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:
- 3x^2+2y^2=5
- 2x^2-y^2-2y=0
- 1x+2y-2z+11=0
- 7*
- Plot:
- 230+14050∙e^(-51∙t)+e^(-25∙t)∙0.042∙cos0.568∙t+1636∙sin0.568∙t
- (x-1)^2+y^2=9
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(x+2)/sqrt((-36)*x^2-288*x+16*y^2-160*y-752)=0
- (3*x^2-10*x+3)/(x^2-2*x-3)
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Identical expressions
- 2xy-4yz
- 2xy minus 4yz
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Similar expressions
- 3*x^2+2*y^2-z^2-2*w^2+2*x*y-4*y*z+2*y*w
- 2xy+4yz
2xy-4yz canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Invariants method
Given equation of the surface of 2-order:
$$2 x y - 4 y z = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 1$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = -2$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 0$$
$$I_{3} = \left|\begin{matrix}0 & 1 & 0\\1 & 0 & -2\\0 & -2 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 1 & 0 & 0\\1 & 0 & -2 & 0\\0 & -2 & 0 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 1 & 0\\1 & - \lambda & -2\\0 & -2 & - \lambda\end{matrix}\right|$$
$$I_{1} = 0$$
$$I_{2} = -5$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 5 \lambda$$
$$K_{2} = 0$$
$$K_{3} = 0$$
Because
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
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
$$\lambda^{3} - 5 \lambda = 0$$
$$\lambda_{1} = - \sqrt{5}$$
$$\lambda_{2} = \sqrt{5}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$- \sqrt{5} \tilde x^{2} + \sqrt{5} \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{5^{\frac{3}{4}}}{5}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{5^{\frac{3}{4}}}{5}\right)^{2}} = 0$$
this equation is fora type two intersecting planes
- reduced to canonical form
$$2 x y - 4 y z = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 1$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = -2$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 0$$
|0 1| |0 -2| |0 0|
I2 = | | + | | + | |
|1 0| |-2 0 | |0 0|$$I_{3} = \left|\begin{matrix}0 & 1 & 0\\1 & 0 & -2\\0 & -2 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 1 & 0 & 0\\1 & 0 & -2 & 0\\0 & -2 & 0 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 1 & 0\\1 & - \lambda & -2\\0 & -2 & - \lambda\end{matrix}\right|$$
|0 0| |0 0| |0 0|
K2 = | | + | | + | |
|0 0| |0 0| |0 0| |0 1 0| |0 -2 0| |0 0 0|
| | | | | |
K3 = |1 0 0| + |-2 0 0| + |0 0 0|
| | | | | |
|0 0 0| |0 0 0| |0 0 0|$$I_{1} = 0$$
$$I_{2} = -5$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 5 \lambda$$
$$K_{2} = 0$$
$$K_{3} = 0$$
Because
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
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
$$\lambda^{3} - 5 \lambda = 0$$
$$\lambda_{1} = - \sqrt{5}$$
$$\lambda_{2} = \sqrt{5}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$- \sqrt{5} \tilde x^{2} + \sqrt{5} \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{5^{\frac{3}{4}}}{5}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{5^{\frac{3}{4}}}{5}\right)^{2}} = 0$$
this equation is fora type two intersecting planes
- reduced to canonical form