Answer :
Given the focus of a parabola as (0,9) and the directrix is y=-9
[tex](x_0,y_0)\text{ be any point on the parabola}[/tex]Let us find the distance between
[tex](x_0,y_0)\text{ and the focus}[/tex]Then we will find the distance between
[tex](x_0,y_0)\text{ and the directrix}[/tex]We will then equate these two distance equations and the simplified equation in
[tex]x_0andy_0_{}[/tex]is the equation of the parabola
Step 1:
[tex]\text{distance betw}een(x_0,y_0)\text{ and (0,9)}[/tex][tex]=\sqrt[]{(x_0}-0)^2+(y_0-9)^2[/tex][tex]\begin{gathered} \text{distance betwe}en((x_0,y_0)\text{ } \\ \text{and the directrix, y=-9} \end{gathered}[/tex][tex]\begin{gathered} \lvert y_0--9\rvert \\ \lvert y_{0_{}}+9\rvert \end{gathered}[/tex]Equating the two distance expressions and square on both sides
[tex](_{}\sqrt[]{(x_0}-0)^2+(y_0-9)^2)^2=(\lvert y_{0_{}}+9\rvert)^2[/tex][tex]\begin{gathered} (x_0-0)^2+(y_0-9)^2)^{}=(y_{0_{}}+9)^2 \\ x^2_0+y^2_0-18y_0+81=y^2_{0_{}}+18y_0+81 \\ x^2_0+y^2_0-y^2_0-18y_0-18y_0+81-81=0 \\ x^2_0-36y_0=0 \end{gathered}[/tex]Let us express the equation in terms of
[tex]\begin{gathered} y_0 \\ x^2_0-36y_0=0 \\ x^2_0=36y_0 \\ 36y_0=x^2_0 \\ y_0=\frac{x^2_0}{36} \end{gathered}[/tex]Hence the equation of the parabola is
[tex]undefined[/tex]