General equation of a curve of the second order. Canonical equation of a curve of the second order.
The set of points in the plane $R^2,$ satisfying the condition $$\sum\limits_{i,j=1}^2a_{i,j}x_ix_j+2\sum\limits_{k=1}^nb_kx_k+c=0,$$ s called a curve of the second order. The canonical equation of a curve of the second order can take one of the following forms:
$$1)\,\, \lambda_1x^2+\lambda_2y^2+c=0\,\,\, (\lambda_1\lambda_2\neq 0);$$
$$2)\,\, \lambda_1x^2+by=0\qquad(\lambda_1\neq 0);$$
$$3)\,\, \lambda_1x^2+c=0\qquad(\lambda_1\neq 0).$$
Example.
1. Write the canonical equation of the curve of the second order, determine its type, and find the canonical coordinate system.
$$9x^2-4xy+6y^2+16x-8y-2=0.$$
Solution.
The matrix of the quadratic part of the polynomial of the second degree is given by:
$$\begin{pmatrix}9&-2\\-2&6\end{pmatrix}.$$
Let's find its eigenvalues:
$$A-\lambda E=\begin{pmatrix}9&-2\\-2&6\end{pmatrix}-\lambda\begin{pmatrix}1&0\\0&1\end{pmatrix}=\begin{pmatrix}9-\lambda&-2\\-2&6-\lambda\end{pmatrix}.$$
$$det(A-\lambda E)=\begin{vmatrix}9-\lambda&-2\\-2&6-\lambda\end{vmatrix}=(9-\lambda)(6-\lambda)-(-2)\cdot(-2)=$$ $$=\lambda^2-15\lambda+40=0.$$
$D=15^2-4\cdot40=225-200=25;$
$\lambda_1=\frac{15+5}{2}=10;\qquad\quad \lambda_2=\frac{15-5}{2}=5.$
Next, we find the eigenvectors:
Eigenvector for the eigenvalue$\lambda_1=10$ Let us find it from the system $$(A-\lambda E)X=0, X\neq 0, \Rightarrow (A-10E)X=0, X\neq 0$$
$$(A-10E)X=\begin{pmatrix}9-10&-2\\-2&6-10\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}-x_1-2x_2\\-2x_1-4x_2\end{pmatrix}=0.$$
We'll solve the homogeneous system of equations:
$$\left\{\begin{array}{lcl}-x_1-2x_2=0\\ -2x_1-4x_2=0\end{array}\right.$$
We solve the homogeneous system of equations:
Compute the rank of the coefficient matrix $A=\begin{pmatrix}-1&-2\\-2&-4\end{pmatrix}$ using the method of border minors:
Fix a non-zero minor of the second order $M_2=\begin{vmatrix}-1&-2\\-2&-4\end{vmatrix}=4-4=0$.
Thus, the rank of the matrix $A$ is equal to one.
Choose the basic minor $M=\begin{vmatrix}-1\end{vmatrix}=-1\neq 0.$ Then, assuming $x_2=c$, we obtain: $$-x_1-2c=0\Rightarrow x_1=-2c.$$
Thus, the general solution of the system is $X(c)=\begin{pmatrix}-2c\\c\end{pmatrix}.$
From the general solution, we find the fundamental solution set: $E=X(1)=\begin{pmatrix}-2\\1\end{pmatrix}.$
The corresponding orthonormalized eigenvector is: $$e_1'=\left(\frac{-2}{\sqrt{4+1}},\frac{1}{\sqrt{4+1}}\right)=\left(\frac{-2}{\sqrt 5},\frac{1}{\sqrt 5}\right).$$
The eigenvector for the eigenvalue $\lambda_2=5$ is found from the system $$(A-\lambda E)X=0, X\neq 0, \Rightarrow (A-5E)X=0, X\neq 0$$
$$(A-10E)X=\begin{pmatrix}9-5&-2\\-2&6-5\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}4x_1-2x_2\\-2x_1+x_2\end{pmatrix}=0.$$
Let's solve the homogeneous system of equations:
$$\left\{\begin{array}{lcl}4x_1-2x_2=0\\ -2x_1+x_2=0\end{array}\right.$$
Let's compute the rank of the coefficient matrix $A=\begin{pmatrix}4&-2\\-2&1\end{pmatrix}$ using the method of bordering minors:
We fix a non-zero minor of order 2: $M_2=\begin{vmatrix}4&-2\\-2&1\end{vmatrix}=4-4=0$.
Thus, the rank of matrix $A$ is one.
We choose the minor $M=\begin{vmatrix}4\end{vmatrix}=4\neq 0$ as the basis minor. Then, assuming $x_2=c$, we obtain:
$$4x_1-2c=0\Rightarrow x_1=c/2$$
Thus, the general solution of the system is $X(c)=\begin{pmatrix}c/2\\c\end{pmatrix}$.
From the general solution, we obtain the fundamental solution set: $E=X(1)=\begin{pmatrix}1/2\\1\end{pmatrix}$.
