[tex](y-5)^2\text{ = 12(x - 4)}[/tex]Explanation:
Vertex = (4, 5)
focuse = (7, 5)
The vertex and focus lie on the same horizontal line; y = 5
So we will be applying the equation of horizontal parabola
[tex]\begin{gathered} \text{Equation of the parabola (horizontal):} \\ (y-k)^2=\text{ 4a(}x\text{ - h)} \end{gathered}[/tex]
The focus lies on the right of the vertex, parabola opens upward as a result, a = positive
[tex]\begin{gathered} As\text{ the y coordinates are the same, }a\text{ = 7 - 4 } \\ a\text{ = 3} \\ \\ \text{Vertex: h = 4, k = 5} \end{gathered}[/tex]
substituting the values:
[tex]\begin{gathered} (y-5)^2\text{ = 4(3)(x - 4)} \\ (y-5)^2\text{ = 12(x - 4)} \end{gathered}[/tex]
