Parabolas from String Art (Part 4)

Recently, I announced that my paper Parabolic Properties from Pieces of String had been published in the magazine Math Horizons. The article had multiple aims; in chronological order of when I first started thinking about them:

  • Prove that string art from two line segments traces a parabola.
  • Prove that a quadratic polynomial satisfies the focus-directrix property of a parabola, which is the reverse of the usual logic when students learn conic sections.
  • Prove the reflective property of parabolas.
  • Accomplish all of the above without using calculus.

While I’m generally pleased with the final form of the article, the necessity of publication constraints somewhat abbreviated the original goal of this project: determining a pedagogically sound way of convincing a bright Algebra I student that string art unexpectedly produces a parabola. While all the necessary mathematics is in the article, I think the article is somewhat lacking on how to sell the idea to students. So, in this series of posts, I’d like to expand on the article with some pedagogical thoughts about connecting string art to parabolas.

As discussed previous posts, we begin our explorations with string art connecting evenly spaced points on line segments \overline{AB} and \overline{BC} with endpoints A(0,8), B(8,0), and C(16,8). We will call these colored line segments “strings.” We then found the string with the largest y-coordinate at x = 2, 4, 6, \dots, 14, resulting in the following picture:

However, perhaps it’s clearer to plot these points on a separate graph, without the clutter of the strings:

These points are definitely following some kind of curve. In the previous post, we established that the curve is a parabola by using the vertex form of a parabola y = a(x-h)^2+k.

In this post, we use the other general form. If the curve is a parabola, then the equation of the curve must be y = ax^2 + bx + c for some values of a, b, and c. Since there are three unknowns, we need to have three equations to solve for them. This can be done by plugging in three (x,y) pairs into this equation. While we can pick any three pairs that we wish, it seems convenient to use the points (0,8), (8,4) and (16,8):

a(0)^2+b(0)+c = 8

a(8)^2 + b(8) + c = 4

a(16)^2 + b(16) + c =8

This simplifies to the 3\times 3 system of linear equations

c = 8



In general 3\times 3 systems of linear equations can be challenging for students to solve. However, while this is technically a 3\times 3 system, it’s clear that c =8, and so this reduces to a 2\times 2 system









In algebra, students are taught multiple ways of solving 2\times 2 systems of linear equations, and any of these techniques can be used at this point to solve for a and b. Perhaps the easiest next step is subtracting the two equations:

(16a + 2b) - (16a + b) = -1 - 0

b = -1

Substituting into 16a+b=0, we see that

16a - 1 = 0

16a = 1

a =\displaystyle \frac{1}{16}.

We conclude that a = \displaystyle \frac{1}{16}, b = -1, and c = 8, so that, if the points lie on a parabola, the equation of the parabola must be

y = \displaystyle \frac{x^2}{16} - x + 8.

By construction, this parabola passes through (0,8), (8,4), and (16,8). To show that this actually works, we can substitute the other six values of x:

At x =2: y = \displaystyle \frac{(2)^2}{16} - 2 + 8 = 6.25

At x =4: y = \displaystyle \frac{(4)^2}{16} - 4 + 8 = 5

At x =6: y = \displaystyle \frac{(6)^2}{16} - 6 + 8 = 4.25

At x =10: y = \displaystyle \frac{(10)^2}{16} - 10 + 8 = 4.25

At x =12: y = \displaystyle \frac{(12)^2}{16} - 12 + 8 = 5

At x =14: y = \displaystyle \frac{(14)^2}{16} - 14 + 8 = 6.25

Therefore, the nine points in the above picture all lie on the parabola y = \displaystyle \frac{x^2}{16} - x + 8.

In the next post, we’ll discuss a third way of convincing students that the points lie on this parabola.

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