Mister Exam
x^2+y^2+5x+2y+4=0 canonical form
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2 2 4 + x + y + 2*y + 5*x = 0
$$x^{2} + 5 x + y^{2} + 2 y + 4 = 0$$
x^2 + 5*x + y^2 + 2*y + 4 = 0
Detail solution
Given line equation of 2-order:
$$x^{2} + 5 x + y^{2} + 2 y + 4 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = \frac{5}{2}$$
$$a_{22} = 1$$
$$a_{23} = 1$$
$$a_{33} = 4$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & 1\end{matrix}\right|$$
$$\Delta = 1$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
substitute coefficients
$$x_{0} + \frac{5}{2} = 0$$
$$y_{0} + 1 = 0$$
then
$$x_{0} = - \frac{5}{2}$$
$$y_{0} = -1$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = \frac{5 x_{0}}{2} + y_{0} + 4$$
$$a'_{33} = - \frac{13}{4}$$
then equation turns into
$$x'^{2} + y'^{2} - \frac{13}{4} = 0$$
Given equation is circle
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{2}{13} \sqrt{13}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{2}{13} \sqrt{13}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$x^{2} + 5 x + y^{2} + 2 y + 4 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = \frac{5}{2}$$
$$a_{22} = 1$$
$$a_{23} = 1$$
$$a_{33} = 4$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & 1\end{matrix}\right|$$
$$\Delta = 1$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
substitute coefficients
$$x_{0} + \frac{5}{2} = 0$$
$$y_{0} + 1 = 0$$
then
$$x_{0} = - \frac{5}{2}$$
$$y_{0} = -1$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = \frac{5 x_{0}}{2} + y_{0} + 4$$
$$a'_{33} = - \frac{13}{4}$$
then equation turns into
$$x'^{2} + y'^{2} - \frac{13}{4} = 0$$
Given equation is circle
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{2}{13} \sqrt{13}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{2}{13} \sqrt{13}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(-5/2, -1)
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
Invariants method
Given line equation of 2-order:
$$x^{2} + 5 x + y^{2} + 2 y + 4 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = \frac{5}{2}$$
$$a_{22} = 1$$
$$a_{23} = 1$$
$$a_{33} = 4$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
substitute coefficients
$$I_{1} = 2$$
$$I_{3} = \left|\begin{matrix}1 & 0 & \frac{5}{2}\\0 & 1 & 1\\\frac{5}{2} & 1 & 4\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 2$$
$$I_{2} = 1$$
$$I_{3} = - \frac{13}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 2 \lambda + 1$$
$$K_{2} = \frac{3}{4}$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} < 0$$
then by line type:
this equation is of type : circle
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 2 \lambda + 1 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = 1$$
then the canonical form of the equation will be
$$\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0$$
or
$$\tilde x^{2} + \tilde y^{2} - \frac{13}{4} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{2}{13} \sqrt{13}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{2}{13} \sqrt{13}}\right)^{2}} = 1$$
- reduced to canonical form
$$x^{2} + 5 x + y^{2} + 2 y + 4 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = \frac{5}{2}$$
$$a_{22} = 1$$
$$a_{23} = 1$$
$$a_{33} = 4$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
|a11 a12|
I2 = | |
|a12 a22|$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
|a11 a13| |a22 a23|
K2 = | | + | |
|a13 a33| |a23 a33|substitute coefficients
$$I_{1} = 2$$
|1 0|
I2 = | |
|0 1|$$I_{3} = \left|\begin{matrix}1 & 0 & \frac{5}{2}\\0 & 1 & 1\\\frac{5}{2} & 1 & 4\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|$$
| 1 5/2| |1 1|
K2 = | | + | |
|5/2 4 | |1 4|$$I_{1} = 2$$
$$I_{2} = 1$$
$$I_{3} = - \frac{13}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 2 \lambda + 1$$
$$K_{2} = \frac{3}{4}$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} < 0$$
then by line type:
this equation is of type : circle
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 2 \lambda + 1 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = 1$$
then the canonical form of the equation will be
$$\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0$$
or
$$\tilde x^{2} + \tilde y^{2} - \frac{13}{4} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{2}{13} \sqrt{13}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{2}{13} \sqrt{13}}\right)^{2}} = 1$$
- reduced to canonical form