Mister Exam
Other calculators
- 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:
- (2)x^2+(-2)xy+(2)y^2+(-2)x+(-2)y+1=0
- 25x^2-120xy+144y^2-242x-298y+491=0
- 11x^2+24xy+4y^2+42x+64y+51=0
- 3x^2+3y^2-18x-18y+51=0
- Plot:
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-0.47y-1.02y^3+0.5ln((1+y)/(1-y))=x(1-exp(2.2*x/2))
- z=x^2+y^2/4
- y*(x)=x+4/x^2
- (7/10)*x^2+(y-sqrt(|x|))^2=10
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Identical expressions
- two y^2−2xy−2xz+2yz+ six √ three *x− six √ three *z− nine = zero .
- 2y squared −2xy−2xz plus 2yz plus 6√3 multiply by x−6√3 multiply by z−9 equally 0.
- two y squared −2xy−2xz plus 2yz plus six √ three multiply by x− six √ three multiply by z− nine equally zero .
- 2y2−2xy−2xz+2yz+6√3*x−6√3*z−9=0.
- 2y²−2xy−2xz+2yz+6√3*x−6√3*z−9=0.
- 2y to the power of 2−2xy−2xz+2yz+6√3*x−6√3*z−9=0.
- 2y^2−2xy−2xz+2yz+6√3x−6√3z−9=0.
- 2y2−2xy−2xz+2yz+6√3x−6√3z−9=0.
- 2y^2−2xy−2xz+2yz+6√3*x−6√3*z−9=O.
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Similar expressions
- 2y^2−2xy−2xz+2yz-6√3*x−6√3*z−9=0.
- 2y^2−2xy−2xz-2yz+6√3*x−6√3*z−9=0.
2y^2−2xy−2xz+2yz+6√3*x−6√3*z−9=0. canonical form
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The solution
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2 ___ ___ -9 + 2*y - 6*z*\/ 3 - 2*x*y - 2*x*z + 2*y*z + 6*x*\/ 3 = 0
$$- 2 x y - 2 x z + 6 \sqrt{3} x + 2 y^{2} + 2 y z - 6 \sqrt{3} z - 9 = 0$$
-2*x*y - 2*x*z + 6*sqrt(3)*x + 2*y^2 + 2*y*z - 6*sqrt(3)*z - 9 = 0
Invariants method
Given equation of the surface of 2-order:
$$- 2 x y - 2 x z + 6 \sqrt{3} x + 2 y^{2} + 2 y z - 6 \sqrt{3} z - 9 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = -1$$
$$a_{13} = -1$$
$$a_{14} = 3 \sqrt{3}$$
$$a_{22} = 2$$
$$a_{23} = 1$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = - 3 \sqrt{3}$$
$$a_{44} = -9$$
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
$$I_{1} = 2$$
$$I_{3} = \left|\begin{matrix}0 & -1 & -1\\-1 & 2 & 1\\-1 & 1 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & -1 & -1 & 3 \sqrt{3}\\-1 & 2 & 1 & 0\\-1 & 1 & 0 & - 3 \sqrt{3}\\3 \sqrt{3} & 0 & - 3 \sqrt{3} & -9\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & -1 & -1\\-1 & 2 - \lambda & 1\\-1 & 1 & - \lambda\end{matrix}\right|$$
$$I_{1} = 2$$
$$I_{2} = -3$$
$$I_{3} = 0$$
$$I_{4} = 108$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 2 \lambda^{2} + 3 \lambda$$
$$K_{2} = -72$$
$$K_{3} = -27$$
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
$$\lambda^{3} - 2 \lambda^{2} - 3 \lambda = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = -1$$
$$\lambda_{3} = 0$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$3 \tilde x^{2} - \tilde y^{2} + 12 \tilde z = 0$$
and
$$3 \tilde x^{2} - \tilde y^{2} - 12 \tilde z = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{2} - \frac{\tilde y^{2}}{6}\right) = 0$$
and
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{2} - \frac{\tilde y^{2}}{6}\right) = 0$$
this equation is fora type hyperbolic paraboloid
- reduced to canonical form
$$- 2 x y - 2 x z + 6 \sqrt{3} x + 2 y^{2} + 2 y z - 6 \sqrt{3} z - 9 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = -1$$
$$a_{13} = -1$$
$$a_{14} = 3 \sqrt{3}$$
$$a_{22} = 2$$
$$a_{23} = 1$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = - 3 \sqrt{3}$$
$$a_{44} = -9$$
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
$$I_{1} = 2$$
|0 -1| |2 1| |0 -1|
I2 = | | + | | + | |
|-1 2 | |1 0| |-1 0 |$$I_{3} = \left|\begin{matrix}0 & -1 & -1\\-1 & 2 & 1\\-1 & 1 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & -1 & -1 & 3 \sqrt{3}\\-1 & 2 & 1 & 0\\-1 & 1 & 0 & - 3 \sqrt{3}\\3 \sqrt{3} & 0 & - 3 \sqrt{3} & -9\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & -1 & -1\\-1 & 2 - \lambda & 1\\-1 & 1 & - \lambda\end{matrix}\right|$$
| ___| | ___|
| 0 3*\/ 3 | |2 0 | | 0 -3*\/ 3 |
K2 = | | + | | + | |
| ___ | |0 -9| | ___ |
|3*\/ 3 -9 | |-3*\/ 3 -9 | | ___ |
| ___| |2 1 0 | | 0 -1 3*\/ 3 |
| 0 -1 3*\/ 3 | | | | |
| | | ___| | ___|
K3 = | -1 2 0 | + |1 0 -3*\/ 3 | + | -1 0 -3*\/ 3 |
| | | | | |
| ___ | | ___ | | ___ ___ |
|3*\/ 3 0 -9 | |0 -3*\/ 3 -9 | |3*\/ 3 -3*\/ 3 -9 |
$$I_{1} = 2$$
$$I_{2} = -3$$
$$I_{3} = 0$$
$$I_{4} = 108$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 2 \lambda^{2} + 3 \lambda$$
$$K_{2} = -72$$
$$K_{3} = -27$$
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
$$\lambda^{3} - 2 \lambda^{2} - 3 \lambda = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = -1$$
$$\lambda_{3} = 0$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$3 \tilde x^{2} - \tilde y^{2} + 12 \tilde z = 0$$
and
$$3 \tilde x^{2} - \tilde y^{2} - 12 \tilde z = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{2} - \frac{\tilde y^{2}}{6}\right) = 0$$
and
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{2} - \frac{\tilde y^{2}}{6}\right) = 0$$
this equation is fora type hyperbolic paraboloid
- reduced to canonical form