Solution:
(A) Given Focus (5,1) and Vertex (2,1)
We are required to find the equation of the parabola
Substituting the values of h, k and f, we have
[tex]\begin{gathered} x=\frac{1}{4(5-2)}(y-1)^2+2 \\ \\ x=\frac{1}{4(3)}(y-1)^2+2 \\ \\ x=\frac{1}{12}(y-1)^2+2 \\ \\ x=\frac{(y-1)^2}{12}+2-------Vertex\text{ form} \end{gathered}[/tex]
Or in standard form, we have
[tex]x=\frac{y^2}{12}-\frac{y}{6}+\frac{25}{12}[/tex]
