MathematicsMedium57×since 2003Q3957The foci of the ellipse x216+y2b2=1{{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 116x2+b2y2=1 and the hyperbola x2144−y281=125{{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}}144x2−81y2=251 coincide. Then the value of b2{b^2}b2 is :A999B111C555D777Check answerSkip