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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.

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