Conic equation: ${4x^{2}−4xy+7y^{2}+12x+6y−9=0}$ Converting this conic to "standard form" 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
 A: Let $2x-y=z$, then the given equation becomes:
\begin{align}
&z^2+6y^2+6z+12y-9=0\\
\implies &(z+3)^2+6(y+1)^2-9-6-9=0\\
\implies &(z+3)^2+6(y+1)^2=24\\
\implies&\boxed{\dfrac{(2x-y+3)^2}{24}+\dfrac{(y+1)^2}{4}=1}
\end{align}
This is the required standard form (note that the major axis is not parallel to the $x$-axis in this case).
A: Let $f(x,y) =4x^{2}−4xy+7y^{2}+12x+6y−9$. The center of the ellipse is $(-2,-1)$, obtained from $f’_x=f’_y=0$. Then, its major and minor axes can  be obtained as follows
$$\frac{f’_y}{f’_x}= \frac{y-(-1)}{x-(-2)}\implies 
y= \frac12x,\>\>\>y =-2x-5$$
They are at an angle $\theta=\tan^{-1} \frac12 $ with the $x$-axis. Thus, with $\cos\theta =\frac2{\sqrt5}$ and $\sin\theta =\frac1{\sqrt5}$, apply the transformation below
$$ x=-2+\frac2{\sqrt5} u + \frac1{\sqrt5} v,\>\>\>\>\>
 y=-1+\frac1{\sqrt5} u - \frac2{\sqrt5} v$$
to convert $f(x,y)=0$ to the standard form
$$ \frac{u^2}8 +\frac{v^2}3=1$$
A: The conic can be written as $P(x,y)=\begin{bmatrix}
x & y 
\end{bmatrix}\begin{bmatrix}
4 &-2 \\ 
-2 & 7
\end{bmatrix}\begin{bmatrix}
x\\ 
y
\end{bmatrix}+2\begin{bmatrix}
6 & 3 
\end{bmatrix}\begin{bmatrix}
x\\ 
y
\end{bmatrix}-9=0$
We want to find a matrix $P \in SO(2)$ such that $P^\top AP$, where $A=\begin{bmatrix}
4  &-2 \\ 
-2 & 7 
\end{bmatrix}$. So you calculate the characteristic polynomial of $A$, which is $c_A(t)=t^2-11t+24$. The eigenvalues are $t_1=8$ and $t_2=3$. With some calculations you will find that the eigenspaces are: $V_8=\mathcal{L}\begin{Bmatrix}
\begin{bmatrix}
\frac{1}{\sqrt{5}}\\ 
\frac{-2}{\sqrt{5}}
\end{bmatrix}\\ 
\end{Bmatrix}$ and $V_3=\mathcal{L}\begin{Bmatrix}
\begin{bmatrix}
\frac{2}{\sqrt{5}}\\ 
\frac{1}{\sqrt{5}}
\end{bmatrix}\\ 
\end{Bmatrix}$. The matrix we were looking for is $P=\begin{bmatrix}
\frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ 
\frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}}
\end{bmatrix}$. We can change coordinates as following $\begin{bmatrix}
x\\ 
y\end{bmatrix}=P\begin{bmatrix}
x'\\ 
y'\end{bmatrix}$. In this way the ellipse will have its axis parallel to the $x$ and $y$ axis of the coordinate plane. Doing some calculations, you will find that the rotated ellipse has equation: $P'(x,y)=8(x')^2+3(y')^2+6\sqrt5y'-9=0$. Its major axis lies on the $y-$axis. If you want it to lie on the $x-$axis you have to rotate the conic we found of an angle of $\pi/2$. Eventually, by completing the square, you can translate the ellipse such that its center corresponds with the origin $(0,0)$. This last two passages should not be too difficult.
