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 $$\sum _{i,j=1}^2{a}_{i,j}{x}_i{x}_j+2\sum _{k=1}^n{b}_k{x}_k+c=0,$$ 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:

$$1){\lambda }_1{x}^2+{\lambda }_2{y}^2+c=0({\lambda }_1{\lambda }_2\ne 0);$$

$$2){\lambda }_1{x}^2+by=0({\lambda }_1\ne 0);$$

$$3){\lambda }_1{x}^2+c=0({\lambda }_1\ne 0).$$

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:

$$\left(\begin{array}{cc}9& -2\\ -2& 6\end{array}\right).$$

Let's find its eigenvalues:

$$A-\lambda E=\left(\begin{array}{cc}9& -2\\ -2& 6\end{array}\right)-\lambda \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}9-\lambda & -2\\ -2& 6-\lambda \end{array}\right).$$

$$det(A-\lambda E)=\left|\begin{array}{cc}9-\lambda & -2\\ -2& 6-\lambda \end{array}\right|=(9-\lambda )(6-\lambda )-(-2)\cdot (-2)=$$ $$={\lambda }^2-15\lambda +40=0.$$

$D={15}^2-4\cdot 40=225-200=25;$

${\lambda }_1=\frac{15+5}{2}=10;{\lambda }_2=\frac{15-5}{2}=5.$

Next, we find the eigenvectors:

Eigenvector for the eigenvalue${\lambda }_1=10$ Let us find it from the system $$(A-\lambda E)X=0,X\ne 0,\Rightarrow (A-10E)X=0,X\ne 0$$

$$(A-10E)X=\left(\begin{array}{cc}9-10& -2\\ -2& 6-10\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{c}-x_1-2x_2\\ -2x_1-4x_2\end{array}\right)=0.$$

We'll solve the homogeneous system of equations:

$$\{\begin{array}{l}-x_1-2x_2=0\\ -2x_1-4x_2=0\end{array}$$

We solve the homogeneous system of equations:

Compute the rank of the coefficient matrix $A=\left(\begin{array}{cc}-1& -2\\ -2& -4\end{array}\right)$ using the method of border minors:

Fix a non-zero minor of the second order $M_2=\left|\begin{array}{cc}-1& -2\\ -2& -4\end{array}\right|=4-4=0$.

Thus, the rank of the matrix $A$ is equal to one.

Choose the basic minor $M=\left|\begin{array}{c}-1\end{array}\right|=-1\ne 0.$ Then, assuming $x_2=c$, we obtain: $$-x_1-2c=0\Rightarrow x_1=-2c.$$

Thus, the general solution of the system is $X(c)=\left(\begin{array}{c}-2c\\ c\end{array}\right).$

From the general solution, we find the fundamental solution set: $E=X(1)=\left(\begin{array}{c}-2\\ 1\end{array}\right).$

The corresponding orthonormalized eigenvector is: $$e_1^{\prime }=(\frac{-2}{\sqrt{4+1}},\frac{1}{\sqrt{4+1}})=(\frac{-2}{\sqrt{5}},\frac{1}{\sqrt{5}}).$$

The eigenvector for the eigenvalue ${\lambda }_2=5$ is found from the system $$(A-\lambda E)X=0,X\ne 0,\Rightarrow (A-5E)X=0,X\ne 0$$

$$(A-10E)X=\left(\begin{array}{cc}9-5& -2\\ -2& 6-5\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{c}4x_1-2x_2\\ -2x_1+x_2\end{array}\right)=0.$$

Let's solve the homogeneous system of equations:

$$\{\begin{array}{l}4x_1-2x_2=0\\ -2x_1+x_2=0\end{array}$$

Let's compute the rank of the coefficient matrix $A=\left(\begin{array}{cc}4& -2\\ -2& 1\end{array}\right)$ using the method of bordering minors:

We fix a non-zero minor of order 2: $M_2=\left|\begin{array}{cc}4& -2\\ -2& 1\end{array}\right|=4-4=0$.

Thus, the rank of matrix $A$ is one.

We choose the minor $M=\left|\begin{array}{c}4\end{array}\right|=4\ne 0$ 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 $X(c)=\left(\begin{array}{c}c/2\\ c\end{array}\right)$.

From the general solution, we obtain the fundamental solution set: $E=X(1)=\left(\begin{array}{c}1/2\\ 1\end{array}\right)$.

