Q 1226812771.     If `e_1` is the eccentricity of the ellipse `x^2/16+y^2/25=1` and `e_2` is the eccentricity of the hyperbola passing through the foci of the ellipse and `e_1 e_2 = 1`, then equation of the hyperbola is

JEE 2006
A

`x^2/9−y^2/25=1`

B

`x^2/16−y^2/9=−1`

C

`x^2/25−y^2/9=1`

D

`x^2/9−y^2/16=1`

HINT

`e = sqrt(1−a^2 / b^2)`
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