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How to use it?
- Canonical form:
- 2y^2-x=0
- 9x^2+9y^2-6x-54y=46=0
-
ax
-
3x^2-6x+4y^2+2y-1=0
- Plot:
- x^2+(y-x^(2/3))^2=1
- x^2+y^2=1
- z=(x^2+y)*sqrt(e^y)
- y*(x^3-1)=x^4
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Identical expressions
- 9x^ two +9y^ two -6x-54y= forty-six = zero
- 9x squared plus 9y squared minus 6x minus 54y equally 46 equally 0
- 9x to the power of two plus 9y to the power of two minus 6x minus 54y equally forty minus six equally zero
- 9x2+9y2-6x-54y=46=0
- 9x²+9y²-6x-54y=46=0
- 9x to the power of 2+9y to the power of 2-6x-54y=46=0
- 9x^2+9y^2-6x-54y=46=O
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Similar expressions
- 9x^2+9y^2-6x+54y=46=0
- 9x^2-9y^2-6x-54y=46=0
- 9x^2+9y^2+6x-54y=46=0
9x^2+9y^2-6x-54y=46=0 canonical form
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The solution
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2 2 -46 - 54*y - 6*x + 9*x + 9*y = 0
$$9 x^{2} - 6 x + 9 y^{2} - 54 y - 46 = 0$$
9*x^2 - 6*x + 9*y^2 - 54*y - 46 = 0
Detail solution
Given line equation of 2-order:
$$9 x^{2} - 6 x + 9 y^{2} - 54 y - 46 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -3$$
$$a_{22} = 9$$
$$a_{23} = -27$$
$$a_{33} = -46$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 0\\0 & 9\end{matrix}\right|$$
$$\Delta = 81$$
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
$$9 x_{0} - 3 = 0$$
$$9 y_{0} - 27 = 0$$
then
$$x_{0} = \frac{1}{3}$$
$$y_{0} = 3$$
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
$$a'_{33} = - 3 x_{0} - 27 y_{0} - 46$$
$$a'_{33} = -128$$
then equation turns into
$$9 x'^{2} + 9 y'^{2} - 128 = 0$$
Given equation is circle
$$\frac{\tilde x^{2}}{\left(\frac{1}{3 \frac{\sqrt{2}}{16}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{3 \frac{\sqrt{2}}{16}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$9 x^{2} - 6 x + 9 y^{2} - 54 y - 46 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -3$$
$$a_{22} = 9$$
$$a_{23} = -27$$
$$a_{33} = -46$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 0\\0 & 9\end{matrix}\right|$$
$$\Delta = 81$$
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
$$9 x_{0} - 3 = 0$$
$$9 y_{0} - 27 = 0$$
then
$$x_{0} = \frac{1}{3}$$
$$y_{0} = 3$$
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
$$a'_{33} = - 3 x_{0} - 27 y_{0} - 46$$
$$a'_{33} = -128$$
then equation turns into
$$9 x'^{2} + 9 y'^{2} - 128 = 0$$
Given equation is circle
$$\frac{\tilde x^{2}}{\left(\frac{1}{3 \frac{\sqrt{2}}{16}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{3 \frac{\sqrt{2}}{16}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(1/3, 3)
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:
$$9 x^{2} - 6 x + 9 y^{2} - 54 y - 46 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -3$$
$$a_{22} = 9$$
$$a_{23} = -27$$
$$a_{33} = -46$$
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
$$I_{1} = 18$$
$$I_{3} = \left|\begin{matrix}9 & 0 & -3\\0 & 9 & -27\\-3 & -27 & -46\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 0\\0 & 9 - \lambda\end{matrix}\right|$$
$$I_{1} = 18$$
$$I_{2} = 81$$
$$I_{3} = -10368$$
$$I{\left(\lambda \right)} = \lambda^{2} - 18 \lambda + 81$$
$$K_{2} = -1566$$
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
$$\lambda^{2} - 18 \lambda + 81 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 9$$
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
$$9 \tilde x^{2} + 9 \tilde y^{2} - 128 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{3 \frac{\sqrt{2}}{16}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{3 \frac{\sqrt{2}}{16}}\right)^{2}} = 1$$
- reduced to canonical form
$$9 x^{2} - 6 x + 9 y^{2} - 54 y - 46 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -3$$
$$a_{22} = 9$$
$$a_{23} = -27$$
$$a_{33} = -46$$
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
$$I_{1} = 18$$
|9 0|
I2 = | |
|0 9|$$I_{3} = \left|\begin{matrix}9 & 0 & -3\\0 & 9 & -27\\-3 & -27 & -46\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 0\\0 & 9 - \lambda\end{matrix}\right|$$
|9 -3 | | 9 -27|
K2 = | | + | |
|-3 -46| |-27 -46|$$I_{1} = 18$$
$$I_{2} = 81$$
$$I_{3} = -10368$$
$$I{\left(\lambda \right)} = \lambda^{2} - 18 \lambda + 81$$
$$K_{2} = -1566$$
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
$$\lambda^{2} - 18 \lambda + 81 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 9$$
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
$$9 \tilde x^{2} + 9 \tilde y^{2} - 128 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{3 \frac{\sqrt{2}}{16}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{3 \frac{\sqrt{2}}{16}}\right)^{2}} = 1$$
- reduced to canonical form