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How to use it?
- Canonical form:
- 2x^2+y^2-z^2-6y=2z+80
- x^2+y^2–z^2–4x–6y+4z+10=0
-
7y^2+3x-28y
- 4x^2-12xy+9y^2+z^2+4xz-6yz+4x-6y+2z-5=0
- Plot:
- x^2+(y-3*sqrt(x)^2)^2=1
-
4*y^2=x^5+4*x^2*y
- z=4*(x-y)-x^2-y^2
- (x^2+y^2-1)^3=e^x^2*y^3
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Identical expressions
- x^ two + six *x*y+y^ two + six *x+ two *y- three = zero
- x squared plus 6 multiply by x multiply by y plus y squared plus 6 multiply by x plus 2 multiply by y minus 3 equally 0
- x to the power of two plus six multiply by x multiply by y plus y to the power of two plus six multiply by x plus two multiply by y minus three equally zero
- x2+6*x*y+y2+6*x+2*y-3=0
- x²+6*x*y+y²+6*x+2*y-3=0
- x to the power of 2+6*x*y+y to the power of 2+6*x+2*y-3=0
- x^2+6xy+y^2+6x+2y-3=0
- x2+6xy+y2+6x+2y-3=0
- x^2+6*x*y+y^2+6*x+2*y-3=O
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Similar expressions
- x^2+6*x*y-y^2+6*x+2*y-3=0
- x^2+6*x*y+y^2-6*x+2*y-3=0
- x^2-6*x*y+y^2+6*x+2*y-3=0
- x^2+6*x*y+y^2+6*x-2*y-3=0
- x^2+6*x*y+y^2+6*x+2*y+3=0
x^2+6*x*y+y^2+6*x+2*y-3=0 canonical form
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2 2 -3 + x + y + 2*y + 6*x + 6*x*y = 0
$$x^{2} + 6 x y + 6 x + y^{2} + 2 y - 3 = 0$$
x^2 + 6*x*y + 6*x + y^2 + 2*y - 3 = 0
Detail solution
Given line equation of 2-order:
$$x^{2} + 6 x y + 6 x + y^{2} + 2 y - 3 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 3$$
$$a_{13} = 3$$
$$a_{22} = 1$$
$$a_{23} = 1$$
$$a_{33} = -3$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 3\\3 & 1\end{matrix}\right|$$
$$\Delta = -8$$
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
$$x_{0} + 3 y_{0} + 3 = 0$$
$$3 x_{0} + y_{0} + 1 = 0$$
then
$$x_{0} = 0$$
$$y_{0} = -1$$
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
$$a'_{33} = 3 x_{0} + y_{0} - 3$$
$$a'_{33} = -4$$
then equation turns into
$$x'^{2} + 6 x' y' + y'^{2} - 4 = 0$$
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
$$\cot{\left(2 \phi \right)} = 0$$
then
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
substitute coefficients
$$x' = \frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}$$
$$y' = \frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}$$
then the equation turns from
$$x'^{2} + 6 x' y' + y'^{2} - 4 = 0$$
to
$$\left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right)^{2} + 6 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) + \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 4 = 0$$
simplify
$$4 \tilde x^{2} - 2 \tilde y^{2} - 4 = 0$$
$$- 4 \tilde x^{2} + 2 \tilde y^{2} + 4 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{1} - \frac{\tilde y^{2}}{2} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$x^{2} + 6 x y + 6 x + y^{2} + 2 y - 3 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 3$$
$$a_{13} = 3$$
$$a_{22} = 1$$
$$a_{23} = 1$$
$$a_{33} = -3$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 3\\3 & 1\end{matrix}\right|$$
$$\Delta = -8$$
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
$$x_{0} + 3 y_{0} + 3 = 0$$
$$3 x_{0} + y_{0} + 1 = 0$$
then
$$x_{0} = 0$$
$$y_{0} = -1$$
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
$$a'_{33} = 3 x_{0} + y_{0} - 3$$
$$a'_{33} = -4$$
then equation turns into
$$x'^{2} + 6 x' y' + y'^{2} - 4 = 0$$
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
$$\cot{\left(2 \phi \right)} = 0$$
then
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
substitute coefficients
$$x' = \frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}$$
$$y' = \frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}$$
then the equation turns from
$$x'^{2} + 6 x' y' + y'^{2} - 4 = 0$$
to
$$\left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right)^{2} + 6 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) + \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 4 = 0$$
simplify
$$4 \tilde x^{2} - 2 \tilde y^{2} - 4 = 0$$
$$- 4 \tilde x^{2} + 2 \tilde y^{2} + 4 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{1} - \frac{\tilde y^{2}}{2} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, -1)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
Invariants method
Given line equation of 2-order:
$$x^{2} + 6 x y + 6 x + y^{2} + 2 y - 3 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 3$$
$$a_{13} = 3$$
$$a_{22} = 1$$
$$a_{23} = 1$$
$$a_{33} = -3$$
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
$$I_{1} = 2$$
$$I_{3} = \left|\begin{matrix}1 & 3 & 3\\3 & 1 & 1\\3 & 1 & -3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 3\\3 & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 2$$
$$I_{2} = -8$$
$$I_{3} = 32$$
$$I{\left(\lambda \right)} = \lambda^{2} - 2 \lambda - 8$$
$$K_{2} = -16$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 2 \lambda - 8 = 0$$
$$\lambda_{1} = 4$$
$$\lambda_{2} = -2$$
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
$$4 \tilde x^{2} - 2 \tilde y^{2} - 4 = 0$$
$$\frac{\tilde x^{2}}{1} - \frac{\tilde y^{2}}{2} = 1$$
- reduced to canonical form
$$x^{2} + 6 x y + 6 x + y^{2} + 2 y - 3 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 3$$
$$a_{13} = 3$$
$$a_{22} = 1$$
$$a_{23} = 1$$
$$a_{33} = -3$$
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 3|
I2 = | |
|3 1|$$I_{3} = \left|\begin{matrix}1 & 3 & 3\\3 & 1 & 1\\3 & 1 & -3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 3\\3 & 1 - \lambda\end{matrix}\right|$$
|1 3 | |1 1 |
K2 = | | + | |
|3 -3| |1 -3|$$I_{1} = 2$$
$$I_{2} = -8$$
$$I_{3} = 32$$
$$I{\left(\lambda \right)} = \lambda^{2} - 2 \lambda - 8$$
$$K_{2} = -16$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 2 \lambda - 8 = 0$$
$$\lambda_{1} = 4$$
$$\lambda_{2} = -2$$
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
$$4 \tilde x^{2} - 2 \tilde y^{2} - 4 = 0$$
$$\frac{\tilde x^{2}}{1} - \frac{\tilde y^{2}}{2} = 1$$
- reduced to canonical form