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How to use it?
- Canonical form:
- x^2+y^2+z^2+20y
- x^2-4x+9y^2+18y-4z^2+24z+6=0
- x^2+3xy+4y^2
- x²-z²=1
- Plot:
- |x+2/2-x|
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|3*y-x|+2=|y-4|
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|x-y+2|=1
- z=x^3+3*x^2-x*y^2+y^2
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Identical expressions
- x^ two +3xy+4y^ two
- x squared plus 3xy plus 4y squared
- x to the power of two plus 3xy plus 4y to the power of two
- x2+3xy+4y2
- x²+3xy+4y²
- x to the power of 2+3xy+4y to the power of 2
-
Similar expressions
- x^2-3xy+4y^2
- x^2+3xy-4y^2
x^2+3xy+4y^2 canonical form
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2 2 x + 4*y + 3*x*y = 0
$$x^{2} + 3 x y + 4 y^{2} = 0$$
x^2 + 3*x*y + 4*y^2 = 0
Detail solution
Given line equation of 2-order:
$$x^{2} + 3 x y + 4 y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = \frac{3}{2}$$
$$a_{13} = 0$$
$$a_{22} = 4$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & \frac{3}{2}\\\frac{3}{2} & 4\end{matrix}\right|$$
$$\Delta = \frac{7}{4}$$
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
$$x_{0} + \frac{3 y_{0}}{2} = 0$$
$$\frac{3 x_{0}}{2} + 4 y_{0} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = 0$$
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
$$a'_{33} = 0$$
$$a'_{33} = 0$$
then equation turns into
$$x'^{2} + 3 x' y' + 4 y'^{2} = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = -1$$
then
$$\phi = - \frac{\pi}{8}$$
$$\sin{\left(2 \phi \right)} = - \frac{\sqrt{2}}{2}$$
$$\cos{\left(2 \phi \right)} = \frac{\sqrt{2}}{2}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}$$
$$y' = - \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} + \tilde y \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}$$
then the equation turns from
$$x'^{2} + 3 x' y' + 4 y'^{2} = 0$$
to
$$4 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} + \tilde y \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}\right)^{2} + 3 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} + \tilde y \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}\right) + \left(\tilde x \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}\right)^{2} = 0$$
simplify
$$- \frac{3 \sqrt{2} \tilde x^{2}}{4} - 3 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + \frac{5 \tilde x^{2}}{2} - 6 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + \frac{3 \sqrt{2} \tilde x \tilde y}{2} + 3 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + \frac{3 \sqrt{2} \tilde y^{2}}{4} + \frac{5 \tilde y^{2}}{2} = 0$$
$$- \frac{3 \sqrt{2} \tilde x^{2}}{2} + \frac{5 \tilde x^{2}}{2} + \frac{3 \sqrt{2} \tilde y^{2}}{2} + \frac{5 \tilde y^{2}}{2} = 0$$
Given equation is degenerate ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{\frac{5}{2} - \frac{3 \sqrt{2}}{2}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\frac{3 \sqrt{2}}{2} + \frac{5}{2}}}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}\right)$$
$$\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}, \ \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}\right)$$
$$x^{2} + 3 x y + 4 y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = \frac{3}{2}$$
$$a_{13} = 0$$
$$a_{22} = 4$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & \frac{3}{2}\\\frac{3}{2} & 4\end{matrix}\right|$$
$$\Delta = \frac{7}{4}$$
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
$$x_{0} + \frac{3 y_{0}}{2} = 0$$
$$\frac{3 x_{0}}{2} + 4 y_{0} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = 0$$
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
$$a'_{33} = 0$$
$$a'_{33} = 0$$
then equation turns into
$$x'^{2} + 3 x' y' + 4 y'^{2} = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = -1$$
then
$$\phi = - \frac{\pi}{8}$$
$$\sin{\left(2 \phi \right)} = - \frac{\sqrt{2}}{2}$$
$$\cos{\left(2 \phi \right)} = \frac{\sqrt{2}}{2}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}$$
