The foci of the ellipse x216+y2b2=1 and the hyperbola x2144-y281=125 coincide. Then the value of b2 is

x2144-y281=125a=14425,b=8125,e=1+81144=1512=54∴   Foci =(±3,0) foci of ellipse = foci of hyperbola ∴   for ellipse ae=3 but a=4,∴  e=34 Then b2=a21-e2 ⇒b2=161-916=7