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How to use it?
- Canonical form:
- 6*x-7*y-4*z-2
- 3x^2-36x+3y^2-56y+85=0
- 4x^2+32x+y^2-4y+52=0
- 4x^2+4y^2-12x+4y+3=0
- Plot:
- d*(x^2*y^2+3*x^2*y^2-3*x*y^2)=0
-
1/(sqrt(-x^2-y^2/4+1))=0
- y=x^(3)+3*x^(2)-4*x+1
-
y=x/10
-
Identical expressions
- 3x^ two -36x+3y^ two -56y+ eighty-five = zero
- 3x squared minus 36x plus 3y squared minus 56y plus 85 equally 0
- 3x to the power of two minus 36x plus 3y to the power of two minus 56y plus eighty minus five equally zero
- 3x2-36x+3y2-56y+85=0
- 3x²-36x+3y²-56y+85=0
- 3x to the power of 2-36x+3y to the power of 2-56y+85=0
- 3x^2-36x+3y^2-56y+85=O
-
Similar expressions
- 3x^2-36x+3y^2-56y-85=0
- 3x^2-36x-3y^2-56y+85=0
- 3x^2+36x+3y^2-56y+85=0
- 3x^2-36x+3y^2+56y+85=0
3x^2-36x+3y^2-56y+85=0 canonical form
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2 2 85 - 56*y - 36*x + 3*x + 3*y = 0
$$3 x^{2} - 36 x + 3 y^{2} - 56 y + 85 = 0$$
3*x^2 - 36*x + 3*y^2 - 56*y + 85 = 0
Detail solution
Given line equation of 2-order:
$$3 x^{2} - 36 x + 3 y^{2} - 56 y + 85 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -18$$
$$a_{22} = 3$$
$$a_{23} = -28$$
$$a_{33} = 85$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 0\\0 & 3\end{matrix}\right|$$
$$\Delta = 9$$
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
$$3 x_{0} - 18 = 0$$
$$3 y_{0} - 28 = 0$$
then
$$x_{0} = 6$$
$$y_{0} = \frac{28}{3}$$
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
$$a'_{33} = - 18 x_{0} - 28 y_{0} + 85$$
$$a'_{33} = - \frac{853}{3}$$
then equation turns into
$$3 x'^{2} + 3 y'^{2} - \frac{853}{3} = 0$$
Given equation is circle
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{853} \sqrt{2559}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{853} \sqrt{2559}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$3 x^{2} - 36 x + 3 y^{2} - 56 y + 85 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -18$$
$$a_{22} = 3$$
$$a_{23} = -28$$
$$a_{33} = 85$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 0\\0 & 3\end{matrix}\right|$$
$$\Delta = 9$$
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
$$3 x_{0} - 18 = 0$$
$$3 y_{0} - 28 = 0$$
then
$$x_{0} = 6$$
$$y_{0} = \frac{28}{3}$$
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
$$a'_{33} = - 18 x_{0} - 28 y_{0} + 85$$
$$a'_{33} = - \frac{853}{3}$$
then equation turns into
$$3 x'^{2} + 3 y'^{2} - \frac{853}{3} = 0$$
Given equation is circle
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{853} \sqrt{2559}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{853} \sqrt{2559}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(6, 28/3)
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:
$$3 x^{2} - 36 x + 3 y^{2} - 56 y + 85 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -18$$
$$a_{22} = 3$$
$$a_{23} = -28$$
$$a_{33} = 85$$
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
$$I_{1} = 6$$
$$I_{3} = \left|\begin{matrix}3 & 0 & -18\\0 & 3 & -28\\-18 & -28 & 85\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0\\0 & 3 - \lambda\end{matrix}\right|$$
$$I_{1} = 6$$
$$I_{2} = 9$$
$$I_{3} = -2559$$
$$I{\left(\lambda \right)} = \lambda^{2} - 6 \lambda + 9$$
$$K_{2} = -598$$
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
$$\lambda^{2} - 6 \lambda + 9 = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = 3$$
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
$$3 \tilde x^{2} + 3 \tilde y^{2} - \frac{853}{3} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{853} \sqrt{2559}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{853} \sqrt{2559}}\right)^{2}} = 1$$
- reduced to canonical form
$$3 x^{2} - 36 x + 3 y^{2} - 56 y + 85 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -18$$
$$a_{22} = 3$$
$$a_{23} = -28$$
$$a_{33} = 85$$
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
$$I_{1} = 6$$
|3 0|
I2 = | |
|0 3|$$I_{3} = \left|\begin{matrix}3 & 0 & -18\\0 & 3 & -28\\-18 & -28 & 85\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0\\0 & 3 - \lambda\end{matrix}\right|$$
| 3 -18| | 3 -28|
K2 = | | + | |
|-18 85 | |-28 85 |$$I_{1} = 6$$
$$I_{2} = 9$$
$$I_{3} = -2559$$
$$I{\left(\lambda \right)} = \lambda^{2} - 6 \lambda + 9$$
$$K_{2} = -598$$
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
$$\lambda^{2} - 6 \lambda + 9 = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = 3$$
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
$$3 \tilde x^{2} + 3 \tilde y^{2} - \frac{853}{3} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{853} \sqrt{2559}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{853} \sqrt{2559}}\right)^{2}} = 1$$
- reduced to canonical form