Strange parabolic calculations  need simplification help.by Axecutioner Tags: calculations, parabolic, simplification, strange 

#1
Mar3112, 07:22 PM

P: 32

I'm trying to find the equation of a parabola that goes through any three given points in the xy plane denoted by (a,b), (c,d), and (e,f), which would be considered givens in these calculations. I need the form of the equation y = Ax^2 + Bx + C where A, B, and C are unknown.
I have gotten as far as to find C, which means that with those original 6 givens I know the C term of any parabolic equation: C = [itex]\frac{a^2(cfde) + c^2(beaf) + e^2(adbc)}{a^2(ce) + c^2(ea) + e^2(ac)}[/itex] Which is screaming to be simplified but I can't get it any further. I know this is correct by checking with known parabolic functions and test points but getting the general function is proving difficult. Any ideas? I think it's weird that there's a^2, c^2, and e^2 terms on the top and bottom, and in the numerator there's exactly two of each letter in the parenthesis, one in a negative term and one in a positive term. And that in the denominator the squared terms match some of the letters with the squared terms in the numerator, eg: a^2(cfde) vs a^2(ce) etc Thanks 



#2
Mar3112, 10:16 PM

P: 688

Perhaps a simpler way to solve the problem is to substitute your three points (x,y) into the equation, and obtain three equations in the three unknowns A,B,C; then solve this system.
If a 3equation system gives you problems, you could try subtracting one equation from the other two, to obtain two equations in the two unknowns A,B. Be aware that the assumption is that your equation has the given form; that is, the parabola's axis is always vertical, not rotated to one side. 



#3
Mar3112, 11:03 PM

P: 32





#4
Mar3112, 11:09 PM

P: 688

Strange parabolic calculations  need simplification help.As for your three points, yes, you should always find a solution as long as the three points are different (no repeated points). If they are on a line, you'll just get A=0. And yes, vertical lines are forbidden (horizontals will just produce A=B=0 and C=y). 



#5
Mar3112, 11:15 PM

P: 771

You could easily approach this as a linear regression problem. Least squares will generate a quadratic polynomial that fits through all three points.




#6
Apr112, 12:54 AM

P: 32

I have no idea what linear regression is. :P
and that's interesting about the Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 formula. But given 3 points and 6 unknowns, wouldn't it be impossible to solve? Meaning there are many different parabolas that fit those 3 points given variable axis slope? 



#7
Apr112, 02:20 AM

Mentor
P: 4,499

The problem is that the general quadratic form also produces hyperbolas and ellipses




#8
Apr112, 03:27 AM

P: 688

Here is a worked out example, as I'm not sure exactly where are you stuck. If you are given the points (1,7), (3,11), (5,13), for example, you set up the system[tex]\begin{align*} A+B+C &= 7 \\ 9A + 3B + C &= 11 \\ 25A + 5B + C &= 13 \end{align*}[/tex]Subtracting the first equation from the other two, you obtain the smaller system[tex]\begin{align*} 8A + 2B &= 4 \\ 24A + 4B &= 6 \end{align*}[/tex]Divide the second equation by 2, making it [itex]12A + 2B = 3[/itex], and subtracting them the [itex]2B[/itex] term is cancelled, leaving [itex]4A = 1[/itex] or [itex]A = \frac 1 4[/itex]. Substituting this value in the first equation gives you [itex]2B = 4  8(\frac 1 4)[/itex] or [itex]B = 3[/itex]; and finally, using one of the equations from the first system (pick the simplest one), you get [itex]C = \frac {17} 4[/itex]. 



#9
Apr312, 08:56 PM

P: 32

Dodo, I know how to do that given points. I'm trying to find an equation for the parabola given any three points as I stated in the OP. I know how to do it. All I'm asking for is simplifying the C= equation I put in the OP if it can be done and nobody has posted anything related to what I originally asked.




#10
Apr312, 11:47 PM

P: 972

is the best that I can do to try to answer your specific question. That (ea) in the numerator worries me. My result equals yours, But it spoils the pattern. Maybe it should be ea in the denominator and the whole thing needs a minus sign. That would make the pattern more uniform. There is still a great deal of symmetry in my result, but I haven't found any way to reduce it further. If you prefer you can use your original numerator with my denominator, the result is the same. 



#11
Apr412, 09:36 AM

HW Helper
P: 3,436

There is no reason that that expression can be simplified. Just because it seems long and follows a nice pattern in both the numerator and denominator, doesn't mean anything.
Many students have also made the mistake of cancelling the x out of [tex]\frac{x}{x+a}[/tex] because they felt it was simpler 



#12
Apr412, 04:44 PM

P: 32

Bill Simpson, thanks! I'm surprised the denominator could factor into (ac)(ce)(ae) but it does work. How'd you figure that out? I like the numerator you set up too, will play around with it and see what works out.
Mentallic, is there a way to separate that fraction out, in any way? Also: If you start with 3 points: (a,b) (c,d) (e,f) The slopes of the lines between the first two and last two are: [itex]\frac{db}{ca}[/itex] and [itex]\frac{fd}{ec}[/itex] Therefore the change in slope between those two lines is: [itex]\frac{\frac{fd}{ec}  \frac{db}{ca}}{ea}[/itex], correct? So given that change in slope, and knowing that a parabola has a constant change in slope (second derivative is constant), then that change in slope formula must be the A term of Ax^2 + Bx + C? Edit: Oh, no, maybe not. Because when differentiating Ax^2 + Bx + C you get a 2 multiplied into the A term by power rule. So A would be half that formula? Or is it right? 



#13
Apr412, 04:56 PM

P: 972

Perhaps someone else can contribute the title or other titles on how symmetries can be used to understand some problems much more easily. 



#14
Apr412, 05:50 PM

P: 32

Interesting. If you ever find the title or author please let me know.
Here's a different version of the C equation: [itex]\frac{ade^2  bce^2 + bc^2e  a^2de  ac^2f + a^2cf}{ae^2  ce^2 + c^2e  a^2e  ac^2 + a^2c}[/itex] I wonder if the top could factor out similarly to the way you got it to in the bottom. It'd sure make things easier... 



#15
Apr512, 02:23 AM

P: 972

If you have not already done so then you might think about why there is this symmetry in the variables. It may be superficial and unimportant or there might be levels of enlightenment when you look at the reasons and consequences of the symmetry. I suppose there might even be a possibility of discovering other forms and perhaps even simpler forms to be found as a result. I'm not seeing any simpler forms yet. 



#16
Apr512, 04:19 AM

P: 32

I went ahead and used your original suggestion to solve for A using: C=(fac(ac)+bce(ce)+dea(ea))/((ac)(ce)(ae))
And let me tell ya, it's amazing how complicated it got, then how simple the answer is. I'll scan my work tomorrow and post it  way too long to type out. Again, thank you for that :D Was a big help. 



#17
Apr512, 04:56 AM

Sci Advisor
HW Helper
PF Gold
P: 12,016

Axecutioner:
That the "C" term should be complicated is not so strange. I believe you would get simpler coefficents for the general expression if you afterwards rearrange them according to the following general formula: A(x+b)^2+c In this case, "c" also has a neat interpretation as an extremal value, "b" as the symmetry axis, and "A" has its standard interpretation. 


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