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JEE Advanced2006MathematicsHyperbolaActual

If e₁ is the eccentricity of the ellipse x^2 16 + y^2 25 =1 and e₂ is the eccentricity of the hyperbola passing through the focii of the ellipse and e₁ e₂=1 , then equation of the hyperbola is

Options

  1. Ax^2 9 - y^2 16 =1
  2. Bx^2 16 - y^2 9 =-1
  3. Cx^2 9 - y^2 25 =1
  4. DNone of these

Correct answer

B. x^2 16 - y^2 9 =-1

Step-by-step solution

The eccentricity of x^2 16 + y^2 25 =1 is e₁= 1- 16 25 = 3 5 e₂= 5 3 Foci of ellipse (0, 3) Equation of hyperbola is x^2 16 - y^2 9 =-1 Hence (b) is the correct answer.

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