If the foci x2a2−y2b2=1coincide with the foci of x225+y29=1 and the eccentricity of the hyperbola is 2, then

For the ellipse,    a=5 and e=25−925=45∴     ae=4Hence, the foci ate (-4, 0) and (4, 0)'For the hyperbola,    ae=4,e=2∴     a=2    b2=4(4−1)=12 or     b=12