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If vertex and focus of hyperbola are (2,3) and (6,3) respectively and eccentricity e of the hyperbola is 2 then equation of the hyperbola is

(A)(x+1)216(y3)248=1(B)(x+2)29(y3)227=1(C)(x+2)216(y3)248=1(D)noneofthese

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