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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:
- x^2-10*x*y+7*y^2+2*x-10*y=a
- 4x^2-3y^2-2xy+z^2+6z+5=0
- 8x^2+6xy-5sqrt(10)x-3sqrt(10)y-22=0
- (y+3)^2=4x-2
- Plot:
- y^2-4*x+x^2=0
- z=/sqrt(3*x+2*y+2)
- x^4-y^4=x*y
-
sin(x+y)-1.2*x-0.1
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Identical expressions
- 4x^ two -3y^ two - two xy+z^2+6z+ five = zero
- 4x squared minus 3y squared minus 2xy plus z squared plus 6z plus 5 equally 0
- 4x to the power of two minus 3y to the power of two minus two xy plus z squared plus 6z plus five equally zero
- 4x2-3y2-2xy+z2+6z+5=0
- 4x²-3y²-2xy+z²+6z+5=0
- 4x to the power of 2-3y to the power of 2-2xy+z to the power of 2+6z+5=0
- 4x^2-3y^2-2xy+z^2+6z+5=O
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Similar expressions
- 4x^2-3y^2-2xy-z^2+6z+5=0
- 4x^2+3y^2-2xy+z^2+6z+5=0
- 4x^2-3y^2+2xy+z^2+6z+5=0
- 4x^2-3y^2-2xy+z^2+6z-5=0
- 4x^2-3y^2-2xy+z^2-6z+5=0
4x^2-3y^2-2xy+z^2+6z+5=0 canonical form
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2 2 2 5 + z - 3*y + 4*x + 6*z - 2*x*y = 0
$$4 x^{2} - 2 x y - 3 y^{2} + z^{2} + 6 z + 5 = 0$$
4*x^2 - 2*x*y - 3*y^2 + z^2 + 6*z + 5 = 0
Invariants method
Given equation of the surface of 2-order:
$$4 x^{2} - 2 x y - 3 y^{2} + z^{2} + 6 z + 5 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = -1$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = -3$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 1$$
$$a_{34} = 3$$
$$a_{44} = 5$$
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
$$I_{1} = 2$$
$$I_{3} = \left|\begin{matrix}4 & -1 & 0\\-1 & -3 & 0\\0 & 0 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}4 & -1 & 0 & 0\\-1 & -3 & 0 & 0\\0 & 0 & 1 & 3\\0 & 0 & 3 & 5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & -1 & 0\\-1 & - \lambda - 3 & 0\\0 & 0 & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 2$$
$$I_{2} = -12$$
$$I_{3} = -13$$
$$I_{4} = 52$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 2 \lambda^{2} + 12 \lambda - 13$$
$$K_{2} = 1$$
$$K_{3} = -69$$
Because
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
$$\lambda^{3} - 2 \lambda^{2} - 12 \lambda + 13 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = \frac{1}{2} - \frac{\sqrt{53}}{2}$$
$$\lambda_{3} = \frac{1}{2} + \frac{\sqrt{53}}{2}$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\tilde x^{2} + \tilde y^{2} \left(\frac{1}{2} - \frac{\sqrt{53}}{2}\right) + \tilde z^{2} \left(\frac{1}{2} + \frac{\sqrt{53}}{2}\right) - 4 = 0$$
this equation is fora type one-sided hyperboloid
- reduced to canonical form
$$4 x^{2} - 2 x y - 3 y^{2} + z^{2} + 6 z + 5 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = -1$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = -3$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 1$$
$$a_{34} = 3$$
$$a_{44} = 5$$
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
$$I_{1} = 2$$
|4 -1| |-3 0| |4 0|
I2 = | | + | | + | |
|-1 -3| |0 1| |0 1|$$I_{3} = \left|\begin{matrix}4 & -1 & 0\\-1 & -3 & 0\\0 & 0 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}4 & -1 & 0 & 0\\-1 & -3 & 0 & 0\\0 & 0 & 1 & 3\\0 & 0 & 3 & 5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & -1 & 0\\-1 & - \lambda - 3 & 0\\0 & 0 & 1 - \lambda\end{matrix}\right|$$
|4 0| |-3 0| |1 3|
K2 = | | + | | + | |
|0 5| |0 5| |3 5| |4 -1 0| |-3 0 0| |4 0 0|
| | | | | |
K3 = |-1 -3 0| + |0 1 3| + |0 1 3|
| | | | | |
|0 0 5| |0 3 5| |0 3 5|$$I_{1} = 2$$
$$I_{2} = -12$$
$$I_{3} = -13$$
$$I_{4} = 52$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 2 \lambda^{2} + 12 \lambda - 13$$
$$K_{2} = 1$$
$$K_{3} = -69$$
Because
I3 != 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
$$\lambda^{3} - 2 \lambda^{2} - 12 \lambda + 13 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = \frac{1}{2} - \frac{\sqrt{53}}{2}$$
$$\lambda_{3} = \frac{1}{2} + \frac{\sqrt{53}}{2}$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\tilde x^{2} + \tilde y^{2} \left(\frac{1}{2} - \frac{\sqrt{53}}{2}\right) + \tilde z^{2} \left(\frac{1}{2} + \frac{\sqrt{53}}{2}\right) - 4 = 0$$
2 2 2
\tilde x \tilde z \tilde y
--------- + ------------------------ - -------------------------- = 1
2 2 2
/ 1\ / 1 \ / 1 \
\2 / |---------------------| |-----------------------|
| ____________ | | ______________ |
| / ____ | | / ____ |
| / 1 \/ 53 | | / 1 \/ 53 |
| / - + ------ *1/2| | / - - + ------ *1/2|
\\/ 2 2 / \\/ 2 2 / this equation is fora type one-sided hyperboloid
- reduced to canonical form