## Identifying the Nature of a Quadratic Curve

Given a quadratic equation of the form  {jatex options:inline}Ax_1^2+Bx_1x_2 +Cx^2_2=k{/jatex}, we can identify the nature of the curve (hyperbola, ellipse, parabola) by writing the equation in matrix form as  {jatex options:inline}(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 {/jatex}
We can find the eigenvalues and eigenvectors of the matrix above and express the conic in the canonical form  {jatex options:inline}A'y_1^2+B'y_2^2=C'{/jatex}.
From this we can identity the nature of the curve.
Example: Write the conic  {jatex options:inline}5x_1^2+4x_1x_2+8x_2^2=9{/jatex}  in canonical form.
We find the eigenvalues and eigenvectors of the matrix  {jatex options:inline}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) {/jatex}
{jatex options:inline}\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}{/jatex}
Hence  {jatex options:inline}\lambda=4, \: 9{/jatex}
{jatex options:inline}\lambda=4{/jatex}:
{jatex options:inline}\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}{/jatex}
We can take  {jatex options:inline}x_1=-2, \: x_2=1{/jatex}
Normalising gives  {jatex options:inline}x_1=-\frac{2}{\sqrt{5}}, \: x_2=\frac{1}{\sqrt{5}}{/jatex}
The first eigenvector is  {jatex options:inline}\begin{pmatrix}-\frac{2}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}\end{pmatrix}{/jatex}
{jatex options:inline}\lambda=9{/jatex}:
{jatex options:inline}\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}{/jatex}
We can take  {jatex options:inline}x_1=1, \: x_2=2{/jatex}
Normalising gives  {jatex options:inline}x_1=-\frac{1}{\sqrt{5}}, \: x_2=\frac{2}{\sqrt{5}}{/jatex}
The second eigenvector is  {jatex options:inline}\begin{pmatrix}\frac{1}{\sqrt{5}}\\ \frac{2}{\sqrt{5}}\end{pmatrix}{/jatex}
The matrix of eigenvectors is  {jatex options:inline}\left( \begin{array}{cc} -\frac{2}{\sqrt{5}} & -\frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \end{array} \right){/jatex}
Now define the transformation  {jatex options:inline}\mathbf{x}=P \mathbf{y}{/jatex}  then  {jatex options:inline}\mathbf{x}^TA \mathbf{x}=9{/jatex}  becomes  {jatex options:inline}(P \mathbf{y})^TA(P \mathbf{x})=\mathbf{y}^TP^TAP=\mathbf{y}D\mathbf{y}=9{/jatex}
We have  {jatex options:inline}(y_1.y_2)\left( \begin{array}{cc} 4 & 0 \\ 0 & 9 \end{array} \right)\begin{pmatrix}y_1\\y_2\end{pmatrix}=4y_1^2+9y_2^2=9 {/jatex}
The curve is an ellipse.