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How to use it?
- Canonical form:
- 4x^2+8x-16y+16=0
- x^2+y^2-2x+1=0
- 5x^2–6y^2–20x–36y–64=0
- 6*x-7*y-4*z-2
- Plot:
- z=x^2+y^2+3*a*x*y
- z=x^2+2*y^2+x*y-x
- z=x+y
- z=sqrt(x^2+25*y^2-25)
-
Identical expressions
- 64x^(two)+81y^(two)-768x+972y+ thirty-six = zero
- 64x to the power of (2) plus 81y to the power of (2) minus 768x plus 972y plus 36 equally 0
- 64x to the power of (two) plus 81y to the power of (two) minus 768x plus 972y plus thirty minus six equally zero
- 64x(2)+81y(2)-768x+972y+36=0
- 64x2+81y2-768x+972y+36=0
- 64x^2+81y^2-768x+972y+36=0
- 64x^(2)+81y^(2)-768x+972y+36=O
-
Similar expressions
- 64x^(2)-81y^(2)-768x+972y+36=0
- 64x^(2)+81y^(2)-768x-972y+36=0
- 64x^(2)+81y^(2)-768x+972y-36=0
- 64x^(2)+81y^(2)+768x+972y+36=0
64x^(2)+81y^(2)-768x+972y+36=0 canonical form
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2 2 36 - 768*x + 64*x + 81*y + 972*y = 0
$$64 x^{2} + 81 y^{2} - 768 x + 972 y + 36 = 0$$
64*x^2 - 768*x + 81*y^2 + 972*y + 36 = 0
Detail solution
Given line equation of 2-order:
$$64 x^{2} + 81 y^{2} - 768 x + 972 y + 36 = 0$$
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
$$a_{11} = 64$$
$$a_{12} = 0$$
$$a_{13} = -384$$
$$a_{22} = 81$$
$$a_{23} = 486$$
$$a_{33} = 36$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}64 & 0\\0 & 81\end{matrix}\right|$$
$$\Delta = 5184$$
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
$$64 x_{0} - 384 = 0$$
$$81 y_{0} + 486 = 0$$
then
$$x_{0} = 6$$
$$y_{0} = -6$$
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
$$a'_{33} = - 384 x_{0} + 486 y_{0} + 36$$
$$a'_{33} = -5184$$
then The equation is transformed to
$$64 x'^{2} + 81 y'^{2} - 5184 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{9^{2}} + \frac{\tilde y^{2}}{8^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$64 x^{2} + 81 y^{2} - 768 x + 972 y + 36 = 0$$
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
$$a_{11} = 64$$
$$a_{12} = 0$$
$$a_{13} = -384$$
$$a_{22} = 81$$
$$a_{23} = 486$$
$$a_{33} = 36$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}64 & 0\\0 & 81\end{matrix}\right|$$
$$\Delta = 5184$$
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
$$64 x_{0} - 384 = 0$$
$$81 y_{0} + 486 = 0$$
then
$$x_{0} = 6$$
$$y_{0} = -6$$
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
$$a'_{33} = - 384 x_{0} + 486 y_{0} + 36$$
$$a'_{33} = -5184$$
then The equation is transformed to
$$64 x'^{2} + 81 y'^{2} - 5184 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{9^{2}} + \frac{\tilde y^{2}}{8^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(6, -6)
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:
$$64 x^{2} + 81 y^{2} - 768 x + 972 y + 36 = 0$$
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
$$a_{11} = 64$$
$$a_{12} = 0$$
$$a_{13} = -384$$
$$a_{22} = 81$$
$$a_{23} = 486$$
$$a_{33} = 36$$
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
$$I_{1} = 145$$
$$I_{3} = \left|\begin{matrix}64 & 0 & -384\\0 & 81 & 486\\-384 & 486 & 36\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 64 & 0\\0 & - \lambda + 81\end{matrix}\right|$$
$$I_{1} = 145$$
$$I_{2} = 5184$$
$$I_{3} = -26873856$$
$$I{\left(\lambda \right)} = \lambda^{2} - 145 \lambda + 5184$$
$$K_{2} = -378432$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} < 0$$
then by line type:
this equation is of type : ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - 145 \lambda + 5184 = 0$$
Solve this equation
$$\lambda_{1} = 81$$
$$\lambda_{2} = 64$$
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
$$81 \tilde x^{2} + 64 \tilde y^{2} - 5184 = 0$$
$$\frac{\tilde x^{2}}{8^{2}} + \frac{\tilde y^{2}}{9^{2}} = 1$$
- reduced to canonical form
$$64 x^{2} + 81 y^{2} - 768 x + 972 y + 36 = 0$$
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
$$a_{11} = 64$$
$$a_{12} = 0$$
$$a_{13} = -384$$
$$a_{22} = 81$$
$$a_{23} = 486$$
$$a_{33} = 36$$
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
$$I_{1} = 145$$
|64 0 |
I2 = | |
|0 81|$$I_{3} = \left|\begin{matrix}64 & 0 & -384\\0 & 81 & 486\\-384 & 486 & 36\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 64 & 0\\0 & - \lambda + 81\end{matrix}\right|$$
| 64 -384| |81 486|
K2 = | | + | |
|-384 36 | |486 36 |$$I_{1} = 145$$
$$I_{2} = 5184$$
$$I_{3} = -26873856$$
$$I{\left(\lambda \right)} = \lambda^{2} - 145 \lambda + 5184$$
$$K_{2} = -378432$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} < 0$$
then by line type:
this equation is of type : ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - 145 \lambda + 5184 = 0$$
Solve this equation
$$\lambda_{1} = 81$$
$$\lambda_{2} = 64$$
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
$$81 \tilde x^{2} + 64 \tilde y^{2} - 5184 = 0$$
$$\frac{\tilde x^{2}}{8^{2}} + \frac{\tilde y^{2}}{9^{2}} = 1$$
- reduced to canonical form
The graph