The corresponding orthonormalized eigenvector is: $$e_1^{\prime }=(\frac{1}{\sqrt{4+1}},\frac{2}{\sqrt{4+1}})=(\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}).$$

Thus, we have found the vectors

$$e_1^{\prime }=(\frac{-2}{\sqrt{5}},\frac{1}{\sqrt{5}});$$

$$e_2^{\prime }=(\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}).$$

Performing the transformation

$$x=\frac{-2}{\sqrt{5}}{x}^{\prime }+\frac{1}{\sqrt{5}}{y}^{\prime },y=\frac{1}{\sqrt{5}}{x}^{\prime }+\frac{2}{\sqrt{5}}{y}^{\prime },$$ We obtain:

$$9(\frac{-2}{\sqrt{5}}{x}^{\prime }+\frac{1}{\sqrt{5}}{y}^{\prime }{)}^2-4(\frac{-2}{\sqrt{5}}{x}^{\prime }+\frac{1}{\sqrt{5}}{y}^{\prime })(\frac{1}{\sqrt{5}}{x}^{\prime }+\frac{2}{\sqrt{5}}{y}^{\prime })+$$ $$+6(\frac{1}{\sqrt{5}}{x}^{\prime }+\frac{2}{\sqrt{5}}{y}^{\prime }{)}^2+16(\frac{-2}{\sqrt{5}}{x}^{\prime }+\frac{1}{\sqrt{5}}{y}^{\prime })-8(\frac{1}{\sqrt{5}}{x}^{\prime }+\frac{2}{\sqrt{5}}{y}^{\prime })-2=$$ $$={x^{\prime }}^2(\frac{36}{5}+\frac{8}{5}+\frac{6}{5})+{y^{\prime }}^2(\frac{9}{5}-\frac{8}{5}+\frac{24}{5})+$$ $$+x^{\prime }{y}^{\prime }(-\frac{36}{5}-\frac{4}{5}+\frac{16}{5}+\frac{24}{5})+x^{\prime }\left(\frac{-40}{\sqrt{5}}\right)-2=$$ $$=10{x^{\prime }}^2+5{y^{\prime }}^2-\frac{40}{\sqrt{5}}{x}^{\prime }-2=0$$

Let's complete the square with respect to the variable $x^{\prime }$.

$$10{x^{\prime }}^2-\frac{40}{\sqrt{5}}{x}^{\prime }=10({x^{\prime }}^2-\frac{4}{\sqrt{5}}+\frac{4}{5})-8=10{(x^{\prime }-\frac{2}{\sqrt{5}})}^2-8.$$

We perform a variable substitution:

$$x^{″}=x^{\prime }-\frac{2}{\sqrt{5}},y^{″}=y^{\prime }$$(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:

$$x=\frac{-2}{\sqrt{5}}(x^{″}+\frac{2}{\sqrt{5}})+\frac{1}{\sqrt{5}}{y}^{″}=\frac{-2}{\sqrt{5}}{x}^{″}+\frac{1}{\sqrt{5}}{y}^{″}-\frac{4}{5},$$

$$y=\frac{1}{\sqrt{5}}(x^{″}+\frac{2}{\sqrt{5}})+\frac{2}{\sqrt{5}}{y}^{″}=\frac{1}{\sqrt{5}}{x}^{″}+\frac{2}{\sqrt{5}}{y}^{″}+\frac{2}{5},$$ and the canonical coordinate system $(O^{\prime },e_1^{\prime },e_2^{\prime }),$ где $O=(-\frac{4}{5},\frac{2}{5}),$

$e_1^{\prime }=-\frac{2}{\sqrt{5}}i+\frac{1}{\sqrt{5}}j;$ $e_2^{\prime }=\frac{1}{\sqrt{5}}i+\frac{2}{\sqrt{5}}j.$

Answer: The ellipse equation is${x^{″}}^2+\frac{{y^{″}}^2}{2}=1.$ $O=(-\frac{4}{5},\frac{2}{5}),$

$e_1^{\prime }=(\frac{-2}{\sqrt{5}},\frac{1}{\sqrt{5}});$ $e_2^{\prime }=(\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}).$

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 ${y^{\prime }}^2=4\sqrt{2}x.$ $O^{\prime }=(2,1),$

$e_1^{\prime }=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}});$ $e_2^{\prime }=(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}).$

2.$5x^2+12xy-22x-12y-19=0.$

Answer: Hyperbola $\frac{{x^{\prime }}^2}{4}-\frac{{y^{\prime }}^2}{9}=1.$ $O^{\prime }=(1,1),$

$e_1^{\prime }=(\frac{3}{\sqrt{13}},\frac{2}{\sqrt{13}});$ $e_2^{\prime }=(\frac{-2}{\sqrt{13}},\frac{3}{\sqrt{13}}).$

Tags: General equation, curve of the second order