Parabolas from String Art (Part 3)

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. Let’s suppose that the curve is a parabola. The vertex form of a parabola is

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

If the curve is a parabola, then clearly the vertex will be the lowest point on the axis of symmetry. By inspection, this point is (8,4), which is labeled V in the above picture. So, if it’s a parabola, the equation has the form

y = a(x-8)^2+4.

To find the value of a, we note that the point (x, y) = (16, 8) must be on the parabola, so that

8 = a(x-8)^2 + 4

8 = 64a + 4

4 = 64a

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

Therefore, the equation of the conjectured parabola is

y = \displaystyle \frac{1}{16}(x-8)^2 + 4

= \displaystyle \frac{1}{16} (x^2 - 16x + 64) + 4

= \displaystyle \frac{x^2}{16} - x + 4 + 4

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

So, if the curve is a parabola, it must follow the function this curve. By construction, this parabola passes through (8,4) and (16,8). To show that this actually works, we can substitute the other seven values of x:

At x =0: y = \displaystyle \frac{(0)^2}{16} - 0 + 8 = 8

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 couple of posts, we’ll discuss a couple of different ways of establishing that the points lie on this parabola.

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