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How to use it?
- Canonical form:
- 17x^2+12xy+8y^2+0x+0y-20=0
- x^2-2x+10y+21=0
- 9x^2+4xy+6y^2+40x+20y-50=0
- 2x^2-5xy+3y^2=0
- Plot:
- x^2=y
-
x^3(y-x)=x^3+1
- x^2/9-y^2/4=1
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x^6+3*x^4*y^2-3*x^4+3*x^2*y^4-x^2*y^3-6*x^2*y^2+3*x^2+y^6-3*y^4+3*y^2-1=0
-
Identical expressions
- two x^ two -5xy+3y^2= zero
- 2x squared minus 5xy plus 3y squared equally 0
- two x to the power of two minus 5xy plus 3y squared equally zero
- 2x2-5xy+3y2=0
- 2x²-5xy+3y²=0
- 2x to the power of 2-5xy+3y to the power of 2=0
- 2x^2-5xy+3y^2=O
-
Similar expressions
- 2x^2+5xy+3y^2=0
- 2x^2-5xy-3y^2=0
2x^2-5xy+3y^2=0 canonical form
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2 2 2*x + 3*y - 5*x*y = 0
$$2 x^{2} - 5 x y + 3 y^{2} = 0$$
2*x^2 - 5*x*y + 3*y^2 = 0
Detail solution
Given line equation of 2-order:
$$2 x^{2} - 5 x y + 3 y^{2} = 0$$
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
$$a_{11} = 2$$
$$a_{12} = - \frac{5}{2}$$
$$a_{13} = 0$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}2 & - \frac{5}{2}\\- \frac{5}{2} & 3\end{matrix}\right|$$
$$\Delta = - \frac{1}{4}$$
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
$$2 x_{0} - \frac{5 y_{0}}{2} = 0$$
$$- \frac{5 x_{0}}{2} + 3 y_{0} = 0$$
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
$$2 x'^{2} - 5 x' y' + 3 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{1}{5}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{1}{5} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{5 \sqrt{26}}{26}$$
$$\cos{\left(2 \phi \right)} = \frac{\sqrt{26}}{26}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}} + \tilde y \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}}$$
then the equation turns from
$$2 x'^{2} - 5 x' y' + 3 y'^{2} = 0$$
to
$$3 \left(\tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}} + \tilde y \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}}\right)^{2} - 5 \left(\tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}} + \tilde y \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}}\right) + 2 \left(\tilde x \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}}\right)^{2} = 0$$
simplify
$$- 5 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}} \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}} - \frac{\sqrt{26} \tilde x^{2}}{52} + \frac{5 \tilde x^{2}}{2} - \frac{5 \sqrt{26} \tilde x \tilde y}{26} + 2 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}} \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}} + \frac{\sqrt{26} \tilde y^{2}}{52} + 5 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}} \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}} + \frac{5 \tilde y^{2}}{2} = 0$$
$$- \frac{\sqrt{26} \tilde x^{2}}{2} + \frac{5 \tilde x^{2}}{2} + \frac{5 \tilde y^{2}}{2} + \frac{\sqrt{26} \tilde y^{2}}{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{- \frac{5}{2} + \frac{\sqrt{26}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\frac{5}{2} + \frac{\sqrt{26}}{2}}}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}}, \ \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}}\right)$$
$$2 x^{2} - 5 x y + 3 y^{2} = 0$$
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
$$a_{11} = 2$$
$$a_{12} = - \frac{5}{2}$$
$$a_{13} = 0$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}2 & - \frac{5}{2}\\- \frac{5}{2} & 3\end{matrix}\right|$$
$$\Delta = - \frac{1}{4}$$
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
$$2 x_{0} - \frac{5 y_{0}}{2} = 0$$
$$- \frac{5 x_{0}}{2} + 3 y_{0} = 0$$
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
$$2 x'^{2} - 5 x' y' + 3 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{1}{5}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{1}{5} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{5 \sqrt{26}}{26}$$
$$\cos{\left(2 \phi \right)} = \frac{\sqrt{26}}{26}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}} + \tilde y \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}}$$
