Simple parabola problem

1. Feb 19, 2008

CINA

First off this is not a homework or a test problem, I just need some help understanding the problem.

"Find the standard form of the equation of the parabola with vertex (2,1) and focus (5,1)."

I think you use Y=-1/4x^2 and F(0,p) to find the ax^2 part but I'm confused about how to make the adjustments so that the vertex is (2,1)

Any help?

Thanks.

2. Feb 20, 2008

symbolipoint

Knowing the line for the directrix may help. Do you see how the parabola is oriented if the focus and the vertex have the same y value? The axis of the parabola is horizontal. If you could begin to draw a picture, you will see that the directrix is the line x=-1. (make a crude sketch so you see this).

Be aware that standard form of a parabola in this oreintation is x = (y-k)^2 +c;
and the graph has been shifted to have a vertex at (k, c) instead of being in standard position.

Since you have only one point "given" on the parabola, you may only find a variablized result, but that is probably all you need according to your exercise. IF you have another second point on the parabola, then you can fix the contants of k and c. You probably do not need to resort to the distance formula; just use a little bit of simultaneous equations (two of them, actually; one for each point on the parabola).