I'm trying to figure out what the heck I'm supposed to do with this:

x^2=8y.

I'm supposed to solve for vertex, focus, and directx... how do I put this into the y=ax^2+bc+c form??

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- Jul 11th 2009, 02:56 PMKevin GreenParabola problem
I'm trying to figure out what the heck I'm supposed to do with this:

x^2=8y.

I'm supposed to solve for vertex, focus, and directx... how do I put this into the y=ax^2+bc+c form?? - Jul 11th 2009, 03:09 PMskeeter
- Jul 11th 2009, 03:31 PMKevin Green
Thank you.... where do I go from here? The math book I have does a horrible job describing this whole matter. :(

- Jul 11th 2009, 04:08 PMskeeter
- Jul 11th 2009, 05:08 PMyeongil
Parabolas can also be expressed in this form:

$\displaystyle (x - h)^2 = 4p(y - k)$ or $\displaystyle (y - k)^2 = 4p(x - h)$.

The first is a parabola that opens upward or downward, and the second is a parabola that opens left or right. Sometimes this form is preferred because you can see what the vertex is.

Looking at your equation:

$\displaystyle x^2=8y$

h = k = 0, so the vertex is (0, 0). The x is squared, and the y coefficient is positive, so it opens upward. Solve 4p = 8 to get the focal length, which is 2.

Since the parabola opens upward, the focus would be defined by (h, k + p), or (0, 2). The directrix is an equation of a line, in our case, defined by y = k - p, or y = -2.

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