Question
9x^2-24xy+16y^2-20x-15y-50=0
Use axis rotation formulas for x and y to transform the quadratic equation to an equation in (u,v) coordinates with no cross product term. Identify the vertex or vertices in (x,y) coordinates
thank you so so so much
Use axis rotation formulas for x and y to transform the quadratic equation to an equation in (u,v) coordinates with no cross product term. Identify the vertex or vertices in (x,y) coordinates
thank you so so so much
Answers
tan(2θ) = B/(A-C) = 24/7
so sinθ = 0.6 and cosθ = 0.8
Now apply the rotation matrix and you have
(x,y) =
(cosθ -sinθ) (u)
(sinθ cosθ) (v)
A' = Acos^2θ + Bsinθcosθ + Csin^2θ = .64A+.48B+.36C = .64*9-.48*24+.36*16 = 0
B' = (C-A)sin2θ + Bcos2θ = .96(C-A)+.28B = .96*7-.28*24 = 0
C' = Asin^2θ - Bsinθcosθ + Ccos^2θ = .36A-.48B+.64C = .36*9+.48*24+.64*16 = 25
D' = Dcosθ + Esinθ = .8D+.6E = -.8*20-.6*15 = 25
E' = -Dsinθ + Ecosθ = -.6D+.8E = .6*20-.8*15 = 0
F' = F = -50
and you end up with
25v^2+25u - 50 = 0
u = 2-v^2
see the graph at
https://www.wolframalpha.com/input/?i=9x%5E2-24xy%2B16y%5E2-20x-15y-50%3D0
Hmmm. I seem to have forgotten a - sign somewhere. The graph shows u = v^2-2 rotated clockwise, rather than the usual counter-clockwise. Maybe you can spot the mistake.
so sinθ = 0.6 and cosθ = 0.8
Now apply the rotation matrix and you have
(x,y) =
(cosθ -sinθ) (u)
(sinθ cosθ) (v)
A' = Acos^2θ + Bsinθcosθ + Csin^2θ = .64A+.48B+.36C = .64*9-.48*24+.36*16 = 0
B' = (C-A)sin2θ + Bcos2θ = .96(C-A)+.28B = .96*7-.28*24 = 0
C' = Asin^2θ - Bsinθcosθ + Ccos^2θ = .36A-.48B+.64C = .36*9+.48*24+.64*16 = 25
D' = Dcosθ + Esinθ = .8D+.6E = -.8*20-.6*15 = 25
E' = -Dsinθ + Ecosθ = -.6D+.8E = .6*20-.8*15 = 0
F' = F = -50
and you end up with
25v^2+25u - 50 = 0
u = 2-v^2
see the graph at
https://www.wolframalpha.com/input/?i=9x%5E2-24xy%2B16y%5E2-20x-15y-50%3D0
Hmmm. I seem to have forgotten a - sign somewhere. The graph shows u = v^2-2 rotated clockwise, rather than the usual counter-clockwise. Maybe you can spot the mistake.
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