Let a and b respectively be the semi-transverse and semi-conjugate axes of a standard hyperbola whose…

Let a and b respectively be the semi-transverse and semi-conjugate axes of a standard hyperbola whose eccentricity satisfies the equation 9e2-18e+5=0. If S5, 0 is a focus and 5x=9 is the corresponding directrix of this hyperbola, then a2-b2 is equal to
  1. -7
  2. -5
  3. 5
  4. 7

Solution

Given that focus of hyperbola is S5, 0

ae=5......i

Therefore, Directrix x=ae

 ae=95  ......ii

From i and ii, we get   a2=9

a=3 and e=53    

(It satisfies 9e2-18e+5=0 )

We also know that for an ellipse,

b2=a2e2-1

b2=16

a2-b2=9-16= -7

Asked in: JEE Main 2016 (09 Apr Online)

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