1. Introduction
Consider the equation of the form
\[
ax^2+by^2+cxy+dx+ey+f=0
\]
where $a,b,\ldots ,f$ are real numbers, and at least one of the numbers $a,b,c$ is not zero. An equation of this type is called a quadratic equation in $x$ and $y$, and
\[
ax^2+by^2+cxy
\]
is called the
associated quadratic form.
Graphs of quadratics equations are known as conics, or conic sections. The most important conics are ellipses, circles, hyperbolas and parabolas; these are called nondegenerate conics. The remaining conics are called degenerated and include single points and pairs of lines.
A conic is said to be in standard position relative to the coordinate axes if its equation can be expressed in one of the following forms:
\begin{align}
& \bullet \;\;\dfrac{x^2}{k^2}+\dfrac{y^2}{l^2}=1; \;\;k,l>0, \nonumber \\ % no number is shown
& \bullet \;\;\dfrac{x^2}{k^2}\dfrac{y^2}{l^2}=1\;\; \text{or}\;\; \dfrac{x^2}{k^2}\dfrac{y^2}{l^2}=1; \;\;k,l>0, \label{conics}\\ % there is a number
& \bullet\;\; x^2=ky\;\; \text{or}\;\; y^2=kx;\;\; k\neq 0. \nonumber % no number
\end{align}

Simulation 1. Click on the image (or link) to explore nondegenerate conics in standard position.
Conics in standard position, click
Here

Observe that no conic in standard position has an $xy$term (that is, a crossproduct term) in its equation; the presence of an $xy$term in the equation of a nondegenerate conic indicates that the conic is rotated out of standard position (Figure 1).

Figure 1. Rotated 
Also no conic in standard position has both an $x^2$ and a $x$ term or both $y^2$ and a $y$ term. If there is no crossproduct term, the occurrence of either of these pairs in the equation of a nondegenerate conic indicates that the conic is translated out of standard position (Figure 2).

Figure 2. Translated. 
The occurrence of either of these pairs and a crossproduct term usually indicates that the conic is both rotated and translated out of standard position (Figure 3).

Figure 3. Rotated and translated. 
2. Identifying conics
One technique for identifying the graph of a nondegenerate conic that is not in standard position consists of rotating and translating the $xy$coordinate axes to obtain an $uv$coordinate system relative to which the conic is in standard position. Once this is done, the equation of the conic in the $uv$system will have one of the forms given in (\ref{conics}) and can then easily be identified.

Simulation 2. Click on the image (or link below) to explore the rotation and translation of conics.
Rotation and translation of conics, click Here

