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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 p1 and p2 and two lines l1 and l2, there is a fold that places p1 onto l1 and p2 onto l2) 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 P1 and P2 have the coordinates (a,1) and (c,b), respectively. Also let two lines L1 and L2 have the equations y+1=0 and x+c=0, respectively. Fold a line placing P1 onto L1 and placing P2 onto L2, 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 P1 and the directrix L1. 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 p1 at (x1,y1), and (x1-a)^2=4y1. Because the crease has the equation (x1-a)(x-x1)=2(y-y1), we get t=(x1-a)/2 and u=y1-x1(x1-a)/2. From these equations, we get u=-t2-at."
Specifically, I do not understand where the equations (x1-a)^2=4y1 and (x1-a)(x-x1)=2(y-y1) are coming from and w hat they mean.
I would greatly appreciate someone helping to explain.
[this explanation comes from K's Origami : Origami Construction if you want a look at the entire thing]
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 p1 and p2 and two lines l1 and l2, there is a fold that places p1 onto l1 and p2 onto l2) 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 P1 and P2 have the coordinates (a,1) and (c,b), respectively. Also let two lines L1 and L2 have the equations y+1=0 and x+c=0, respectively. Fold a line placing P1 onto L1 and placing P2 onto L2, 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 P1 and the directrix L1. 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 p1 at (x1,y1), and (x1-a)^2=4y1. Because the crease has the equation (x1-a)(x-x1)=2(y-y1), we get t=(x1-a)/2 and u=y1-x1(x1-a)/2. From these equations, we get u=-t2-at."
Specifically, I do not understand where the equations (x1-a)^2=4y1 and (x1-a)(x-x1)=2(y-y1) are coming from and w hat they mean.
I would greatly appreciate someone helping to explain.
[this explanation comes from K's Origami : Origami Construction if you want a look at the entire thing]