JEE Main 09 April 2016 Paper

Section: Mathematics

Question : 81 of 90

Marks: +1, -0

Let a and b respectively be the semi-transverse semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation 9e2−18e+5=0. If S(5,0) is a focus and 5x=9 is the corresponding directrix of this hyperbola, then a2−b2 is equal to :

[Main 9 Apr 2016]

7
− 7
5
− 5

Validate

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