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How to use it?
- Canonical form:
- 4x^2+4xy+y^2+16x-2y+13=0
- 2x^2+2(sqrt8)xy-5y^2
- 3x^2+y^2-6x-4y-2=0
- 5x^2+6y^2+7z^2-4xy+4yz-10x+8y+14z-6=0
- Plot:
- 5*y^2-3*x=0
- z=x^3+y^3-3*x*y+5
- x^2/16+y^2/9=1
-
log(x)+sin(2*x+y/x)=4
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Identical expressions
- 4x^ two -9y^ two -8x+36y- sixty-eight = zero
- 4x squared minus 9y squared minus 8x plus 36y minus 68 equally 0
- 4x to the power of two minus 9y to the power of two minus 8x plus 36y minus sixty minus eight equally zero
- 4x2-9y2-8x+36y-68=0
- 4x²-9y²-8x+36y-68=0
- 4x to the power of 2-9y to the power of 2-8x+36y-68=0
- 4x^2-9y^2-8x+36y-68=O
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Similar expressions
- 4x^2-9y^2+8x+36y-68=0
- 4x^2+9y^2-8x+36y-68=0
- 4x^2-9y^2-8x-36y-68=0
- 4x^2-9y^2-8x+36y+68=0
4x^2-9y^2-8x+36y-68=0 canonical form
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The solution
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2 2 -68 - 9*y - 8*x + 4*x + 36*y = 0
$$4 x^{2} - 9 y^{2} - 8 x + 36 y - 68 = 0$$
4*x^2 - 8*x - 9*y^2 + 36*y - 68 = 0
Detail solution
Given line equation of 2-order:
$$4 x^{2} - 9 y^{2} - 8 x + 36 y - 68 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = -9$$
$$a_{23} = 18$$
$$a_{33} = -68$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & -9\end{matrix}\right|$$
$$\Delta = -36$$
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
$$4 x_{0} - 4 = 0$$
$$- 9 y_{0} + 18 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = 2$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - 4 x_{0} + 18 y_{0} - 68$$
$$a'_{33} = -36$$
then The equation is transformed to
$$4 x'^{2} - 9 y'^{2} - 36 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{9} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$4 x^{2} - 9 y^{2} - 8 x + 36 y - 68 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = -9$$
$$a_{23} = 18$$
$$a_{33} = -68$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & -9\end{matrix}\right|$$
$$\Delta = -36$$
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
$$4 x_{0} - 4 = 0$$
$$- 9 y_{0} + 18 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = 2$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - 4 x_{0} + 18 y_{0} - 68$$
$$a'_{33} = -36$$
then The equation is transformed to
$$4 x'^{2} - 9 y'^{2} - 36 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{9} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(1, 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:
$$4 x^{2} - 9 y^{2} - 8 x + 36 y - 68 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = -9$$
$$a_{23} = 18$$
$$a_{33} = -68$$
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
$$I_{1} = -5$$
$$I_{3} = \left|\begin{matrix}4 & 0 & -4\\0 & -9 & 18\\-4 & 18 & -68\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 4 & 0\\0 & - \lambda - 9\end{matrix}\right|$$
$$I_{1} = -5$$
$$I_{2} = -36$$
$$I_{3} = 1296$$
$$I{\left(\lambda \right)} = \lambda^{2} + 5 \lambda - 36$$
$$K_{2} = 0$$
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:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} + 5 \lambda - 36 = 0$$
Solve this equation
$$\lambda_{1} = 4$$
$$\lambda_{2} = -9$$
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
$$4 \tilde x^{2} - 9 \tilde y^{2} - 36 = 0$$
$$\frac{\tilde x^{2}}{9} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
$$4 x^{2} - 9 y^{2} - 8 x + 36 y - 68 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = -9$$
$$a_{23} = 18$$
$$a_{33} = -68$$
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
$$I_{1} = -5$$
|4 0 |
I2 = | |
|0 -9|$$I_{3} = \left|\begin{matrix}4 & 0 & -4\\0 & -9 & 18\\-4 & 18 & -68\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 4 & 0\\0 & - \lambda - 9\end{matrix}\right|$$
|4 -4 | |-9 18 |
K2 = | | + | |
|-4 -68| |18 -68|$$I_{1} = -5$$
$$I_{2} = -36$$
$$I_{3} = 1296$$
$$I{\left(\lambda \right)} = \lambda^{2} + 5 \lambda - 36$$
$$K_{2} = 0$$
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:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} + 5 \lambda - 36 = 0$$
Solve this equation
$$\lambda_{1} = 4$$
$$\lambda_{2} = -9$$
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
$$4 \tilde x^{2} - 9 \tilde y^{2} - 36 = 0$$
$$\frac{\tilde x^{2}}{9} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
The graph