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- Canonical form of a elliptical paraboloid
- Canonical form of a double hyperboloid
- Canonical form of a imaginary ellipsoid
- Canonical form of a degenerate ellipse
- Canonical form of a parabola
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How to use it?
- Canonical form:
- 3x^2+0xy+4y^2-6x+16y+7=0
- y=-3-sqrt(-3x-21)
- 3*x-4*y+4=0
- x^2+4*y^2+2*x-32*y+64=0
- Plot:
- (x^2+y^2-1)^3-x^2*y^3=0
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y^2=x(1-x^2)^3
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y^2=(x^4+83*x^2-408+60*x+10*x^3)/(21+2*x^2+10*x-y^2)
- y=-0,8x;y-2,5x;y=0,4x
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Identical expressions
- 7x^ two +7y^ two +4 zero z^ two -4xy+20xz+20yz-4x-4y+2z=0
- 7x squared plus 7y squared plus 40z squared minus 4xy plus 20xz plus 20yz minus 4x minus 4y plus 2z equally 0
- 7x to the power of two plus 7y to the power of two plus 4 zero z to the power of two minus 4xy plus 20xz plus 20yz minus 4x minus 4y plus 2z equally 0
- 7x2+7y2+40z2-4xy+20xz+20yz-4x-4y+2z=0
- 7x²+7y²+40z²-4xy+20xz+20yz-4x-4y+2z=0
- 7x to the power of 2+7y to the power of 2+40z to the power of 2-4xy+20xz+20yz-4x-4y+2z=0
- 7x^2+7y^2+40z^2-4xy+20xz+20yz-4x-4y+2z=O
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Similar expressions
- 7x^2+7y^2+40z^2+4xy+20xz+20yz-4x-4y+2z=0
- 7x^2+7y^2-40z^2-4xy+20xz+20yz-4x-4y+2z=0
- 7x^2+7y^2+40z^2-4xy+20xz+20yz-4x+4y+2z=0
- 7x^2+7y^2+40z^2-4xy+20xz+20yz+4x-4y+2z=0
- 7x^2+7y^2+40z^2-4xy+20xz+20yz-4x-4y-2z=0
- 7x^2-7y^2+40z^2-4xy+20xz+20yz-4x-4y+2z=0
- 7x^2+7y^2+40z^2-4xy-20xz+20yz-4x-4y+2z=0
- 7x^2+7y^2+40z^2-4xy+20xz-20yz-4x-4y+2z=0
7x^2+7y^2+40z^2-4xy+20xz+20yz-4x-4y+2z=0 canonical form
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2 2 2 -4*x - 4*y + 2*z + 7*x + 7*y + 40*z - 4*x*y + 20*x*z + 20*y*z = 0
$$7 x^{2} - 4 x y + 20 x z - 4 x + 7 y^{2} + 20 y z - 4 y + 40 z^{2} + 2 z = 0$$
7*x^2 - 4*x*y + 20*x*z - 4*x + 7*y^2 + 20*y*z - 4*y + 40*z^2 + 2*z = 0
Invariants method
Given equation of the surface of 2-order:
$$7 x^{2} - 4 x y + 20 x z - 4 x + 7 y^{2} + 20 y z - 4 y + 40 z^{2} + 2 z = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0$$
where
$$a_{11} = 7$$
$$a_{12} = -2$$
$$a_{13} = 10$$
$$a_{14} = -2$$
$$a_{22} = 7$$
$$a_{23} = 10$$
$$a_{24} = -2$$
$$a_{33} = 40$$
$$a_{34} = 1$$
$$a_{44} = 0$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$
substitute coefficients
$$I_{1} = 54$$
$$I_{3} = \left|\begin{matrix}7 & -2 & 10\\-2 & 7 & 10\\10 & 10 & 40\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}7 & -2 & 10 & -2\\-2 & 7 & 10 & -2\\10 & 10 & 40 & 1\\-2 & -2 & 1 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}7 - \lambda & -2 & 10\\-2 & 7 - \lambda & 10\\10 & 10 & 40 - \lambda\end{matrix}\right|$$
