Question

Find the standard form of the equation of the parabola with a focus at (3, 0) and a directrix at x = -3.

Answer 1

so... any ideas on where the vertex may be?

Answer 2

it would be at 3,0 right?

Answer 3

thought that was the focus point?

Answer 4

oh then not really no

Answer 5

i know i need to use the equation 4py=x^2 but im not sure how

Answer 6

well, notice, the directrix is x = -3, meaning is vertical one
also bear in mind that, the vertex is right between the directrix, and the focus point
thus

Answer 7

well. I should have said, the vertex is half-way between the directrix and focus point
so, the focus point is a "p" distance from the vertex, and so is the directrix, a same "p" distance from it

so, where would that put it anyway?

Answer 8

on 0,0

Answer 9

yeap

notice the distance "p" from the focus point and the directrix, is 3, thus "p" is 3
or p is +3, because the parabola is opening towards the right

so, now we know the vertex coordinates and "p" is +3
the parabola is opening horizontally, thus the squared variable is the "y" then
thus \(\begin{array}{llll}
(y-{\color{blue}{ k}})^2=4{\color{purple}{ p}}(x-{\color{brown}{ h}}) \\
(x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\
\end{array}

\qquad
\begin{array}{llll}
vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\\
{\color{purple}{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}
\\ \quad \\ \quad \\
(y-{\color{blue}{ 0}})^2=4{\color{purple}{ (3)}}(x-{\color{brown}{ 0}})\)