Let a hyperbola passes through the focus of the ellipse x225+y216=1. The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse. Also, the product of the eccentricities of the given ellipse and hyperbola is 1. Then,

For the given ellipsex225+y216=1we have  e=1−1625=35Hence, the eccentricity of the hyperbola is 5/3.Let the hyperbola bex2A2−y2B2=1Then B2=A2e2−1=A2259−1=169A2Therefore, the equation of the hyperbola is x2A2−9y216A2=1As it passes through (3, 0), we get A2=9 and B2 =16' The equation isx29−y216=1The foci of the hyperbola are (±ae,0) = (±5,0).The vertex of the hyperbola is (±3, 0)