Example: Completing the square and translating.
Since the quadratic equation
\[
2x^2+y^212x4y+18=0
\]
contains $x^2$, $x$, $y^2$, and $y$terms but no crossproduct term, its graph is a conic that is translated out of standard position but not rotated. This conic can be brought into standard position by suitably translating coordinate axes. To do this, first collect $x$terms and $y$terms. This yields
\[
(2x^212x)+(y^24y)+18=0 \quad \text{or}\quad 2(x^26x)+(y^24y)=18.
\]
By completing the squares on the two expressions in parentheses, we obtain
\begin{eqnarray}\label{uno}
2(x^26x+9)+(y^24y+4)=18+18+4 \quad \text{or}\quad 2(x3)^2+(y2)^2=4.
\end{eqnarray}
If we translate the coordinate axes by means of the translation equations
\[
u=x3,\quad v=y2
\]
then (\ref{uno}) becomes
\[
2u^2+v^2=4\quad \text{or}\quad \frac{u^2}{2}+\frac{v^2}{4}=1
\]
which is the equation of an ellipse in standard position in the $uv$system.
2.1 Eliminating the crossproduct term
We shall now show how to identify conics that are rotated out of standard position. Now consider a conic $C$ whose equation in $xy$coordinates is $ax^2+by^2+cxy+dx+ye+f=0$. This equation can always be written in matrix form:
\[
\big(x\;\;y\big)
\Bigg(
\begin{matrix}
a & c/2 \\
c/2 & b
\end{matrix}\Bigg)
\Bigg(
\begin{matrix}
x \\
y
\end{matrix}\Bigg)
+
\big(d\;\;e\big)
\Bigg(
\begin{matrix}
x \\
y
\end{matrix}\Bigg)
+f=0
\]
or
\[
\mathbf{x}^{T}A\mathbf{x}+K\mathbf{x}+f=0
\]
where
\[
A=\Bigg(
\begin{matrix}
a & c/2 \\
c/2 & b
\end{matrix}\Bigg),\quad \mathbf{x}= \Bigg(
\begin{matrix}
x \\
y
\end{matrix}\Bigg) \quad \text{and}\quad K=\big(d\;\;e\big).
\]
We would like to rotate the $xy$coordinate axes so that the equation of the conic in the new $uv$coordinate system has no crossproduct term. This can be done as follows.
Step 1. Find a matrix
\[
P=\Bigg(
\begin{matrix}
p_{11} & p_{12} \\
p_{21} & p_{22}
\end{matrix}\Bigg)
\]
that orthogonally diagonalises the matrix $A$.
Step 2. Interchange the columns of $P$, if necessary, to make $\text{det}(P) = 1$. This ensures that the orthogonal coordinate transformation
\begin{eqnarray}\label{dos}
\mathbf{x}=P\mathbf{v},\quad \text{that is}\quad \Bigg(
\begin{matrix}
x \\
y
\end{matrix}\Bigg)=P \Bigg(
\begin{matrix}
u \\
v
\end{matrix}\Bigg)
\end{eqnarray}
is a rotation.
Step 3. To obtain the equation for $C$ in the $uv$system, substitute (\ref{dos}) into the matrix form of the equation. This yields
\[
\left(P\mathbf{v} \right)^{T}A\left(P\mathbf{v} \right)+K\left(P\mathbf{v} \right)+f=0
\]
or
\begin{eqnarray}\label{tres}
\mathbf{v} ^{T}\left(P^TAP\right)\mathbf{v} +\left(KP \right)\mathbf{v}+f=0.
\end{eqnarray}
Since $P$ orthogonally diagonalises $A$,
\[
P^TAP=\Bigg(
\begin{matrix}
\lambda_1 & 0 \\
0 & \lambda_2
\end{matrix}\Bigg)
\]
where $\lambda_1$ and $\lambda_2$ are eigenvalues of $A$. Thus (\ref{tres}) can be rewritten as
\[
\big(u\;\;v\big)
\Bigg(
\begin{matrix}
\lambda_1 & 0 \\
0 & \lambda_2
\end{matrix}\Bigg)
\Bigg(
\begin{matrix}
u \\
v
\end{matrix}\Bigg)
+
\big(d\;\;e\big)\Bigg(
\begin{matrix}
p_{11} & p_{12} \\
p_{21} & p_{22}
\end{matrix}\Bigg)
\Bigg(
\begin{matrix}
u \\
v
\end{matrix}\Bigg)
+f=0
\]
or
\[
\lambda_1u^2+\lambda_2u^2+d^{\prime}u+e^{\prime}v+f=0
\]
(where $d^{\prime}=dp_{11}+ep_{21}$ and $e^{\prime}=dp_{12}+ep_{22}$). This equation has no crossproduct term.
Example: Eliminating the CrossProduct Term.
Consider the conic $C$ whose equation is $5x^24xy+8y^236=0$.
To describe this conic, first we write the quadratic equation in its matrix form. That is
\begin{eqnarray}\label{cuatro}
\mathbf{x}^{T}A\mathbf{x}36=0
\end{eqnarray}
where
$$A=\Bigg(
\begin{matrix}
5 & 2 \\
2 & 8
\end{matrix}\Bigg)$$
The characteristic equation of $A$ is
$$\text{det}(A\lambda I)=\left \begin{matrix}
5\lambda & 2 \\
2 & 8\lambda
\end{matrix}\right=(\lambda9)(\lambda4)=0$$
so the eigenvalues of $A$ are $\lambda=4$ and $\lambda=9$. We leave it for the reader to show that orthonormal bases for the eigenvalues are
\[
\lambda=4:\quad \mathbf{v}_1= \Bigg(
\begin{matrix}
2/\sqrt{5} \\
1/\sqrt{5}
\end{matrix}\Bigg),\qquad
\lambda=9:\quad \mathbf{v}_2= \Bigg(
\begin{matrix}
1/\sqrt{5} \\
2/\sqrt{5}
\end{matrix}\Bigg)
\]
Thus
$$P=\Bigg(
\begin{matrix}
2/\sqrt{5} & 1/\sqrt{5} \\
1/\sqrt{5} & 2/\sqrt{5}
\end{matrix}\Bigg)$$
orthogonally diagonalises $A$. Moreover, $\text{det}(P)=1$, and thus the orthogonal coordinate transformation
\begin{eqnarray}\label{cinco}
\mathbf{x} =P\mathbf{v}
\end{eqnarray}
is a rotation. Substituting (\ref{cinco}) into (\ref{cuatro}) yields
\[
(P\mathbf{v})^{T}A(P\mathbf{v})36=0\quad\text{or}\quad \mathbf{v}^T(P^TAP)\mathbf{v}36=0
\]
Since
\[
P^TAP=\Bigg(
\begin{matrix}
4 & 0 \\
0 & 9
\end{matrix}\Bigg)
\]
this equation can be written as
\[
\big(u\;\;v\big)
\Bigg(
\begin{matrix}
4 & 0 \\
0 & 9
\end{matrix}\Bigg)
\Bigg(
\begin{matrix}
u \\
v
\end{matrix}\Bigg)
36=0
\]
or
\[
4u^2+9v^236=0\quad\text{or}\quad \frac{u^2}{9}+\frac{v^2}{4}=1
\]
which is the equation of the ellipse in standard form.
3. Exercise
Translate and rotate the coordinate axes, if necessary, to put the conic in standard position. Name the conic and give its equation in the final coordinate system.
1. $9x^24xy+6y^210x20y5=0$
2. $3x^28xy12y^230x64y=0$
3. $2x^24xyy^24x8y+14=0$
4. $21x^2+6xy+13y^2114x+34y+73=0$

Simulation 3. You can check your results in the following simulation. Click on the image (or link below) to access the simulation.
Classification of conics, click
Here
