\[Ax_1^2+Bx_1x_2 +Cx^2_2=k\]
to find directions of the principal axes, 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
\[\lambda_1 , \: \lambda_2\]
and corresponding eigenvectors of the matrix above and express the conic in the canonical form \[\lambda_1 y_1^2+ \lambda_2 y_2^2=C'\]
.The lengths of the axes will be
\[\sqrt{C'/ \lambda_1}\]
and \[\sqrt{C'/ \lambda_2}\]
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
\[\begin{pmatrix}-\frac{2}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}\end{pmatrix}\]
The length of this axis is
\[\sqrt{9/4}=3/2\]
.\[\lambda=9\]
:\[\begin{equation} \begin{aligned} \left( \begin{array}{cc} 5-9 & 2 \\ 2 & 8-9 \end{array} \right) \begin{pmatrix}x_1\\x_2\end{pmatrix} &= \left( \begin{array}{cc} -4 & 2 \\ 2 & -1 \end{array} \right) \begin{pmatrix}x_1\\x_2\end{pmatrix} \\ &= \begin{pmatrix}-4x_1+2x_2\\2x_1-x_2\end{pmatrix} \\ &= \begin{pmatrix}0\\0\end{pmatrix}\end{aligned} \end{equation}\]
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
\[\left( \begin{array}{cc} -\frac{2}{\sqrt{5}} & -\frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \end{array} \right)\]
The length of this axis is
\[\sqrt{9/9}=1\]
.