The corresponding orthonormalized eigenvector is: $$e_1'=\left(\frac{1}{\sqrt{4+1}},\frac{2}{\sqrt{4+1}}\right)=\left(\frac{1}{\sqrt 5},\frac{2}{\sqrt 5}\right).$$
Thus, we have found the vectors
$$e_1'=\left(\frac{-2}{\sqrt 5},\frac{1}{\sqrt 5}\right);$$
$$e_2'=\left(\frac{1}{\sqrt 5},\frac{2}{\sqrt 5}\right).$$
Performing the transformation
$$x=\frac{-2}{\sqrt 5}x '+\frac{1}{\sqrt 5}y',\qquad\quad y=\frac{1}{\sqrt 5}x'+\frac{2}{\sqrt 5}y',$$ We obtain:
$$9(\frac{-2}{\sqrt 5}x '+\frac{1}{\sqrt 5}y')^2-4(\frac{-2}{\sqrt 5}x '+\frac{1}{\sqrt 5}y')(\frac{1}{\sqrt 5}x'+\frac{2}{\sqrt 5}y')+$$ $$+6(\frac{1}{\sqrt 5}x'+\frac{2}{\sqrt 5}y')^2+16(\frac{-2}{\sqrt 5}x '+\frac{1}{\sqrt 5}y')-8(\frac{1}{\sqrt 5}x'+\frac{2}{\sqrt 5}y')-2=$$ $$={x'}^2\left(\frac{36}{5}+\frac{8}{5}+\frac{6}{5}\right)+{y'}^2\left(\frac{9}{5}-\frac{8}{5}+\frac{24}{5}\right)+$$ $$+x'y'\left(-\frac{36}{5}-\frac{4}{5}+\frac{16}{5}+\frac{24}{5}\right)+x'\left(\frac{-40}{\sqrt 5}\right)-2=$$ $$=10{x'}^2+5{y'}^2-\frac{40}{\sqrt 5}x'-2=0$$
Let's complete the square with respect to the variable $x'$.
$$10{x'}^2-\frac{40}{\sqrt 5}x'=10\left({x'}^2-\frac{4}{\sqrt 5}+\frac{4}{5}\right)-8=10\left(x'-\frac{2}{\sqrt 5}\right)^2-8.$$
We perform a variable substitution:
$$x''=x'-\frac{2}{\sqrt 5}, \qquad\quad y''=y'$$(The variable substitution corresponds to a shift along the $x$-axis.) Thus, we obtain: $$10{x''}^2+5{y''}^2-10=0\Rightarrow {x''}^2+\frac{{y''}^2}{2}=1.$$ This is an equation of an ellipse.
The resulting coordinate transformation is given by:
$$x=\frac{-2}{\sqrt 5}\left(x ''+\frac{2}{\sqrt 5}\right)+\frac{1}{\sqrt 5}y''=\frac{-2}{\sqrt 5}x''+\frac{1}{\sqrt 5}y''-\frac{4}{5},$$
$$ y=\frac{1}{\sqrt 5}\left(x ''+\frac{2}{\sqrt 5}\right)+\frac{2}{\sqrt 5}y''=\frac{1}{\sqrt 5}x''+\frac{2}{\sqrt 5}y''+\frac{2}{5},$$ and the canonical coordinate system $(O', e_1', e_2'),$ где $O=\left(-\frac{4}{5}, \frac{2}{5}\right),$
$e_1'=-\frac{2}{\sqrt 5}i+\frac{1}{\sqrt 5}j;$ $e_2'=\frac{1}{\sqrt 5}i+\frac{2}{\sqrt 5}j.$
Answer: The ellipse equation is${x''}^2+\frac{{y''}^2}{2}=1.$ $O=\left(-\frac{4}{5}, \frac{2}{5}\right),$
$e_1'=\left(\frac{-2}{\sqrt 5},\frac{1}{\sqrt 5}\right);\,\,$ $e_2'=\left(\frac{1}{\sqrt 5},\frac{2}{\sqrt 5}\right).$
Homework:
Write the canonical equation of the second-order curve, determine its type, and find the canonical coordinate system.
1. $x^2-2xy+y^2-10x-6y+25=0.$
Answer: Parabola ${y'}^2=4\sqrt 2 x.$ $O'=\left(2, 1\right),$
$e_1'=\left(\frac{1}{\sqrt 2},\frac{1}{\sqrt 2}\right);\,\,$ $e_2'=\left(-\frac{1}{\sqrt 2},\frac{1}{\sqrt 2}\right).$
2.$5x^2+12xy-22x-12y-19=0.$
Answer: Hyperbola $ \frac{{x'}^2}{4}-\frac{{y'}^2}{9}=1.$ $O'=\left(1, 1\right),$
$e_1'=\left(\frac{3}{\sqrt {13}},\frac{2}{\sqrt {13}}\right);\,\,$ $e_2'=\left(\frac{-2}{\sqrt {13}},\frac{3}{\sqrt {13}}\right).$
