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February 5th, 2014, 07:01 AM   #1
mms
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Solve for coefficients in quadratic equation

I had written a computer program a while back that has this code:

'solve for coefficients in quadratic equation
Ca = 1 + m ^ 2
Cb = -((x2 * 2) + (2 * (y2 - B) * m))
Cc = x2 ^ 2 + (y2 - B) ^ 2 - l ^ 2


I have been searching the net with [color=#FF0000]solve for coefficients in quadratic equation[/color] hoping to find a site
that discusses the formulas to calculate these coefficients, but can find nothing.

Could someone point me to a site that shows how to solve for these 3 coefficients, as I'm sure I didn't come up with these on my own.

I am particularly interested in what the [color=#FF0000]l[/color] is in the last formula.
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February 5th, 2014, 09:17 AM   #2
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Re: Solve for coefficients in quadratic equation

What had been given about these parabola? the extremum and a point, 3 points, something else?
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February 6th, 2014, 09:34 AM   #3
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Re: Solve for coefficients in quadratic equation

It's not for a parabola.

x,y co-ordinates of start and end points of a line are known.
I then was calculating start and end points of two lines perpendicular to that line at it's third points.
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February 6th, 2014, 10:39 AM   #4
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Re: Solve for coefficients in quadratic equation

Quote:
Originally Posted by mms
I then was calculating start and end points of two lines perpendicular to that line at it's third points.
Can you explain what d'heck that means...
"it's third points": more than one "third point"?
Can you provide a representative diagram?
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February 6th, 2014, 12:12 PM   #5
mms
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Re: Solve for coefficients in quadratic equation

The attached image shows two light blue lines; these are perpendicular to the purple line, and located at the third points on that line.

I will always know the start and end co-ordinates of the purple line.
Attached Images
File Type: png bezier.png (5.1 KB, 380 views)
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February 6th, 2014, 07:25 PM   #6
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Re: Solve for coefficients in quadratic equation

Boy oh boy, you're sure being "mysterious"

So, as far as I can see:
Code:
         E        F


A        C        D        B
- given are the coordinates of the ends of a straight line AB, this line being a chord of a circle
- points C and D are on AB, such that AC = CD = DB
- points E and F are on the circle's circumference, such that EC and FD are perpendicular to AB
- and, of course, EC = FD and AE = BF

You are trying to determine the coordinates of points C, D, E and F : RIGHT?

It is easy enough to get C's and D's coordinates, and the equations of CE and DF.
BUT we are given ONLY 2 points on the circle: A and B.
A minimum of 3 points is required to define a circle.
So we don't have a circle equation for poor CE and DF to intersect

Did I miss anything?
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February 7th, 2014, 04:08 AM   #7
mms
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Re: Solve for coefficients in quadratic equation

I wasn't intentionally being mysterious.

You are correct for the most part.

I am trying to determine the co-ordinates of C, D, E and F.
The curve is not a circle, but rather a Bézier curve defined by end points A and B,
and control points E and F. That is why the curve does not pass through points E or F.
Points E and F determine what the Bézier will look like.
Different E and F will cause Bézier to look differently (see new attachment).

Points E and F can be wherever I want, so I decided to place them perpendicular to the base line (the chord) at third points C and D,
at a predetermined distance away from chord line. This distance away is 10% of chord length.

I think I have explained myself this time.
Attached Images
File Type: png Bezier2.png (3.9 KB, 356 views)
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February 7th, 2014, 10:38 AM   #8
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Re: Solve for coefficients in quadratic equation

Beyond me, but I'm sure our Sir MathBalarka will step in and handle this diligently for you...
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February 7th, 2014, 11:16 AM   #9
mms
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Re: Solve for coefficients in quadratic equation

I think I've described this more difficult than it is (the curve is irrelative)

Given:
co-ordinates of A and B

Required:
1. co-ordinates of C and D such that AC, CD, and DB are equal lengths
2. co-ordinates of E such that the length of CE is 10% the length of AB, and CE is perpendicular to AB
3. co-ordinates of F such that the length of DF is 10% the length of AB, and DF is perpendicular to AB

New diagram attached.
Attached Images
File Type: png bezier3.png (6.8 KB, 335 views)
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February 7th, 2014, 01:43 PM   #10
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Do you know how to calculate the length of AB?
Do you understand these equations: C = (2A + B)/3, D = (A + 2B)/3?
Do you know on which side of AB points E and F lie?
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