Expert: Sherry Wallin - 6/10/2011

Question
Hello Sherry -

I am designing a parabolic solar reflecting panel.  I know two points and the focus, how can I determine the equation?

For example, with the focus at (0, 0) one point is at (10, -6) and the other at (20, 6).

I last studied this subject in 1963, so have forgotten a bit since then.

Thanks much,
Bill

Answer
Hi Bill~

It is not clear to me what information you really have. I understand in this particular instance that the focus is at (0,0) and that there are two points on the parabola (10, -6) and (20, 6). I can show you or explain to you how to find the equation of this particular parabola but somehow I think you want more than that even though you haven't actually said so. So for now I will tell you one way to determine the equation with the given information.

You can choose to orient your parabola but since parabolas are usually taught opening up or down I will choose the orientation so that the line x = 0 (the y-axis) is the line of symmetry. The vertex is at (h,k) and the focus is (h, k+a) so the directrix is y = k-a. The basic/standard equation for a parabola that has an axis of symmetry parallel to the y-axis and opens up is (x-h)^2 = 4a(y-k). You know the focus is (0,0) so that means that h = 0. Substitute h = 0 into this basic equation:

(x-0)^2 = 4a(y-k) => x^2 = 4a(y-k). Now you gave me 2 points that are on the parabola, substitute them into this basic equation and solve each for a:

First use (10,-6) in x^2 = 4a(y-k) =>
10^2 = 4a(-6-k) =>
100 = 4a(-6-k) =>
25= a(y-k) =>
25/(-6-k) = a =>
a = -25/(k+6)

Now use (20,6) in x^2 = 4a(y-k) =>
20^2 = 4a(6-k) =>
400 = 4a(6-k) =>
100 = a(6-k) =>
100/(6-k) = a =>
a = -100/(k-6)

since a = a you have  -25/(k+6) = -100/(k-6)

drop the negative signs or divide both sides by -1
25/(k+6) = 100/(k-6)

cross multiply:
25(k-6) = 100(k+6) =>

divide by 25 on both sides
k-6 = 4(k+6) =>
k-6 = 4k + 24 =>
subtract k from both sides
-6 = 3k + 24 =>
subtract 24 from both sides
-30 = 3k =>
divide by 3 both sides
-10 = k

substitute k = -10 into  x^2 = 4a(y-k) =>
x^2 = 4a(y-(-10) =>
x^2 = 4a(y+10)  solve for a by using one of the points, say (10,-6)
10^2 = 4a(-6+10) =>
100 = 4a(4) =>
100/16 = a =>
25/4 = a

now go back to the basic equation: x^2 = 4a(y-k) and substitute a =  25/4
x^2 = 4(25/4)(y+10) =>
x^2 = 25(y+10)   => solve for y
x^2/25 = y + 10   subtract 10 from both sides
x^2/25 -10 = y =>
y =(1/25)x^2 -10  is the equation you are seeking in the form y = ax^2 + bx + c, where a = 1/25, b = 0, and c = -10

Math Prof