My Math Forum Solve for coefficients in quadratic equation

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 February 5th, 2014, 08:01 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, 10:17 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, 10:34 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 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.
February 6th, 2014, 11:39 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?

February 6th, 2014, 01:12 PM   #5
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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
 bezier.png (5.1 KB, 380 views)

 February 6th, 2014, 08:25 PM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,788 Thanks: 970 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?
February 7th, 2014, 05:08 AM   #7
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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
 Bezier2.png (3.9 KB, 356 views)

 February 7th, 2014, 11:38 AM #8 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,788 Thanks: 970 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, 12:16 PM   #9
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From: Hamilton, Ontario

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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
 bezier3.png (6.8 KB, 335 views)

 February 7th, 2014, 02:43 PM #10 Global Moderator   Joined: Dec 2006 Posts: 20,095 Thanks: 1905 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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### how do u solve m=2my-5¡Â2

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