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Thread: Playing with parabolas

20040201, 22:26 #1
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Playing with parabolas
Not a puzzle, but in need of a bit of advice here. If this is not the right board, one of our moderators ( <img src=/S/bravo.gif border=0 alt=bravo width=16 height=30> <img src=/S/cheers.gif border=0 alt=cheers width=30 height=16> <img src=/S/clapping.gif border=0 alt=clapping width=19 height=23> <img src=/S/salute.gif border=0 alt=salute width=15 height=20>) might want to move this to a more appropriate area.
It began with a student having an argument with a textbook. The book said that only one parabola could pass through 3 given points and demonstrated by example. The lateral thinking student found another  a horizontal one. Yes, textbooks still teach vertical parabolas throughout until it comes to the chapter on conic sections, where they all mystically lay down on their sides. Students aren't supposed to remember this, except for problems on conics of course, but this one did! <img src=/S/grin.gif border=0 alt=grin width=15 height=15>
I proceeded to dig myself a hole by introducing him to the general (Cartesian) conic equation and showing him how a whole family of oblique parabolas could pass through the 3 given points. He now wants to see an analytical solution for the equations describing all members of such a family. I dug myself in deeper, saying that it would not be hard to write a computer program that could do the plotting for 3 userspecified points.
So here I sit, surrounded by dirt, with just a glimmer of daylight above me. OK, so it's not that bad and I know what the task involves, but it's the best method of attacking it that I'm interested in. I've attached a .PDF with some initial investigating, and a few issues come to mind:

20040201, 22:56 #2
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Re: Playing with parabolas
Totally beyoind me Alan <img src=/S/smile.gif border=0 alt=smile width=15 height=15> but I agree that this is the forum most likely to turn up some help for you.
StuartR

20040202, 01:46 #3
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Re: Playing with parabolas
No bright ideas here. Where/how do you want to plot the parabolas? Do you have mathematics software such as MathCAD, Mathematica or MatLab?

20040202, 05:07 #4
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Re: Playing with parabolas
Hi Hans
The only way I've tackled doing this sort of thing in the past was using <img src=/S/fanfare.gif border=0 alt=fanfare width=31 height=23> QBasic! Not even Quick Basic, but QBasic. I just wrote my code to fill pixels on a 640 x 480 screen with various colours for each curve. And the results were "coarse" to say the least. I don't have any plotting software with any real power, and what I was thinking about is actually recalculating the equation parameters for each 1

20040202, 09:13 #5
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Re: Playing with parabolas
Drawing a parabola for each 1

20040202, 13:10 #6
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Re: Playing with parabolas
Not sure I understand completely. I agree with the textbook: If you are given 3 x,y points this defines a parabola. If you rotate the triangle you are getting different x,y points and the parabola will be a different one.
If you consider them 3dimensional (x,y,z) points the plane of the triangle may be shifted up and down to keep the same x,y points but have different y values. (this could be done thru infinite dimensions. But the equation in the plane are each identical, the equations only differ in that the Z is defined.
I do not think that "rotating" the triangle on the same plane, (or even rotating the triangle in another plane) is "keeping the same points".
The textbook assumes that you have a coord system and you have coordinates of a 3 points. Only 1 parabola will go thru those 3 points. Rotating the paper also rotates the Axis system if you want them to be the "same points" before you rotate.
1) I don't thinnk polar coords should be introduced until the students have at least have a good understanding of trig functions (your call on their readiness)
2) You can draw smooth curves with excel, just plot a lot of points (2550 are more than sufficient for a circle) and make the lines smooth
Steve

20040202, 13:27 #7
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Re: Playing with parabolas
Steve,
If by parabola, you mean the graph of a quadratic function y = ax

20040202, 13:54 #8
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Re: Playing with parabolas
I think that a parabola must have an axis of symmetry. It is the line that goes thru the focus and the vertex, perpendicular to the directrix. If it does not have an axis of symmetry it is not a parabola.
Steve

20040202, 14:06 #9
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Re: Playing with parabolas
Steve,
I think we are talking at crosspurposes at the moment. <img src=/S/smile.gif border=0 alt=smile width=15 height=15>
I fully agree that a parabola has an axis of symmetry. But the axis does not have to be vertical (i.e. parallel to the y axis.) There is only one parabola with a vertical axis of symmetry through three given points, but there is an infinite number of parabolas through the same points if you don't require the axis of symmetry to be vertical.

20040202, 14:20 #10
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Re: Playing with parabolas
The parabola x^2 + 2xy + y^2 = 0 has an axis of symmetry, but it's not vertical  it's the line y = x. In other words, this is a vertical parabola rotated by 45

20040202, 14:32 #11
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Re: Playing with parabolas
Hi Steve
Yes, I think your interpretation is true for 3 noncolinear distinct points, but the single vertical parabola is not the only parabola that can pass through these 3 points. The idea of rotating the triangle was part of the method I proposed, as an "easy" way of fitting a parabola  fitting a vertical one, then rotating it back to where it "should be" seemed easier than dealing with all the terms of the conic equation, using the 3 given (x,y) values.
The thought of using polar coordinates was really just a possibility in implementing the mechanics of my solution i.e. it may have made the rotations simpler... but just a rough look suggests swapping into and out of polar coordinates would be a lot more work (and mess) overall.
I'll have a look at XL's "curve" plotting capabilities once I get to that stage. Thanks for the tip.
Alan

20040202, 14:46 #12
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Re: Playing with parabolas
I guess I had always assumed (probably a bad idea) that the 3 points were not colinear. If they are colinear I can see see the infinite parabolas. If they are not colinear, I can only see 1 solution.
In other words: 3 non colinear points define a parabola, where 3 colinear points can be an infinite number of parabolas.
Steve

20040202, 14:48 #13
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Re: Playing with parabolas
Thanks Hans. My further look at a real example indicates that achieving an even, uncluttered distribution of plots might be quite a task in itself. Noses congregate on each side of the triangle, while the arms tend to crowd up as they asymptote along the lines formed by extending each side. I'd really need a diagram  the object of the exercise <img src=/S/grin.gif border=0 alt=grin width=15 height=15>. I think I'd probably need 5 or 6 along each side to illustrate the pattern of how the family evolves. Anyway, that's final stage stuff, so I may defer that detail until I've got the equations and parameters nutted out.
Alan

20040202, 17:57 #14
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Re: Playing with parabolas
<img src=/S/pirate.gif border=0 alt=pirate width=22 height=18>My only comment is that when my wife and I go skiing on our carving skis you have a pair of bowlies on a pair of parabolies <img src=/S/duck.gif border=0 alt=duck width=23 height=23> <img src=/S/pirate.gif border=0 alt=pirate width=22 height=18> <img src=/S/evilgrin.gif border=0 alt=evilgrin width=15 height=15>

20040203, 00:45 #15
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Re: Playing with parabolas
Steve,
Here's a better example of what I'm talking about, except that I'll be using 3 points rather than the 4 shown here.
Alan