We will learn how to find a rectangular Cartesian reference system in which the equation of the analytical conical is as simple as possible.
We are going to solve this problem by means of successive reductions, or changes of coordinates so that, after each one, the equation of the conic is formulated simplifying some aspects of the equation in the previous step.
In any case, the target is to see that a rectangular change of coordinates exists: $$$\left\{ \begin{array}{l} x=ax' +by' +e \\ y= cx' +dy' +f\end{array} \right. $$$ that conects the polynomial $$q (x, y)$$ into a polynomial of the form $$$q(ax'+by'+e,cx'+dy'+f)$$$ that has one of the following forms:
- $$\lambda_1x'^2+\lambda_2y'^2+ \mu$$
- $$\lambda_1x'^2+2ey'$$
- $$\lambda_1x'^2+\mu$$
where$$\lambda_1,\lambda_2$$ and $$e$$ are real numbers different from zero and $$\mu$$ is an arbitrary real number. We are going to name the previous expressions reduced forms.
For some reasons (which we will see further on), we will say that they are of the centered, parabolic type and of parallel straight lines.
The first reduction
The first step will be to compute the associated matrix $$A'$$ given the conic with equation $$q (x, y) = 0$$. Then, we will calculate the eigenvalues of the stated matrix, calculating the characteristical polynomial and then the determinant $$det(A'- \lambda I)$$.
As soon as the eigenvalues are found, we can execute a change of coordinates that converts the general equation of the conic into an equation of the following form: $$$\lambda_1x^2+ \lambda_ 2y^2+2dx+2ey+f=0$$$ Notice that if in the general equation of the conic the term $$xy$$ does not exist, this reduction will not be necessary since matrix $$A'$$ will already be diagonal, and what the reduction is doing is diagonalizing the main matrix of the conic.
Next, we are going to give an example of this reduction.
Considering the equation $$q (x, y) =x^2+4xy+2y^2+3=0$$, first we calculate its associated main matrix: $$$A'= \begin{bmatrix} 1 & 2 \\ 2 & 2\end{bmatrix} \Longrightarrow det(A'-\lambda I)=x^2-3x-2$$$ Then, as soon as the characteristical polynomial associated with the main matrix is obtained, we calculate its two roots.
In our case, the roots of the polynomial are $$$\displaystyle\lambda_1=\frac{3+\sqrt{17}}{2} \mbox{ and } \lambda_2=\frac{3-\sqrt{17}}{2}$$$ Therefore, we can effect a change of coordinates to convert the equation of the conic into an equation thus: $$$ \lambda_1x^2+\lambda_ 2y^2+2dx+2ey+f=0$$$
The second reduction
As soon as the first reduction is done, the new equation of the conic has the form $$$\lambda_1 x^2+\lambda_2y^2+2dx+2ey+f=0 $$$ Notice that this equation has two situations to be considered: $$\lambda_1\lambda_2=0$$ and$$\lambda_1\lambda_2 \neq 0$$.
In the first case, there is an eigenvalue equal to zero and another that is not (let's remember that we supposed that matrix $$A$$ is not the zero matrix $$0$$). Exchanging the axes if needed, we can suppose that $$\lambda_1 \neq 0$$ and $$\lambda_2 = 0$$.
We can also suppose that the eigenvalue is positive. At this point, we can differentiate two new cases if $$e$$ is zero or other than zero.
In the first case, the basic tool for continuing is the completion of the squares, $$$ \displaystyle \lambda x^2+2dx=\lambda x'^2+k \mbox{ being } x=x'-\frac{d}{\lambda} \mbox{ and } k=-\frac{d^2}{\lambda}$$$
In the second case, we also complete squares for $$x$$ to obtain an equation of the form $$$\lambda x^2+2ey+f=0$$$ So we obtain an equation without linear term $$x$$.
If we are in the second case, or $$\lambda_1\lambda_2\neq 0$$, then completing squares twice, and the corresponding change of coordinates, allow us to suppose that $$d = y = 0$$, and we obtain a reduced equation of the centered type: $$$ \lambda_1 x'^2+\lambda_2y'^2+\mu=0$$$ In short, if one of the eigenvalues is zero, different changes of coordinates can be enforced to obtain one of the following limited forms:
- $$\lambda_1 x'^2+\lambda_2 y'^2+ \mu$$
- $$\lambda_1 x'^2+2ey'$$
- $$\lambda_1 x'^2+\mu$$
Considering the equation $$$q(x,y)=x^2+2y^2+2x+1=0$$$ we are going to reduce it to obtain one of the three reduced forms. Since we only have a linear term for $$x$$, it will be enough to complete squares for $$x$$.
In our case, doing the proposed change of variables we have $$x' = x + 1$$ , and the equation of the conic becomes $$$q(x,y)=x^2+2x^2=0 $$$ Therefore, this is the first reduced form, in which the two eigenvalues are other than zero.
Canonical equations
Next, we are going to do the last step to be able to classify a conic.
Let $$\lambda_1 x'^2+\lambda_2 y'^2+\mu=0$$ be a reduced equation of the centered type, and we are going to discuss the different possible cases.
