Given: The equation of a parabola below
[tex]12y=(x-1)^2-48[/tex]
To Determine: The vertex, focus, and directrix of the parabola
Let us re-write the given equation
[tex]\begin{gathered} 12y=(x-1)^2-48 \\ (x-1)^2=12y+48 \\ (x-1)^2=12(y+4) \end{gathered}[/tex]
The general equation of the parabola with vertex (h, k) is of this form
[tex](x-h)^2=4p(y-k)[/tex][tex]\begin{gathered} focus=(h,k+p) \\ directrix=y=k-p \\ vertex=(h,k) \end{gathered}[/tex]
Let us compare the general equation with the given equation
[tex]\begin{gathered} (x-1)^2=12(y+4) \\ (x-h)^2=4p(y-k)_{} \\ h=1,k=-4 \end{gathered}[/tex]
Therefore
[tex]\begin{gathered} vertex=(h,k)=(1,-4) \\ vertex=(1,-4) \end{gathered}[/tex][tex]\begin{gathered} 4p=12 \\ p=\frac{12}{4}=3 \\ p=3 \end{gathered}[/tex][tex]\begin{gathered} focus=(h,k+p)=(1,-4+3)=(1,-1) \\ focus=(1,-1) \end{gathered}[/tex][tex]\begin{gathered} directrix=y=k-p,y=-4-3=-7 \\ directrix=y=-7 \end{gathered}[/tex]
Hence,
Vertex = (1, -4)
Focus = (1, -1)
Directrix, y = -7
