The eccentricity of the hyperbola x2a2−y2b2=1 is reciprocal to that of the ellipse x2+4y2=4 . If the hyperbola passes through a foci of the ellipse, then

Given ellipse is x24+y21=1 .  Eccentricity of the ellipse =1−14=32 Focus of the ellipse =(−3,0) Eccentricity of the hyperbola =1+b2a2 So, 1+b2a2=23⇒ba=13 Since the hyperbola passes through the focus of the ellipse, 3a2−0=1⇒a2=3 and b2=1 Thus, the equation of the hyperbola is x23−y21=1 or x2−3y2=3 Focus of hyperbola is (±2,0) .