Answer :
To write the equation of a parabola, we are given the following information;
[tex]\begin{gathered} \text{Focus}=(3,6) \\ \text{Directrix;} \\ y=8 \end{gathered}[/tex]We shall begin by taking any point on the parabola which would be point;
[tex](x_1,y_1)[/tex]The distance between the focus and this random point can be calculated as follows;
[tex]d=\sqrt[]{(x_1-3)^2+(y_1-6)^2}[/tex]Similarly, the distance between the directrix and this random point is;
[tex]|y_1-8|[/tex]Note that the directrix is given as
[tex]\begin{gathered} y=8 \\ \text{Therefore;} \\ The\text{ two y-coordinates would be;} \\ y_1\text{ and} \\ y=8 \\ \text{The distance therefore is, } \\ y_1-8 \end{gathered}[/tex]This is expressed in absolute value because the distance cannot be a negative.
We now equate both distances and we can begin to simplify;
[tex]\begin{gathered} \sqrt[]{(x_1-3)^2+(y_1-6)^2}=|y_1-8| \\ \text{Square both sides of the equation to eliminate the radical;} \\ (x_1-3)^2+(y_1-6)^2=(y_1-8)^2 \\ \end{gathered}[/tex]We can now simplify all parenthesis;
[tex]x^2_1-6x_1+9+y^2_1-12y_1+36=y^2_1-16y_1+64[/tex]We now move all terms to one side and we now have;
[tex]\begin{gathered} x^2_1-6x_1+9-64+y^2_1-y^2_1-12y_1+16y_1=0 \\ x^2_1-6x_1-55+4y_1=0 \end{gathered}[/tex]Next step, we make y the subject of the formula;
[tex]\begin{gathered} 4y_1=-x^2_1+6x_1+55 \\ \text{Divide all through by 4;} \\ y_1=-\frac{1}{4}x^2_1+\frac{3}{2}x_1+\frac{55}{4} \end{gathered}[/tex]We can now re-write for (x, y) as follows;
ANSWER:
[tex]y=-\frac{x^2}{4}+\frac{3x}{2}+\frac{55}{4}[/tex]