Parabolas from String Art (Part 9)

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.

We have shown in the last couple of posts that if the three points that generate the Our explorations of string art led us to consider an arbitrary string \overline{PQ} depicted below. For brevity, this string will be called “string s,” matching the (possibly non-integer) x-coordinate of its left endpoint P. Since P is s units to the right of A, the right endpoint Q must correspondingly be s units to the right of B. Therefore, the x-coordinate of Q is s + 8.

Previously, we established that the equation for string s is

y = -\displaystyle \frac{s^2}{4} + \frac{xs}{4} - x + 8.

We also obtained a bonus result that we obtained using only algebra: string s is tangent to the parabola y = \displaystyle \frac{x^2}{16} - x + 8, which is traced by the strings, when x=2s. Of course, tangent lines are usually obtained using calculus, and so calculus should be able to confirm this result. The derivative of this function is

y' = \displaystyle \frac{x}{8} - 1,

so that the slope of the tangent line when x=2s is m = \displaystyle \frac{s}{4} - 1 = \frac{s-4}{4}. We observe that this matches the slope of line segment \overline{PQ} in the above picture:

slope = \displaystyle \frac{s - (s-8)}{(s+8) - 8} = \frac{2s-8}{8} = \frac{s-4}{4}.

Therefore, to show that \overline{PQ} is the tangent line, it suffices to show that either P or Q is on the tangent line.

At x = 2s, the y-coordinate of where the tangent line intersects the curve is

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

Using the point-slope formula for a line, the equation of the tangent line is thus

y-y_1 = m(x-x_1)

y-\displaystyle \left( \frac{s^2}{4} - 2s + 8 \right) = \frac{s-4}{4} (x-2s)

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

We now check to see if P(s,8-s) is on the tangent line. Substituting x =s, we find

y = \displaystyle \frac{s-4}{4} (s-2s) +  \frac{s^2}{4} - 2s + 8

= \displaystyle \frac{s-4}{4} (-s) +  \frac{s^2}{4} - 2s + 8

= \displaystyle \frac{(s-4)(-s) + s^2}{4} - 2s + 8

= \displaystyle \frac{-s^2+4s + s^2}{4} - 2s + 8

= \displaystyle \frac{4s}{4} - 2s + 8

= s - 2s + 8

= -s + 8

Therefore, the point (s,8-s) is on the tangent line, thus confirming that P is on the tangent line and that \overline{PQ} is the tangent line.

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