Answer
$$\eqalign{
& {\text{Focus: }}\left( {4,3} \right) \cr
& {\text{directrix: }}x = - 2 \cr
& {\text{axis of symmetry: }}y = 3 \cr} $$
Work Step by Step
$$\eqalign{
& {\left( {y - 3} \right)^2} = 12\left( {x - 1} \right) \cr
& {\text{This equation is written in the form }}{\left( {y - k} \right)^2} = 4p\left( {x - h} \right){\text{ }} \cr
& {\text{represents Parabola with horizontal axis of symmetry}} \cr
& {\text{of Symmetry and Vertex }}\left( {h,k} \right). \cr
& \underbrace {{{\left( {y - 3} \right)}^2} = 12\left( {x - 1} \right)}_{{{\left( {y - k} \right)}^2} = 4p\left( {x - h} \right).} \cr
& {\text{Let }}k = 3,\,\,\,\,h = 1 \cr
& 4p = 12 \cr
& p = 3 \cr
& {\text{Focus}}\left( {p + h,k} \right){\text{ and directrix }}x = - p + h \cr
& {\text{Axis of symmetry }}y = k \cr
& {\text{Focus: }}\left( {3 + 1,3} \right) \cr
& {\text{Focus: }}\left( {4,3} \right) \cr
& {\text{directrix: }}x = - 3 + 1 \cr
& {\text{directrix: }}x = - 2 \cr
& {\text{axis of symmetry: }}y = 3 \cr
& \cr
& {\text{Focus: }}\left( {4,3} \right) \cr
& {\text{directrix: }}x = - 2 \cr
& {\text{axis of symmetry: }}y = 3 \cr} $$
