Question

Find the equation of a parabola with a focus of (0,9) and a directrix y = -9

Answer 1

Given the focus of a parabola as (0,9) and the directrix is y=-9

`(x_0,y_0)\text{ be any point on the parabola}`

Let us find the distance between

`(x_0,y_0)\text{ and the focus}`

Then we will find the distance between

`(x_0,y_0)\text{ and the directrix}`

We will then equate these two distance equations and the simplified equation in

`x_0andy_0_{}`

is the equation of the parabola

Step 1:

`\text{distance betw}een(x_0,y_0)\text{ and (0,9)}`
`=\sqrt[]{(x_0}-0)^2+(y_0-9)^2`
`\begin{gathered} \text{distance betwe}en((x_0,y_0)\text{ } \\ \text{and the directrix, y=-9} \end{gathered}`
`\begin{gathered} \lvert y_0--9\rvert \\ \lvert y_{0_{}}+9\rvert \end{gathered}`

Equating the two distance expressions and square on both sides

`(_{}\sqrt[]{(x_0}-0)^2+(y_0-9)^2)^2=(\lvert y_{0_{}}+9\rvert)^2`
`\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}`

Let us express the equation in terms of

`\begin{gathered} y_0 \\ x^2_0-36y_0=0 \\ x^2_0=36y_0 \\ 36y_0=x^2_0 \\ y_0=(x^2_0)/(36) \end{gathered}`

Hence the equation of the parabola is