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I have to write a research paper on a mathematical topic for my class; I chose the above topic.

I understand that a parabola can be formed using a focus and directrix, both created by origami folds, and that Axiom 6 of Origami-Folding (Given two points $P_1$ and $P_2$ and two lines $L_1$ and $L_2$, there is a fold that places $P_1$ onto $L_1$ and $P_2$ onto $L_2$) can be used to solve a cubic equation. But some of this explanation of why confuses me:

Now, let's solve the cubic equation $x^3+ax^2+bx+c=0$ with origami. Let two points $P_1$ and $P_2$ have the coordinates $(a,1)$ and $(c,b)$, respectively. Also let two lines $L_1$ and $L_2$ have the equations $y+1=0$ and $x+c=0$, respectively. Fold a line placing $P_1$ onto $L_1$ and placing $P_2$ onto $L_2$, and the slope of the crease is the solution of $x^3+ax^2+bx+c=0$.

I will explain why. Let p1 be a parabola having the focus $P_1$ and the directrix $L_1$. Since the crease is not parallel to the $y$-axis, we can let the crease have the equation $y=tx+u$. Let the crease be tangent to $P_1$ at $(x_1,y_1)$, and $(x_1 -a)^2=4y_1$. Because the crease has the equation $(x_1 -a)(x-x_1)=2(y-y_1)$, we get $t=\frac {x_1-a}{2}$ and $u= y_1-\frac {x_1(x_1-a)}{2}$. From these equations, we get $u=-t^2-at$.

Specifically, I do not understand where the equation $(x_1 -a)^2=4y_1$ is coming from.

I would greatly appreciate someone helping to explain.

[this explanation comes from http://origami.ousaan.com/library/conste.html if you want a look at the entire thing]

namikazeartemis
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1 Answers1

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The equation $(x_1-1)^2=4y_1$ comes from the assumption that $(x_1,y_1)$ is a point on the parabola with focus at $(a,1)$ and directrix having the equation $y+1=0$ (i.e., the horizontal line $1$ unit below the $x$ axis): The distance from $(x_1,y_1)$ to the focus is $\sqrt{(x_1-a)^2+(y_1-1)^2}$, while the distance from $(x_1,y_1)$ to the directrix is $|y_1+1|$. By the definition of a parabola, these are equal:

$$\sqrt{(x_1-a)^2+(y_1-1)^2}=|y_1+1|$$

Squaring and equating gives

$$(x_1-a)^2+(y_1-1)^2=(y_1+1)^2$$

If you expand out the $(y_1-1)^2$ and $(y_1+1)^2$ terms, stuff cancels and you get the equation you asked about.

Barry Cipra
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