Parabola someone please help me
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Parabola someone please help me

[From: ] [author: ] [Date: 13-01-21] [Hit: ]
so the point of intersection between the directrix and axis of symmetry must be (-3, 0)We also know that the axis of symmetry is y = 0, so any perpendicular to that must be of the form x = k.Since this goes through the point (-3, 0), this line must be x = -3.......

So the focus is at (3, 0)

Remember, if you are dealing with a parabola that opens up or down, this is the y coordinate.

The directrix must be perpendicular to the axis of symmetry,

and since all points on a parabola must be equidistant from the focus and directrix, then the vertex must be the midpoint of the segment from the focus to the point of intersection with the axis of symmetry and the directrix.

We have the focus at (3, 0) and the vertex at (0,0), so the point of intersection between the directrix and axis of symmetry must be (-3, 0)

We also know that the axis of symmetry is y = 0,
so any perpendicular to that must be of the form x = k.

Since this goes through the point (-3, 0), this line must be x = -3.

So now we have vertex at (0,0), axis of symmetry at y = 0, focus at (3,0), and directrix at x = -3

-
Try plotting it like so
http://www.wolframalpha.com/input/?t=crm…

Seeing the answer helps you get a feeling of what the answer should be.
Then you can look to standard forms
http://www.mathwarehouse.com/geometry/pa…

You can do it,

-
(y-0)^2 =12(x-0)

vertex : (0,0)
axis if symmetry : y=0
(y-k)^2 = 4p(x-h)
4p =-12
p=-3

focus : (h+p, k) = (0-3, 0) = (-3,0)
directrix x=k-p
x=0-(-3)
x=3
12
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