then the equation turns from
$$2 x'^{2} - 5 x' y' + 3 y'^{2} = 0$$
to
$$3 \left(\tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}} + \tilde y \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}}\right)^{2} - 5 \left(\tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}} + \tilde y \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}}\right) + 2 \left(\tilde x \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}}\right)^{2} = 0$$
simplify
$$- 5 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}} \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}} - \frac{\sqrt{26} \tilde x^{2}}{52} + \frac{5 \tilde x^{2}}{2} - \frac{5 \sqrt{26} \tilde x \tilde y}{26} + 2 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}} \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}} + \frac{\sqrt{26} \tilde y^{2}}{52} + 5 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}} \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}} + \frac{5 \tilde y^{2}}{2} = 0$$
$$- \frac{\sqrt{26} \tilde x^{2}}{2} + \frac{5 \tilde x^{2}}{2} + \frac{5 \tilde y^{2}}{2} + \frac{\sqrt{26} \tilde y^{2}}{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{- \frac{5}{2} + \frac{\sqrt{26}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\frac{5}{2} + \frac{\sqrt{26}}{2}}}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}}, \ \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}}\right)$$
Invariants method
Given line equation of 2-order:
$$2 x^{2} - 5 x y + 3 y^{2} = 0$$
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
$$a_{11} = 2$$
$$a_{12} = - \frac{5}{2}$$
$$a_{13} = 0$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 5$$
$$I_{3} = \left|\begin{matrix}2 & - \frac{5}{2} & 0\\- \frac{5}{2} & 3 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & - \frac{5}{2}\\- \frac{5}{2} & 3 - \lambda\end{matrix}\right|$$
$$I_{1} = 5$$
$$I_{2} = - \frac{1}{4}$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda - \frac{1}{4}$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} < 0$$
then by line type:
this equation is of type : degenerate hyperbole
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 5 \lambda - \frac{1}{4} = 0$$
$$\lambda_{1} = \frac{5}{2} - \frac{\sqrt{26}}{2}$$
$$\lambda_{2} = \frac{5}{2} + \frac{\sqrt{26}}{2}$$
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
$$\tilde x^{2} \left(\frac{5}{2} - \frac{\sqrt{26}}{2}\right) + \tilde y^{2} \left(\frac{5}{2} + \frac{\sqrt{26}}{2}\right) = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{- \frac{5}{2} + \frac{\sqrt{26}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\frac{5}{2} + \frac{\sqrt{26}}{2}}}\right)^{2}} = 0$$
- reduced to canonical form
$$2 x^{2} - 5 x y + 3 y^{2} = 0$$
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
$$a_{11} = 2$$
$$a_{12} = - \frac{5}{2}$$
$$a_{13} = 0$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 5$$
| 2 -5/2|
I2 = | |
|-5/2 3 |$$I_{3} = \left|\begin{matrix}2 & - \frac{5}{2} & 0\\- \frac{5}{2} & 3 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & - \frac{5}{2}\\- \frac{5}{2} & 3 - \lambda\end{matrix}\right|$$
|2 0| |3 0|
K2 = | | + | |
|0 0| |0 0|$$I_{1} = 5$$
$$I_{2} = - \frac{1}{4}$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda - \frac{1}{4}$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} < 0$$
then by line type:
this equation is of type : degenerate hyperbole
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 5 \lambda - \frac{1}{4} = 0$$
$$\lambda_{1} = \frac{5}{2} - \frac{\sqrt{26}}{2}$$
$$\lambda_{2} = \frac{5}{2} + \frac{\sqrt{26}}{2}$$
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
$$\tilde x^{2} \left(\frac{5}{2} - \frac{\sqrt{26}}{2}\right) + \tilde y^{2} \left(\frac{5}{2} + \frac{\sqrt{26}}{2}\right) = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{- \frac{5}{2} + \frac{\sqrt{26}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\frac{5}{2} + \frac{\sqrt{26}}{2}}}\right)^{2}} = 0$$
- reduced to canonical form