গণিত প্রশ্ন

The foci of the ellipse x216+y2b2=1\dfrac{x^{2}}{16} + \dfrac{y^{2}}{b^{2}} =1 and the hyperbola x2144−y281=125\dfrac{x^{2}}{144} - \dfrac{y^{2}}{81} =\dfrac{1}{25} coincide, then the value of b2b^{2} is: