Quadratic forms
A conic is the set of solutions of a quadratic equation in two variables, that is, an
equation of the form
$$a\,x^2+b\,xy+c\,y^2+d\,x+e\,y+f=0,$$
where $a,b,c,d,e,f\in \mathbb R$ and at least one of the numbers $a,b,c$ is not zero.
The quadratic equation can also be written in matrix form as
\[
\mathbf x^T A \mathbf x+ K \mathbf x+f=0
\]
where
\[
\mathbf x= \left( \begin{array}{c}
x \\
y
\end{array} \right)\quad A=\left( \begin{array}{cc}
a & b/2 \\
b/2 & c \end{array} \right),\quad K=\Big( \begin{array}{cc}
d & e \end{array} \Big).
\]
By means of a rotation of the plane about the origin, a translation of the plane, or both, it is possible to represent every conic in a simplified standard, or canonical, form.
Exemplar
Consider the conic $C$ with equation
$$9x^2-4xy+6y^2-10x-20y-5=0.$$
The matrix form of this equation is
\begin{equation}\label{eq01}
\mathbf x^T A \mathbf x+ K \mathbf x-5=0
\end{equation}
where
\[
A=\left( \begin{array}{cc}
9 & -2 \\
-2 & 6 \end{array} \right)\quad\text{and}\quad K=\Big( \begin{array}{cc}
-10 & -20 \end{array} \Big).
\]
The eigenvalues of $A$ are $\lambda_1=10$ and $\lambda_2=5$. We leave it for the reader to show that orthonormal bases for the eigenspaces are
\[
\lambda_1=10:\; \left( \begin{array}{cc}
\dfrac{2}{\sqrt{5}} \\
\dfrac{-1}{\sqrt{5}} \end{array} \right)\qquad \lambda_2=5:\;\left( \begin{array}{cc}
\dfrac{1}{\sqrt{5}} \\
\dfrac{2}{\sqrt{5}} \end{array} \right)
\]
Thus the matrix that orthogonally diagonalises $A$ is
$$P=\left( \begin{array}{cc}
\dfrac{2}{\sqrt{5}} & \dfrac{1}{\sqrt{5}} \\
\dfrac{-1}{\sqrt{5}} & \dfrac{2}{\sqrt{5}} \end{array} \right).$$
Note that $\text{det}(P)=1$. Thus the orthogonal coordinate transformation
\begin{equation}\label{eq02}
\mathbf x=P \mathbf s=P \left( \begin{array}{c}
s \\
t
\end{array} \right)
\end{equation}
is a rotation. Substituing (\ref{eq02}) in (\ref{eq01}) yields
\begin{equation}\label{eq03}
(P \mathbf s)^TA(P \mathbf s)+K(P \mathbf s)-5=0\quad\text{or}\quad \mathbf s^T (P^TAP)\mathbf s+(KP)\mathbf s-5=0.
\end{equation}
Since
\begin{equation*}
P^TAP=\left( \begin{array}{cc}
10 & 0 \\
0 & 5 \end{array} \right) \quad\text{and}\quad KP=\Big( \begin{array}{cc}
0 & -10\sqrt{5} \end{array} \Big)
\end{equation*}
equation (\ref{eq03}) can be written as
\begin{equation}\label{eq04}
10s^2 +5t^2 -10\sqrt{5}\,t-5=0.
\end{equation}
To bring the conic into standard position, the axes must be translated. We rewrite (\ref{eq04}) as
$$10s^2 +5\left(t^2-2\sqrt{5}\,t\right)=5$$
Completing the squares yields
$$10s^2 +5\left(t^2-2\sqrt{5}\,t+5\right)=5+25$$
or
\begin{equation}\label{eq05}
10s^2+5(t-\sqrt{5})^2=30.
\end{equation}
If we translate the coordinate axes by means of the translation equations
$$u=s\quad \text{and}\quad v=t-\sqrt{5}.$$
then (\ref{eq05}) becomes
$$10u^2+5v^2=30\quad\text{or}\quad \frac{u^2}{3}+\frac{v^2}{6}=1$$
which corresponds to an ellipse.
Observation: Of course we could also rotate to obtain the same ellipse in the form $u^2 + 2v^2 = 6$,
which is just the other standard position.
Simulation
With this simulation you can:
- Observe the rotation and translation of the axes required to put a conic in standard position.
- Check the standard equation of the conic that you obtained analytically, since the simulation provides the equation of the conic in the final coordinate system (with coefficients rounded to two decimal places).
- Check the eigenvalues and eigenvectors.
Things to try:
- Change the values of the coefficients $a,b,c,d,e$ and $f$.
- Drag the sliders to rotate and translate.
- Activate the box Conic equation to show the quadratic equation in Standard or General form.
- Activate the box Check Eigenvectors to input the eigenvectors to the corresponding eigenvalues (rounded to two decimal places), so you can check your calculations.
- Zoom In or Out, if needed.
For instance, from the exemplar discussed above, first we have to define:
a=$9$, b=$-4$, c=$6$, d=$-10$, e=$-20$ and f=$-5$. Then, we input the eigenvectors as follows:
- $\text{Evector}_1$=(0.89,-0.45) for $\lambda_1=10$
- $\text{Evector}_2$=(0.45,0.89) for $\lambda_2=5$
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Note: The simulation shows only one case of the conic section's equation in standard form. The other case can be found by rotating the conic in the oposite direction, or by switching the variables from the equation that you obtained (see the example above).
Worksheet exemplar
The following file contains activities and problems associated with the simulation.
The University of Queensland