IIT-JEE 2006 | Hyperbola Question 21 | Mathematics

IIT-JEE 2006

MCQ (More than One Correct Answer)

-1

If a hyperbola passes through the focus of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ and its transverse and conjugate axes coincide with the major and minor axes of the ellipse, and the product of eccentricities is 1 , then

the equation of hyperbola is $\frac{x^2}{9}-\frac{y^2}{16}=1$

the equation of hyperbola is $\frac{x^2}{9}-\frac{y^2}{25}=1$

focus of hyperbola is $(5,0)$

focus of hyperbola is $(5 \sqrt{3}, 0)$

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