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

@tkhunny

Answer 2

This is a standard question design to see if you know where the Vertex is.

Draw the Focus and Directrix. It should be very easy to find the Vertex. By Definition, it must be equidistant from both. After that, we need the magic "p". That is also very easy if we know the Focus, the Vertex, and the Directrix. What distance IS the Vertex from both? Done.

Go!

Answer 3

so would it be Vertex: (0, 0); Focus: (0, 3); Directrix: y = -3; Focal width: 12?

Answer 4

I thought the focus was given at 3,0?

Answer 5

oh okay! I see

Answer 6

anyhow... keeping in mind that
the directrix is the same distance to the vertex
as the focus is to the vertex

where would that put the vertex at?

Answer 7

also... let us notice the focus and directrix points
is that a vertical or horizontal parabola anyway?

Answer 8

Would the answer then be y = 1/12x

Answer 9

y = 1/12x^2

Answer 10

well.. how did you get that?

Answer 11

Well.. Im not sure but the vertex must be at (0,0) right?

Answer 12

hmmm 0, 0?

Answer 13

notice the directrix
notice the focus point :)

Answer 14

the focus point is at (3,0) ad the directix is at (-3,0)

Answer 15

so the middle of that would be (0,0)?

Answer 16

well, yes
that is correct :)
also notice, the distance from the vertex to either, the directrix or the focus point, is 3

notice something


the parabola opens towards the focus, and thus, is a horizontally opening parabola
thus we'll use then the form with the "y" squared
or \(\begin{array}{llll}
(y-{\color{blue}{ k}})^2=4{\color{purple}{ p}}(x-{\color{brown}{ h}}) \\
\\
\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}\)

so.. what would that give you ?:)

Answer 17

y^2=12x

Answer 18

\(\begin{array}{llll}
(y-{\color{blue}{ 0}})^2=4{\color{purple}{ (3)}}(x-{\color{brown}{ 0}}) \\
\\
\end{array}

\qquad
\begin{array}{llll}
vertex\ ({\color{brown}{ 0}},{\color{blue}{ 0}})\\
{\color{purple}{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}
\\ \quad \\
y^2=12x\implies \cfrac{y^2}{12}=x\)

keep in mind that
because the parabola opens to the "right", the distance "p" or 3, is positive
if the parabola opened to the left, it'd be -3 then

Answer 19

y = 1/12x^2
x = 1/12y^2
-12y=x^2
y2=6x

Answer 20

?

Answer 21

SO out of all my choices would you agree that the 2nd option is correct? @jdoe0001

Answer 22

yes \(\bf \cfrac{y^2}{12}=x\iff y^2\cfrac{1}{12}=x\)

Answer 23

thanks so much!! Could you help me with another question? @jdoe0001

Answer 24

sure, just post anew

Answer 25

A building has an entry the shape of a parabolic arch 96 ft high and 18 ft wide at the base, as shown below.
Find an equation for the parabola if the vertex is put at the origin of the coordinate system.

Answer 26

okay!

Answer 27

Ill post a new one