$$I_{1} = 54$$
$$I_{2} = 405$$
$$I_{3} = 0$$
$$I_{4} = -3645$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 54 \lambda^{2} - 405 \lambda$$
$$K_{2} = -9$$
$$K_{3} = -486$$
Because
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0$$
or
$$\lambda^{3} - 54 \lambda^{2} + 405 \lambda = 0$$
$$\lambda_{1} = 45$$
$$\lambda_{2} = 9$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
and
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$45 \tilde x^{2} + 9 \tilde y^{2} + 6 \tilde z = 0$$
and
$$45 \tilde x^{2} + 9 \tilde y^{2} - 6 \tilde z = 0$$
and
this equation is fora type elliptical paraboloid
- reduced to canonical form
$$7 x^{2} - 4 x y + 20 x z - 4 x + 7 y^{2} + 20 y z - 4 y + 40 z^{2} + 2 z = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0$$
where
$$a_{11} = 7$$
$$a_{12} = -2$$
$$a_{13} = 10$$
$$a_{14} = -2$$
$$a_{22} = 7$$
$$a_{23} = 10$$
$$a_{24} = -2$$
$$a_{33} = 40$$
$$a_{34} = 1$$
$$a_{44} = 0$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
|a11 a12| |a22 a23| |a11 a13|
I2 = | | + | | + | |
|a12 a22| |a23 a33| |a13 a33|$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$
|a11 a14| |a22 a24| |a33 a34|
K2 = | | + | | + | |
|a14 a44| |a24 a44| |a34 a44| |a11 a12 a14| |a22 a23 a24| |a11 a13 a14|
| | | | | |
K3 = |a12 a22 a24| + |a23 a33 a34| + |a13 a33 a34|
| | | | | |
|a14 a24 a44| |a24 a34 a44| |a14 a34 a44|substitute coefficients
$$I_{1} = 54$$
|7 -2| |7 10| |7 10|
I2 = | | + | | + | |
|-2 7 | |10 40| |10 40|$$I_{3} = \left|\begin{matrix}7 & -2 & 10\\-2 & 7 & 10\\10 & 10 & 40\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}7 & -2 & 10 & -2\\-2 & 7 & 10 & -2\\10 & 10 & 40 & 1\\-2 & -2 & 1 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}7 - \lambda & -2 & 10\\-2 & 7 - \lambda & 10\\10 & 10 & 40 - \lambda\end{matrix}\right|$$
|7 -2| |7 -2| |40 1|
K2 = | | + | | + | |
|-2 0 | |-2 0 | |1 0| |7 -2 -2| |7 10 -2| |7 10 -2|
| | | | | |
K3 = |-2 7 -2| + |10 40 1 | + |10 40 1 |
| | | | | |
|-2 -2 0 | |-2 1 0 | |-2 1 0 |$$I_{1} = 54$$
$$I_{2} = 405$$
$$I_{3} = 0$$
$$I_{4} = -3645$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 54 \lambda^{2} - 405 \lambda$$
$$K_{2} = -9$$
$$K_{3} = -486$$
Because
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0$$
or
$$\lambda^{3} - 54 \lambda^{2} + 405 \lambda = 0$$
$$\lambda_{1} = 45$$
$$\lambda_{2} = 9$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
and
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$45 \tilde x^{2} + 9 \tilde y^{2} + 6 \tilde z = 0$$
and
$$45 \tilde x^{2} + 9 \tilde y^{2} - 6 \tilde z = 0$$
2 2 \tilde x \tilde y --------- + --------- + 2*\tilde z = 0 1/15 1/3
and
2 2 \tilde x \tilde y --------- + --------- - 2*\tilde z = 0 1/15 1/3
this equation is fora type elliptical paraboloid
- reduced to canonical form