For the parabola equation in standard form
[tex]y=a(x-h)^2+k[/tex]we know that the focus coordinate is
[tex](h,k+\frac{1}{4a})[/tex]Since our focus is (-2,5), by comparing this coordinate with the last result, we can see that
[tex]\begin{gathered} h=-2 \\ 5=k+\frac{1}{4a} \end{gathered}[/tex]Then, the possible solution has the form
[tex]\begin{gathered} y=a(x-(-2))^2+k \\ or\text{ equivalently} \\ y=a(x+2)^2+k \end{gathered}[/tex]We can see tha the last option has this form. Lets corroborate that this is the correct choice. Then, we have that
[tex]\begin{gathered} a=\frac{1}{12} \\ \text{and} \\ k=2 \end{gathered}[/tex]by substituting these values into the above equation:
[tex]5=k+\frac{1}{4a}[/tex]we have
[tex]5=2+\frac{1}{4(\frac{1}{12})}[/tex]which gives
[tex]\begin{gathered} 5=2+\frac{1}{\frac{1}{3}}=2+3 \\ \text{then} \\ 5=5 \\ \text{which corroborate the result.} \end{gathered}[/tex]Therefore, the answer is the last option:
[tex]y=\frac{1}{12}(x+2)^2+2[/tex]