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How to use it?
- Canonical form:
- x^2+6x+y+11=0
-
x^2-9y^2-4x+36y-41=0
- 4x^2-9y^2+32x-54y-53=0
- 9x2+4y2-36x+16y+16=0.
- Plot:
-
(-x^2)*y^3+(x^2+y^2-1)^3=0
-
sin(x^e)+cos(y^e)=1
- f*(x)=100-x*y-x*y*z
-
x^2+4y^2
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Identical expressions
- x^ two -9y^ two -4x+36y- forty-one = zero
- x squared minus 9y squared minus 4x plus 36y minus 41 equally 0
- x to the power of two minus 9y to the power of two minus 4x plus 36y minus forty minus one equally zero
- x2-9y2-4x+36y-41=0
- x²-9y²-4x+36y-41=0
- x to the power of 2-9y to the power of 2-4x+36y-41=0
- x^2-9y^2-4x+36y-41=O
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Similar expressions
- x^2-9y^2+4x+36y-41=0
- x^2+9y^2-4x+36y-41=0
- x^2-9y^2-4x+36y+41=0
- x^2-9y^2-4x-36y-41=0
x^2-9y^2-4x+36y-41=0 canonical form
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The solution
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2 2 -41 + x - 9*y - 4*x + 36*y = 0
$$x^{2} - 4 x - 9 y^{2} + 36 y - 41 = 0$$
x^2 - 4*x - 9*y^2 + 36*y - 41 = 0
Detail solution
Given line equation of 2-order:
$$x^{2} - 4 x - 9 y^{2} + 36 y - 41 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = -2$$
$$a_{22} = -9$$
$$a_{23} = 18$$
$$a_{33} = -41$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & -9\end{matrix}\right|$$
$$\Delta = -9$$
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
$$x_{0} - 2 = 0$$
$$18 - 9 y_{0} = 0$$
then
$$x_{0} = 2$$
$$y_{0} = 2$$
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
$$a'_{33} = - 2 x_{0} + 18 y_{0} - 41$$
$$a'_{33} = -9$$
then equation turns into
$$x'^{2} - 9 y'^{2} - 9 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{9} - \tilde y^{2} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$x^{2} - 4 x - 9 y^{2} + 36 y - 41 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = -2$$
$$a_{22} = -9$$
$$a_{23} = 18$$
$$a_{33} = -41$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & -9\end{matrix}\right|$$
$$\Delta = -9$$
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
$$x_{0} - 2 = 0$$
$$18 - 9 y_{0} = 0$$
then
$$x_{0} = 2$$
$$y_{0} = 2$$
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
$$a'_{33} = - 2 x_{0} + 18 y_{0} - 41$$
$$a'_{33} = -9$$
then equation turns into
$$x'^{2} - 9 y'^{2} - 9 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{9} - \tilde y^{2} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(2, 2)
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:
$$x^{2} - 4 x - 9 y^{2} + 36 y - 41 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = -2$$
$$a_{22} = -9$$
$$a_{23} = 18$$
$$a_{33} = -41$$
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
$$I_{1} = -8$$
$$I_{3} = \left|\begin{matrix}1 & 0 & -2\\0 & -9 & 18\\-2 & 18 & -41\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda - 9\end{matrix}\right|$$
$$I_{1} = -8$$
$$I_{2} = -9$$
$$I_{3} = 81$$
$$I{\left(\lambda \right)} = \lambda^{2} + 8 \lambda - 9$$
$$K_{2} = 0$$
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
$$\lambda^{2} + 8 \lambda - 9 = 0$$
Solve this equation
$$\lambda_{1} = 1$$
$$\lambda_{2} = -9$$
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
$$\tilde x^{2} - 9 \tilde y^{2} - 9 = 0$$
$$\frac{\tilde x^{2}}{9} - \tilde y^{2} = 1$$
- reduced to canonical form
$$x^{2} - 4 x - 9 y^{2} + 36 y - 41 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = -2$$
$$a_{22} = -9$$
$$a_{23} = 18$$
$$a_{33} = -41$$
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
$$I_{1} = -8$$
|1 0 |
I2 = | |
|0 -9|$$I_{3} = \left|\begin{matrix}1 & 0 & -2\\0 & -9 & 18\\-2 & 18 & -41\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda - 9\end{matrix}\right|$$
|1 -2 | |-9 18 |
K2 = | | + | |
|-2 -41| |18 -41|$$I_{1} = -8$$
$$I_{2} = -9$$
$$I_{3} = 81$$
$$I{\left(\lambda \right)} = \lambda^{2} + 8 \lambda - 9$$
$$K_{2} = 0$$
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
$$\lambda^{2} + 8 \lambda - 9 = 0$$
Solve this equation
$$\lambda_{1} = 1$$
$$\lambda_{2} = -9$$
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
$$\tilde x^{2} - 9 \tilde y^{2} - 9 = 0$$
$$\frac{\tilde x^{2}}{9} - \tilde y^{2} = 1$$
- reduced to canonical form
The graph