If $$\mu \neq 0$$, we write $$$\displaystyle a=\sqrt{\frac{|\mu|}{\lambda_1}} \mbox{ and } b=\sqrt{\frac{|\mu|}{|\lambda_2|}}$$$ then we obtain the equivalent to one of the following three forms: $$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \ \frac{x^2}{a^2}+\frac{y^2}{b^2}=-1, \ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$$ The first one and the third one are the canonical equations of an ellipse and an hyperbole, respectively, with semiaxes $$a$$ and $$b$$.
The second one, which does not have real points, we will call an imaginary ellipse.
On the other hand, if $$ \mu=0$$ then we write $$$ \displaystyle a=\sqrt{\frac{1}{\lambda_1}} \mbox{ and } b=\sqrt{\frac{1}{|\lambda_2|}}$$$ we see that the equation has one of the following forms: $$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=0, \ \frac{x^2}{a^2}-\frac{y^2}{b^2}=0$$$ The second one gives us two real straight lines and the first one gives us two combined imaginary straight lines.
If we consider the limited equation of the parabolic type, $$\lambda_1x'^2+2ey'=0$$, we can suppose that we have the equation $$x^2=2py$$, where $$ \displaystyle p=\frac{e}{\lambda_1}$$.
We can also suppose that $$p> 0$$ (if it was not like that, we would change the sense of the axis $$OY$$).
Therefore, the equation is referring to a parabola with focal parameter $$p$$ and with the axis coinciding with axis $$OY$$.
Finally, if we consider the reduced equation of the type $$\lambda_1x'^2+\mu= 0$$, it is equivalent - if writing $$$\displaystyle k=\sqrt{\frac{|\mu|}{\lambda_1}}$$$ when $$\mu \neq 0$$- to one of the three following equations: $$$x^2=k^2, \ x^2=0, \ x^2= -k^2 (k>0)$$$
The first one gives us two parallel straight lines ($$x = k$$ and $$x =-k$$, $$k$$ being the semidistance between the two straight lines) and the second one gives two coincidental straight lines. We will say that the third one gives us two combined parallel straight lines.
Shortly, this procedure gives us an effective algorithm to go from a general equation of the conic to a canonical equation. To obtain it, we use the following steps:
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Considering the equation of the conic, we calculate its main matrix $$A'$$ and calculate the eigenvalues in order to diagonalize $$A'$$. This step is called the first reduction. When this step is finished, we see that the equation of the conic is as: $$\lambda_1 x^2+\lambda_2y^2+2dx+2ey+f=0$$.
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As soon as the first reduction is done, we look to see if some of the eigenvalues are zero. Then, using the tools for completing squares, we can convert the equation given by the first reduction into one of the following form:
- $$\lambda_1 x'^2+\lambda_2 y'^2+\mu$$
- $$\lambda_1 x'^2+2ey'$$
- $$\lambda_1 x'^2+\mu$$ This step is called the second reduction.
- Finally, depending on the reduced form that we have, by means of a few new changes of coordinates, the different canonical equations are obtained. As soon as the canonical equation is obtained, we will already have classified the conic.
Now, we are going to propose an example that allows us to follow all these steps.
Considering the equation $$$q(x,y)=x^2+y^2+2x+1=0$$$ we first calculate the associate main matrix.
In this case, $$$A'= \begin{bmatrix} 1-x & 1 \\ 1 & 1-x \end{bmatrix} \Longrightarrow det(A-\lambda_1)=x^2-2x=x(x-2)$$$ Therefore it is easily seen that its roots are $$0$$ and $$2$$.
Therefore, we are faced with a case wherein the product of the eigenvalues is zero, and as there is no linear term in $$y$$, we have the third reduced form. So, the polynomial has the form $$$q(x,y)=2x^2+2x+1=0$$$ Completing the squares, we see that $$$\displaystyle q (x, y) =2(x+\frac{1}{2})+\frac{3}{4}=0$$$ and therefore, doing the change of coordinates $$\displaystyle x' = x+\frac{1}{2}$$, we see that the equation can be written as $$$\displaystyle q(x,y)=2x^2+\frac{3}{4}$$$ Finally, dividing the equation by $$2$$, the canonical equation becomes $$$\displaystyle x^2+\frac{3}{8}=0$$$ This is a pair of combined parallel straight lines.
Give an affine classification for the following conic: $$$q (x, y) =3x^2+3y^2-6xy-6x+4y-8=0$$$ We start by calculating its main matrix $$A'$$: $$$A'=\begin{bmatrix} 3 & -3 \\ -3 & 3\end{bmatrix} \Longrightarrow det(A'-xI)=x^2-6x$$$
As soon as we calculated the main matrix of the conic and its associated characteristic polynomial, we are going to calculate its roots (the eigenvalues).
Therefore, the roots of the typical polynomial are: $$\lambda_1=6$$ and $$\lambda_2=0$$.
Therefore, our conic is going to be of the form $$$q(x,y)=6x^2-6x+4y-8=0$$$ Since we only have a quadratic term in $$x$$ and it has linear term also, we will need to complete the squares to eliminate this term.
Then, doing the change of variable $$\displaystyle x'=x-\frac{1}{2}$$, the conical becomes $$$\displaystyle q(x,y)=6x^2-4y-\frac{19}{2}=0$$$ As we only have quadratic term for $$x$$, this is a parabola.