Question 673675
focus is 0,2 and the directrix is y = -4
the distance between the focus and the directrix is equal to (0,2) to (0,-4) which is equal to 6
that makes 2p = 6 which makes p = 3
there are 3 possible equations for this parabola.
they are:
ax^2 + bx + c = 0 (standard form)
y = a(x-h)^2 + k (vertex form)
y = (x-h)^2/4p + k (conics form)
another form of the equation in the conics form is:
4p(y-k) = (x-h)^2
we should be able to work either either one because they are equivalent.
we should be able to solve for all of them.
let's see if we can.
since p = 3, then 4p in the conics form must be equal to 12.
since a = 1/4p, this mean that a in the vertex form must be equal to 1/12.
(h,k) is the vertex of the parabola.
since you know the focus, then you can solve for the vertex because the vertex is halfway between the focus and the directrix.
you're going from (0,2) to (0,-4) which gets you to a vertex of (0,-1) because that is halfway between (0-2) and (0,-4).
with a vertex of (0,-1), we get h = 0 and k = -1.
we may be able to piece together some of the equations now.
we have:
h = 0
k = -1
4p = 12 in the conics form
a = 1/12 in the vertex form
note that a in the vertex form is not the same as a in the standard form.
we'll start with the conics form.
the equation is:
y = (x-h)^2/4p + k (conics form)
substitution gets us:
y = x^2/12 - 1
let's try the vertex form next.
the equation is:
y = a(x-h)^2 + k (vertex form)
substitution gets us:
y = 1/12 * x^2 - 1 which is equivalent to:
y = x^2/12 - 1
this is the same as the conics form so we're doing good so far.
let's try the standard form next.
the equation is:
ax^2 + bx + c = 0 (standard form)
since we know that the equation is y = x^2/12 - 1, we just set y = 0 and we get:
x^2/12 - 1 = 0
in standard form, this means that:
a = 1/12, b = 0, and c = -1
it's not clear to me if the a in the standard form is the same as the a in the vertex.  it does seem to work out in this example so maybe it is.
a graph of this equation of y = x^1/12 - 1 is shown below:
<img src = "http://theo.x10hosting.com/2012/oct302.jpg" alt = "&&&&&&&&&&&" />
from the graph, you can see that the focus is at (0,2) and the vertex is at (0,-1) and the point on the line y = -4 that you would measure to is the point (0,-4) which is in line with the focus and the vertex and lies on the axis of symmetry which is x = 0.
the reference that i used to answer this question is at:
<a href = "http://www.purplemath.com/modules/parabola.htm" target = "_blank">http://www.purplemath.com/modules/parabola.htm</a>
this is an excellent reference and well worth reading to understand this better.
it's a little difficult to follow but just be careful of all the a's and b's and c's and p's and you should be able to figure it out.
be careful though because they talk about the standard equations that have the parabola facing up or down, and the sideways equations that have the parabola facing left or right.
ideally, you should be understand both, but for this problem you only need to understand the up and down equations.
those are the ones i showed you above.
i got them from this reference so they should track with what the reference is saying.
the reference is 4 pages long.  what you see when you click on the link is the first page only.
good luck.