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I was asked to find the focus and diretrix of the given equation: $y=x^2 -4$. This is what I have so far:

Let $F = (0, -\frac{p}{2})$ be the focus, $D = (x, \frac{p}{2})$ and $P = (x,y)$ which reduces to $x^2 = 2py$ for $p>0$. Now I have $x^2 = 2p(x^2 - 4)$ resulting in $ x^2 = \frac{-8p}{(1-2p)}$ I have no clue how to find the focus. I just know that it will be at $(0, -4+\frac{p}{2})$

Can I get help from some college math major? I went to the tutoring center at my high school but no one there understands what I'm talking about.

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The vertex of the parabola (in this case, (0, -4)) should be half way between the focus and the directrix, right? So let f = (0, -4-a) and d = (0, -4+a). For some value of a, the point f will be your focus, and the point d will be a point on your directrix. You don't yet know what a is, so pick a point on your parabola (e.g. (1, -3)) and work out its distance from f and its distance from the directrix. They should be equal, and this will hopefully give you an equation for a. – Billy Mar 21 '13 at 22:14
up vote 1 down vote accepted

I seem to remember the focal distance $p$ satisfies $4ap=1$ where the equation for the parabola is $y = ax^2 + bx + c$. Your focus will be $1/4$ above your vertex, and the directrix will be a horizontal line $1/4$ below your vertex.

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I'm not sure where you got $4ap = 1$. We're supposed to do everything from scratch using a graph – Person Mar 21 '13 at 22:05
What is the definition of the focus/directrix you are starting with? – orlandpm Mar 21 '13 at 22:06
A focus according to my notes is a "fixed point." A directrix is a "fixed line." That is all my teacher wrote on the board. – Person Mar 21 '13 at 22:08
By the way, you can verify what I said here – orlandpm Mar 21 '13 at 22:10
I know where you got your $4ap=1$. I wrote the equations from scratch and verified it. – Person Mar 22 '13 at 4:53

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