Since the coordinates of one focus at \( (0, 4) = (0, \pm be) \), it is a case of the vertical hyperbola. \[ be = 4 \] It is a case of the vertical hyperbola. The equation of the hyperbola is: \[ \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \quad \text{...(1)} \] Length of the transverse axis is 6: \[ 2b = 6 \Rightarrow b = 3 \] Also, \( a^2 = b^2 (e^2 - 1) \): \[ a^2 = b^2 e^2 - b^2 \] \[ a^2 = 7 \] Substituting in Equation (1), we get: \[ \frac{y^2}{9} - \frac{x^2}{7} = 1 \] Hence, the equation of the hyperbola with center at the origin, length of the transverse axis 6, and one focus at \( (0,4) \) is: \[ \frac{y^2}{9} - \frac{x^2}{7} = 1 \]