$$y' = - \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} + \tilde y \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}$$
then the equation turns from
$$x'^{2} + 3 x' y' + 4 y'^{2} = 0$$
to
$$4 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} + \tilde y \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}\right)^{2} + 3 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} + \tilde y \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}\right) + \left(\tilde x \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}\right)^{2} = 0$$
simplify
$$- \frac{3 \sqrt{2} \tilde x^{2}}{4} - 3 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + \frac{5 \tilde x^{2}}{2} - 6 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + \frac{3 \sqrt{2} \tilde x \tilde y}{2} + 3 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + \frac{3 \sqrt{2} \tilde y^{2}}{4} + \frac{5 \tilde y^{2}}{2} = 0$$
$$- \frac{3 \sqrt{2} \tilde x^{2}}{2} + \frac{5 \tilde x^{2}}{2} + \frac{3 \sqrt{2} \tilde y^{2}}{2} + \frac{5 \tilde y^{2}}{2} = 0$$
Given equation is degenerate ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{\frac{5}{2} - \frac{3 \sqrt{2}}{2}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\frac{3 \sqrt{2}}{2} + \frac{5}{2}}}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}\right)$$
$$\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}, \ \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}\right)$$
Invariants method
Given line equation of 2-order:
$$x^{2} + 3 x y + 4 y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = \frac{3}{2}$$
$$a_{13} = 0$$
$$a_{22} = 4$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 5$$
$$I_{3} = \left|\begin{matrix}1 & \frac{3}{2} & 0\\\frac{3}{2} & 4 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & \frac{3}{2}\\\frac{3}{2} & 4 - \lambda\end{matrix}\right|$$
$$I_{1} = 5$$
$$I_{2} = \frac{7}{4}$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda + \frac{7}{4}$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} > 0$$
then by line type:
this equation is of type : degenerate ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 5 \lambda + \frac{7}{4} = 0$$
$$\lambda_{1} = \frac{5}{2} - \frac{3 \sqrt{2}}{2}$$
$$\lambda_{2} = \frac{3 \sqrt{2}}{2} + \frac{5}{2}$$
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
$$\tilde x^{2} \left(\frac{5}{2} - \frac{3 \sqrt{2}}{2}\right) + \tilde y^{2} \left(\frac{3 \sqrt{2}}{2} + \frac{5}{2}\right) = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{\frac{5}{2} - \frac{3 \sqrt{2}}{2}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\frac{3 \sqrt{2}}{2} + \frac{5}{2}}}\right)^{2}} = 0$$
- reduced to canonical form
$$x^{2} + 3 x y + 4 y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = \frac{3}{2}$$
$$a_{13} = 0$$
$$a_{22} = 4$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 5$$
| 1 3/2|
I2 = | |
|3/2 4 |$$I_{3} = \left|\begin{matrix}1 & \frac{3}{2} & 0\\\frac{3}{2} & 4 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & \frac{3}{2}\\\frac{3}{2} & 4 - \lambda\end{matrix}\right|$$
|1 0| |4 0|
K2 = | | + | |
|0 0| |0 0|$$I_{1} = 5$$
$$I_{2} = \frac{7}{4}$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda + \frac{7}{4}$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} > 0$$
then by line type:
this equation is of type : degenerate ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 5 \lambda + \frac{7}{4} = 0$$
$$\lambda_{1} = \frac{5}{2} - \frac{3 \sqrt{2}}{2}$$
$$\lambda_{2} = \frac{3 \sqrt{2}}{2} + \frac{5}{2}$$
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
$$\tilde x^{2} \left(\frac{5}{2} - \frac{3 \sqrt{2}}{2}\right) + \tilde y^{2} \left(\frac{3 \sqrt{2}}{2} + \frac{5}{2}\right) = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{\frac{5}{2} - \frac{3 \sqrt{2}}{2}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\frac{3 \sqrt{2}}{2} + \frac{5}{2}}}\right)^{2}} = 0$$
- reduced to canonical form