The foci of the ellipse x^216+ y^2b^2=1 and the hyperbola x^2144- y^281= 125 coincide. Then, the value of b^2 is

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Correct answer: B. (b) Equation of ellipse, x^216+ y^2b^2=1 Eccentricity, ∴e= 1- b^216= 16-b^24 So, the focus will be ( 16-b^2 ,0) Also, the equation of hyperbola, x^2144- y^281= 125 ⇒ x^2 ( 14425 )- y^2 ( 8125 )=1 ∴e= 1+ b^2a^2= 1+ 8125 14425= 1512 So, the focus will be ( 125 1512 ,0) i .e . , ( 3 ,0 ) On comparing the focus of ellipse with hyperbola, we get 16-b^2=316-b^2=9b^2=7

$The foci of the ellipse \frac{x ^{2}}{16} + \frac{y ^{2}}{b ^{2}} = 1 and the hyperbola \frac{x ^{2}}{144} - \frac{y ^{2}}{81} = \frac{1}{25} coincide. Then, the value of b^{2} is$

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