If the foci of the ellipse \(\dfrac{x^2}{16} + \dfrac{y^2}{b^2}=1\) and the hyperbola \(\dfrac{x^2}{144}-\dfrac{y^2}{81}=\dfrac{1}{25}\) coincide, then the value of b2 is

1
1
2
5
3
7
4
More than one of the above
5
None of the above

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