SOLUTION: The equation of the parabola whose focus is at (0, 5) and directrix at y = -5 is

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Question 917415: The equation of the parabola whose focus is at (0, 5) and directrix at y = -5 is
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
the focus is at (0,5)
the directrix is at y = -5
the vertex is halftway between the focus and the intersection of the directrix with the axis of symmetry.
that would make the vertex at (0,0).

the vertex form of the equation of a parabola is:

y = a * (x-h)^2 + k

the conic form of the equation of a parabola is:

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

p is the distance from the directrix to the vertex and from the vertex to the focus.

that makes p half the distance between the focus and the directrix on the axis of symmetry.

that makes p = 5.

from the conic form of the equation of a parabola, you get:

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

since the vertex is at (0,0), this equation becomes:

4p * y = x^2

since p = 5, this equation becomes:

20 * y = x^2

divide both sides of this equation by 20 and you get:

y = (1/20) * x^2

that's your equation.

in the vertex form of your equation, you have:

y = a(x-h)^2 + k

since h = 0 and k = 0 and a = 1/(4p) and p = 5, you then get:

y = (1/20) * x^2.

the vertex form of the equation and the conic form of the equation can be directly translated into each other with a = 1/(4p) or 4p = 1/a.

the graph of your equation is shown below:

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