Yours can't be put in that form, unless you are talking about a change of variables (e.g. rotation). Can you tell us more about the assignment? What topics are you studying?
I tried a way like this:
1) Identify the conic section using the discriminant: [MATH]{B^{2}-4AC}[/MATH]2) Determine θ using the formula : [MATH]{Tan2θ = \frac{B}{A-C}}[/MATH]3) Calculate [MATH]{A^{'}, B^{'}, C^{'}, D^{'}, E^{'}, F^{'}}[/MATH]4) Rewrite the original equation using [MATH]{A^{'}, B^{'}, C^{'}, D^{'}, E^{'}, F^{'}}[/MATH]
Now,
[MATH]{4x^{2}−4xy+7y^{2}+12x+6y−9=0}[/MATH]
According to the equation x^2, xy, y^2, x, y, and the trailing -9. Their coefficients:
A=4, B=-4, C=7, D=12, E=6, F=-9
1.) [MATH]{B^{2}-4AC <0}[/MATH] ((-4)^2-4x4x7 = 16-112 = -96 < 0) "So this is an ellipse"
2.) [MATH]{Tan2θ = \frac{B}{A-C} = \frac{4}{3} ->θ = 27 "degree"}[/MATH]
3.) There are formulas for [MATH]{A^{'}, B^{'}, C^{'}, D^{'}, E^{'}, F^{'}}[/MATH] Taking B = 0 to get rid of the xy expression, and the others are:
[MATH]{A^{'} = Acos^{2}θ+Bcosθsinθ+Csin^{2}θ}[/MATH] Plug in the angle "θ" 27 degrees and A 'is about 2,96.
[MATH]{B^{'} = 0 }[/MATH][MATH]{C^{'} = Asin^{2}θ+Bsinθcosθ+Ccos^{2}θ}[/MATH] Plug in the angle "θ" 27 degrees and A 'is about 4,73
[MATH]{D^{'} = Dcosθ+Esinθ}[/MATH] Plug in the angle "θ" 27 degrees and A 'is about 13,38
[MATH]{E^{'} = -Dsinθ+Ecosθ}[/MATH] Plug in the angle "θ" 27 degrees and A 'is about -0,06
[MATH]{F^{'} = F = -9}[/MATH]
Since this is an ellipse according to 1, the standard form of the general equation given is:
[MATH]{\frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1}[/MATH]it must be.
I've got this:
[MATH]{2,96x^{2}+4,73y^{2}+13,38x-0,06y=9}[/MATH]
Where could I go wrong? In the formulas A 'B' C 'D' E 'F'?