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Writing a Conic in Canonical Form

Given the equation of a conic of the form  
\[Ax_1^2+Bx_1x_2 +Cx^2_2=k\]
  we can write in matrix form as  
\[(x_1.x_2)\left( \begin{array}{cc} A & B/2 \\ B/2 & C \end{array} \right)\begin{pmatrix}x_1\\x_2\end{pmatrix}=k \]

We can find the eigenvalues and eigenvectors of the matrix above and express the conic in the canonical form  
\[A'y_1^2+B'y_2^2=C'\]
.
Example: Write the conic  
\[5x_1^2+4x_1x_2+8x_2^2=9\]
  in canonical form.
We find the eigenvalues and eigenvectors of the matrix  
\[A=\left( \begin{array}{cc} 5 & 4/2 \\ 4/2 & 8 \end{array} \right)=\left( \begin{array}{cc} 5 & 2 \\ 2 & 8 \end{array} \right) \]

\[\begin{equation} \begin{aligned} \left| \begin{array}{cc} 5- \lambda & 2 \\ 2 & 8- \lambda \end{array} \right| & =(5-- \lambda)(8- \lambda )-2^2 \\ &=\lambda^2-13 \lambda-36 \\ & =(\lambda-4)(\lambda-9)=0 \end{aligned} \end{equation}\]

Hence  
\[\lambda=4, \: 9\]

\[\lambda=4\]
:
\[\begin{equation} \begin{aligned}\left( \begin{array}{cc} 5-4 & 2 \\ 2 & 8-4 \end{array} \right) \begin{pmatrix}x_1\\x_2\end{pmatrix}&=\left( \begin{array}{cc} 1 & 2 \\ 2 & 4 \end{array} \right) \begin{pmatrix}x_1\\x_2\end{pmatrix}\\ &=\begin{pmatrix}x_1+2x_2\\2x_1+4x_2\end{pmatrix} \\ &= \begin{pmatrix}0\\0\end{pmatrix} \end{aligned} \end{equation}\]

We can take  
\[x_1=-2, \: x_2=1\]

Normalising gives  
\[x_1=-\frac{2}{\sqrt{5}}, \: x_2=\frac{1}{\sqrt{5}}\]

The first eigenvector is  
\[\begin{pmatrix}-\frac{2}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}\end{pmatrix}\]

\[\lambda=9\]
:
\[\begin{equation} \begin{aligned} \left( \begin{array}{cc} 5-9 & 2 \\ 2 & 8-9 \end{array} \right) \begin{pmatrix}x_1\\x_2\end{pmatrix} &= \left( \begin{array}{cc} -4 & 2 \\ 2 & -1 \end{array} \right) \begin{pmatrix}x_1\\x_2\end{pmatrix} \\ &= \begin{pmatrix}-4x_1+2x_2\\2x_1-x_2\end{pmatrix} \\ &= \begin{pmatrix}0\\0\end{pmatrix}\end{aligned} \end{equation}\]

We can take  
\[x_1=1, \: x_2=2\]

Normalising gives  
\[x_1=-\frac{1}{\sqrt{5}}, \: x_2=\frac{2}{\sqrt{5}}\]

The second eigenvector is  
\[\begin{pmatrix}\frac{1}{\sqrt{5}}\\ \frac{2}{\sqrt{5}}\end{pmatrix}\]

The matrix of eigenvectors is  
\[\left( \begin{array}{cc} -\frac{2}{\sqrt{5}} & -\frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \end{array} \right)\]

Now define the transformation  
\[\mathbf{x}=P \mathbf{y}\]
  then  
\[\mathbf{x}^TA \mathbf{x}=9\]
  becomes  
\[(P \mathbf{y})^TA(P \mathbf{x})=\mathbf{y}^TP^TAP=\mathbf{y}D\mathbf{y}=9\]

We have  
\[(y_1.y_2)\left( \begin{array}{cc} 4 & 0 \\ 0 & 9 \end{array} \right)\begin{pmatrix}y_1\\y_2\end{pmatrix}=y_1^2+9y_2^2=9 \]

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