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How to use it?
- Canonical form:
- 3x^2+0xy-6y^2-12x-108y-492=0
- sqrt(16-y^2)=sqrt(16-a^2*x^2)
- 0,6666*sqrt(x^2-9)
- 3x^2+5y^2+3z^2+2xy+2yz+2xz
- Plot:
-
xy=4
-
xy=6
-
4x^2+4y^2=25
- xyz
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Identical expressions
- 3x^ two + zero xy-6y^ two -12x-108y- four hundred and ninety-two =0
- 3x squared plus 0xy minus 6y squared minus 12x minus 108y minus 492 equally 0
- 3x to the power of two plus zero xy minus 6y to the power of two minus 12x minus 108y minus four hundred and ninety minus two equally 0
- 3x2+0xy-6y2-12x-108y-492=0
- 3x²+0xy-6y²-12x-108y-492=0
- 3x to the power of 2+0xy-6y to the power of 2-12x-108y-492=0
- 3x^2+0xy-6y^2-12x-108y-492=O
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Similar expressions
- 3x^2+0xy-6y^2-12x+108y-492=0
- 3x^2-0xy-6y^2-12x-108y-492=0
- 3x^2+0xy-6y^2+12x-108y-492=0
- 3x^2+0xy-6y^2-12x-108y+492=0
- 3x^2+0xy+6y^2-12x-108y-492=0
3x^2+0xy-6y^2-12x-108y-492=0 canonical form
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The solution
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2 2 -492 - 108*y - 12*x - 6*y + 3*x = 0
$$3 x^{2} - 12 x - 6 y^{2} - 108 y - 492 = 0$$
3*x^2 - 12*x - 6*y^2 - 108*y - 492 = 0
Detail solution
Given line equation of 2-order:
$$3 x^{2} - 12 x - 6 y^{2} - 108 y - 492 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -6$$
$$a_{22} = -6$$
$$a_{23} = -54$$
$$a_{33} = -492$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 0\\0 & -6\end{matrix}\right|$$
$$\Delta = -18$$
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
$$3 x_{0} - 6 = 0$$
$$- 6 y_{0} - 54 = 0$$
then
$$x_{0} = 2$$
$$y_{0} = -9$$
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
$$a'_{33} = - 6 x_{0} - 54 y_{0} - 492$$
$$a'_{33} = -18$$
then equation turns into
$$3 x'^{2} - 6 y'^{2} - 18 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{6} - \frac{\tilde y^{2}}{3} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$3 x^{2} - 12 x - 6 y^{2} - 108 y - 492 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -6$$
$$a_{22} = -6$$
$$a_{23} = -54$$
$$a_{33} = -492$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 0\\0 & -6\end{matrix}\right|$$
$$\Delta = -18$$
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
$$3 x_{0} - 6 = 0$$
$$- 6 y_{0} - 54 = 0$$
then
$$x_{0} = 2$$
$$y_{0} = -9$$
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
$$a'_{33} = - 6 x_{0} - 54 y_{0} - 492$$
$$a'_{33} = -18$$
then equation turns into
$$3 x'^{2} - 6 y'^{2} - 18 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{6} - \frac{\tilde y^{2}}{3} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(2, -9)
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:
$$3 x^{2} - 12 x - 6 y^{2} - 108 y - 492 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -6$$
$$a_{22} = -6$$
$$a_{23} = -54$$
$$a_{33} = -492$$
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
$$I_{1} = -3$$
$$I_{3} = \left|\begin{matrix}3 & 0 & -6\\0 & -6 & -54\\-6 & -54 & -492\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0\\0 & - \lambda - 6\end{matrix}\right|$$
$$I_{1} = -3$$
$$I_{2} = -18$$
$$I_{3} = 324$$
$$I{\left(\lambda \right)} = \lambda^{2} + 3 \lambda - 18$$
$$K_{2} = -1476$$
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
$$\lambda^{2} + 3 \lambda - 18 = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = -6$$
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
$$3 \tilde x^{2} - 6 \tilde y^{2} - 18 = 0$$
$$\frac{\tilde x^{2}}{6} - \frac{\tilde y^{2}}{3} = 1$$
- reduced to canonical form
$$3 x^{2} - 12 x - 6 y^{2} - 108 y - 492 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -6$$
$$a_{22} = -6$$
$$a_{23} = -54$$
$$a_{33} = -492$$
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
$$I_{1} = -3$$
|3 0 |
I2 = | |
|0 -6|$$I_{3} = \left|\begin{matrix}3 & 0 & -6\\0 & -6 & -54\\-6 & -54 & -492\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0\\0 & - \lambda - 6\end{matrix}\right|$$
|3 -6 | |-6 -54 |
K2 = | | + | |
|-6 -492| |-54 -492|$$I_{1} = -3$$
$$I_{2} = -18$$
$$I_{3} = 324$$
$$I{\left(\lambda \right)} = \lambda^{2} + 3 \lambda - 18$$
$$K_{2} = -1476$$
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
$$\lambda^{2} + 3 \lambda - 18 = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = -6$$
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
$$3 \tilde x^{2} - 6 \tilde y^{2} - 18 = 0$$
$$\frac{\tilde x^{2}}{6} - \frac{\tilde y^{2}}{3} = 1$$
- reduced to canonical form