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If the foci of the ellipse x^2 16 + y^2 b^2 =1 coincide with the foci of the hyperbola x^2 144 - y^2 81 = 1 25 , then b^2 is equal to

Options

  1. A8
  2. B10
  3. C7
  4. D9

Correct answer

C. 7

Step-by-step solution

Given equation of ellipse is x^2 16 + y^2 b^2 =1 aligned & eccentricity =e= 1- b^2 16 & foci: a e= 4 1- b^2 16 aligned Equation of hyperbola is x^2 144 - y^2 81 = 1 25 aligned & x^2 144 25 - y^2 81 25 =1 & eccentricity =e= 1+ 81 25 25 144 = 1+ 81 144 & = 225 144 = 15 12 & foci: a e= 12 5 15 12 = 3 & aligned Since, foci of ellipse and hyperbola coincide & 4 1- b^2 16 = 3 b^2=7

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