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IIT JEE 2009 Mathematics Paper2 Code 1 Solutions

8.     An ellipse intersects the hyperbola 2x2 - 2y2 = 1 orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinates axes, then

        (A)    equation of ellipse is x2 + 2y2 = 2

        (B)    the foci of ellipse are (+1, 0)

        (C)    equation of ellipse is x2 + 2y2 = 4

        (D)    the foci of ellipse are (+√2, 0)

 

Sol.   (A, B)

        Ellipse and hyperbola will have the same focus if

        => (+ ae, 0) ≡ (+1, 0)

        => (+a x 1/√2 , 0) ≡ (+1, 0)

        => a = √2 and e = 1/√2

        => b2 = a2 (1 - e2) => b2 = 1

       => Hence equation of ellipse  x2/2 + y2/1 = 1.

 

    q9-2009-maths

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