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QJEEAdv06Mab282: JEE Advanced 2006 Mathematics - Class 11 JEEAdv06Mab282 Solution: Let a hyperbola passes through the focus of the ellipse . The…

Let a hyperbola passes through the focus of the ellipse x225+y216=1{{{x^2}} \over {25}} + {{{y^2}} \over {16}} = 1 . The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse, also the produced of eccentricities of given ellipse and hyperbola is 11 , then

Multiple Choice Options

  • A the equation of hyperbola is x29+y216=1{{{x^2}} \over 9} + {{{y^2}} \over {16}} = 1
  • B the equation of hyperbola is x29+y225=1{{{x^2}} \over 9} + {{{y^2}} \over {25}} = 1
  • C focus of hyperbola is (5,0)(5, 0)
  • D vertex of hyperbola is (53,0)\left( {5\sqrt 3 ,0} \right)

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