This is a conic equation:
${4x^{2}−4xy+7y^{2}+12x+6y−9=0}$
I tried a way like this:
1.)${B^{2}-4AC <0} --> ((-4)^2-4x4x7 = 16-112 = -96 < 0)$ "So this is an ellipse"
2.)${Tan2θ = \frac{B}{A-C} = \frac{4}{3} ->θ = 27}$ "degree"
3.) (I saw them using these formulas on a site) There are formulas for ${A^{'}, B^{'}, C^{'}, D^{'}, E^{'}, F^{'}}$ Taking B = 0 to get rid of the xy expression, and the others are:
${A^{'} = Acos^{2}θ+Bcosθsinθ+Csin^{2}θ}$ Plug in the angle "θ" 27 degrees and A 'is about 2,96.
${B^{'} = 0 }$
${C^{'} = Asin^{2}θ+Bsinθcosθ+Ccos^{2}θ}$ Plug in the angle "θ" 27 degrees and A 'is about 4,73
${D^{'} = Dcosθ+Esinθ}$ Plug in the angle "θ" 27 degrees and A 'is about 13,38
${E^{'} = -Dsinθ+Ecosθ}$ Plug in the angle "θ" 27 degrees and A 'is about -0,06
${F^{'} = F = -9}$
4.)Since this is an ellipse according to 1, the standard form of the general equation given is:
${\frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1}$
it must be.
Finally,
I've got this:
${2,96x^{2}+4,73y^{2}+13,38x-0,06y=9}$
Well, what should I do now? As a result of my various researches, there are completely different solutions and I am confused.
I request your help