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February 5th, 2014, 07:01 AM  #1 
Member Joined: Nov 2009 From: Hamilton, Ontario Posts: 30 Thanks: 0  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. 
February 5th, 2014, 09:17 AM  #2 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361  Re: Solve for coefficients in quadratic equation
What had been given about these parabola? the extremum and a point, 3 points, something else?

February 6th, 2014, 09:34 AM  #3 
Member Joined: Nov 2009 From: Hamilton, Ontario Posts: 30 Thanks: 0  Re: Solve for coefficients in quadratic equation
It's not for a parabola. x,y coordinates 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. 
February 6th, 2014, 10:39 AM  #4  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,402 Thanks: 829  Re: Solve for coefficients in quadratic equation Quote:
"it's third points": more than one "third point"? Can you provide a representative diagram?  
February 6th, 2014, 12:12 PM  #5 
Member Joined: Nov 2009 From: Hamilton, Ontario Posts: 30 Thanks: 0  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 coordinates of the purple line. 
February 6th, 2014, 07:25 PM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,402 Thanks: 829  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  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? 
February 7th, 2014, 04:08 AM  #7 
Member Joined: Nov 2009 From: Hamilton, Ontario Posts: 30 Thanks: 0  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 coordinates 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. 
February 7th, 2014, 10:38 AM  #8 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,402 Thanks: 829  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... 
February 7th, 2014, 11:16 AM  #9 
Member Joined: Nov 2009 From: Hamilton, Ontario Posts: 30 Thanks: 0  Re: Solve for coefficients in quadratic equation
I think I've described this more difficult than it is (the curve is irrelative) Given: coordinates of A and B Required: 1. coordinates of C and D such that AC, CD, and DB are equal lengths 2. coordinates of E such that the length of CE is 10% the length of AB, and CE is perpendicular to AB 3. coordinates of F such that the length of DF is 10% the length of AB, and DF is perpendicular to AB New diagram attached. 
February 7th, 2014, 01:43 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 18,956 Thanks: 1602 
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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