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Parabolic Equation of an Arch


Date: 05/25/98 at 00:36:49
From: Derek Child
Subject: parabolas

For an assignment in my trig class, I was asked to find the equation 
of the parabola in the McDonald's golden arches. I have searched high 
and low all over the Internet. I don't want a full blown solution, 
just some hints as to how I can come up with a solution. I have tried 
taking a picture of the "M" and putting it on a x-y axis, but this 
doesn't work too well. Any help would be greatly appreciated. Thanks.


Date: 06/08/98 at 15:41:53
From: Doctor Jeremiah
Subject: Re: parabolas

Dear Derek:

To define a parabola requires three (x,y) coordinates. Measure the 
height (I will use the symbol H for the value that you measure) and 
width at the widest part (I will use the symbol W for that value) of 
one of the parabolas in the "M."

If you draw the parabola like this ...

                          y

                          |
                          |
                          |
                          |
                          |
             -W/2         | (0,0)   W/2
   --- -+- ----|--------+++++--------|-------- x
        |            +    |    +
        |          +      |      +
        |         +       |       +
        H        +        |        +
        |                 |
        |       +         |         +
        |                 |
        |                 |
       -+-     +         -+- -H      +
           (-W/2,-H)              (W/2,-H)

               |----------W----------|

... then you have three points: (0,0), (W/2,-H), and (-W/2,-H).

So all you have to do is solve the equation of the parabola with these 
three coordinates.

The equation of a parabola is:   

   y = ax^2 + bx + c

We need to make three equations, because we want to solve for three 
variables: a, b and c. So we take our equation and we substitute in 
each of our coordinates for x and y, and we end up with three 
equations:

    0 = a(0)^2 + b(0) + c
   -H = a(W/2)^2 + b(W/2) + c
   -H = a(-W/2)^2 + b(-W/2) + c

The first obviously makes c = 0.

To solve this problem, we need to find b. We have two equations left 
that just happen to be equal to the same thing (-H), so we can equate
the equations and solve for b:

   a(W/2)^2 + b(W/2) + c = a(-W/2)^2 + b(-W/2) + c

Now that we know a and b, we can substitute them into the second or 
third equation and solve for a.

Now that we have a, b and c, we can change y = ax^2 + bx + c to the 
real equation by replacing the letters with the values of those 
variables.

If you need more details, mail me back.

-Doctor Jeremiah, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
High School Conic Sections/Circles
High School Equations, Graphs, Translations
High School Geometry
High School Practical Geometry

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