- #1
Axecutioner
- 32
- 0
I'm trying to find the equation of a parabola that goes through any three given points in the x-y 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(cf-de) + c^2(be-af) + e^2(ad-bc)}{a^2(c-e) + c^2(e-a) + e^2(a-c)}[/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(cf-de) vs a^2(c-e) etc
Thanks
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(cf-de) + c^2(be-af) + e^2(ad-bc)}{a^2(c-e) + c^2(e-a) + e^2(a-c)}[/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(cf-de) vs a^2(c-e) etc
Thanks