Finding the Standard Form of a Parabola with Given Vertex and Focus

[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.

 

[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).