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How to use it?
- Canonical form:
- 25x^2-30xy+9y^2+68x+19=0
- 4x^2-16y^2-8x+96y-204=0
- z=ln(x+y)
-
9x^2-4y^2-18x+16y-43=0
- Plot:
-
sin(x+y)=1.2*x
-
x^2+y^2-1
-
x^2+(y-cbrt(x^2))^2=1
- x^2+(y-(x^2)^(1/3))^2=1
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Identical expressions
- 4x^ two -16y^ two -8x+96y- two hundred and four = zero
- 4x squared minus 16y squared minus 8x plus 96y minus 204 equally 0
- 4x to the power of two minus 16y to the power of two minus 8x plus 96y minus two hundred and four equally zero
- 4x2-16y2-8x+96y-204=0
- 4x²-16y²-8x+96y-204=0
- 4x to the power of 2-16y to the power of 2-8x+96y-204=0
- 4x^2-16y^2-8x+96y-204=O
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Similar expressions
- 4x^2+16y^2-8x+96y-204=0
- 4x^2-16y^2-8x-96y-204=0
- 4x^2-16y^2-8x+96y+204=0
- 4x^2-16y^2+8x+96y-204=0
4x^2-16y^2-8x+96y-204=0 canonical form
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2 2 -204 - 16*y - 8*x + 4*x + 96*y = 0
$$4 x^{2} - 8 x - 16 y^{2} + 96 y - 204 = 0$$
4*x^2 - 8*x - 16*y^2 + 96*y - 204 = 0
Detail solution
Given line equation of 2-order:
$$4 x^{2} - 8 x - 16 y^{2} + 96 y - 204 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = -16$$
$$a_{23} = 48$$
$$a_{33} = -204$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & -16\end{matrix}\right|$$
$$\Delta = -64$$
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
$$4 x_{0} - 4 = 0$$
$$48 - 16 y_{0} = 0$$
then
$$x_{0} = 1$$
$$y_{0} = 3$$
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
$$a'_{33} = - 4 x_{0} + 48 y_{0} - 204$$
$$a'_{33} = -64$$
then equation turns into
$$4 x'^{2} - 16 y'^{2} - 64 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{16} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$4 x^{2} - 8 x - 16 y^{2} + 96 y - 204 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = -16$$
$$a_{23} = 48$$
$$a_{33} = -204$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & -16\end{matrix}\right|$$
$$\Delta = -64$$
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
$$4 x_{0} - 4 = 0$$
$$48 - 16 y_{0} = 0$$
then
$$x_{0} = 1$$
$$y_{0} = 3$$
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
$$a'_{33} = - 4 x_{0} + 48 y_{0} - 204$$
$$a'_{33} = -64$$
then equation turns into
$$4 x'^{2} - 16 y'^{2} - 64 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{16} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(1, 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:
$$4 x^{2} - 8 x - 16 y^{2} + 96 y - 204 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = -16$$
$$a_{23} = 48$$
$$a_{33} = -204$$
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
$$I_{1} = -12$$
$$I_{3} = \left|\begin{matrix}4 & 0 & -4\\0 & -16 & 48\\-4 & 48 & -204\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0\\0 & - \lambda - 16\end{matrix}\right|$$
$$I_{1} = -12$$
$$I_{2} = -64$$
$$I_{3} = 4096$$
$$I{\left(\lambda \right)} = \lambda^{2} + 12 \lambda - 64$$
$$K_{2} = 128$$
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
$$\lambda^{2} + 12 \lambda - 64 = 0$$
$$\lambda_{1} = 4$$
$$\lambda_{2} = -16$$
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
$$4 \tilde x^{2} - 16 \tilde y^{2} - 64 = 0$$
$$\frac{\tilde x^{2}}{16} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
$$4 x^{2} - 8 x - 16 y^{2} + 96 y - 204 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = -16$$
$$a_{23} = 48$$
$$a_{33} = -204$$
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
$$I_{1} = -12$$
|4 0 |
I2 = | |
|0 -16|$$I_{3} = \left|\begin{matrix}4 & 0 & -4\\0 & -16 & 48\\-4 & 48 & -204\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0\\0 & - \lambda - 16\end{matrix}\right|$$
|4 -4 | |-16 48 |
K2 = | | + | |
|-4 -204| |48 -204|$$I_{1} = -12$$
$$I_{2} = -64$$
$$I_{3} = 4096$$
$$I{\left(\lambda \right)} = \lambda^{2} + 12 \lambda - 64$$
$$K_{2} = 128$$
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
$$\lambda^{2} + 12 \lambda - 64 = 0$$
$$\lambda_{1} = 4$$
$$\lambda_{2} = -16$$
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
$$4 \tilde x^{2} - 16 \tilde y^{2} - 64 = 0$$
$$\frac{\tilde x^{2}}{16} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form