Posted by **sally** on Thursday, November 2, 2006 at 11:04am.

umm suppose a parabola has a vertex at (0,2) and points (1,1)

how would I derive the equation and focus, i've been trying to understand this for so long, I can't get it. Does this parabola have the equation (y-2)^2 = x and a focus of 1/4?? Is that correct??

I searched Google under the key words "

*parabola equation focus*" to get these possible sources:

http://en.wikipedia.org/wiki/Parabola
http://www.tpub.com/math2/13.htm
http://colalg.math.csusb.edu/~devel/precalcdemo/conics/src/parabola.html
http://www.algebralab.org/lessons/lesson.aspx?file=Algebra_conics_directrix.xml
http://mathworld.wolfram.com/Parabola.html
Use the <Find> command to locate keywords within these sites.

I hope this helps. Thanks for asking.

Formats to help you find the equation for a parabola:

(x - h)^2 = 4p(y - k)

Vertex = (h, k)

Focus = (h, k + p)

Directrix: y = k - p

You are given the vertex (0,2) and a point (1,1).

Let's try to find p using the vertex and the point you were given:

(x - h)^2 = 4p(y - k)

(1 - 0)^2 = 4p(1 - 2)

1 = 4p(-1)

1 = -4p

-1/4 = p

Let's try to find the equation now that we know p:

(x - 0)^2 = 4(-1/4)(y - 2)

x^2 = -1(y - 2)

x^2 = -y + 2

Set the equation equal to 0:

x^2 + y - 2 = 0 -->this is the equation for the parabola.

(You can check the point you were given by substituting the values into the equation.)

Focus = (h, k + p)

Therefore: (0, 7/4) is the focus.

Directrix: y = k - p

Therefore: y = 9/4

In this case, the focus is below the directrix; therefore, the parabola opens downward and p is negative.

I hope this helps.

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