, e₂ = 5/3.

The foci of the ellipse are at (0, ±3). The hyperbola passes through (0, ±3), so its transverse axis is along y, meaning it has the form y²/A² - x²/B² = 1.

For this hyperbola, e₂ = c/A where c² = A² + B², and the vertices/foci give A = 3 (since the hyperbola passes through (0,±3), these are the vertices). Then e₂ = 5/3 means c = 5, so B² = c² - A² = 25 - 9 = 16.

The equation is y²/9 - x²/16 = 1, which is the same as x²/16 - y²/9 = -1. This matches option 2.

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