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:
- (x2+y2–1)3–x2y3=0
- sin^2x=1
- ax^2+ay^2+z^2+2axy=0
-
5x^2 + 5y^2 + 6xy – 6x – 10y – 3 = 0
- Plot:
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y=x^4-26x^2+25
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xy=0
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xy=6
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4x^2+4y^2=25
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Identical expressions
- (x2+y2– one) three –x2y3= zero
- (x2 plus y2–1)3–x2y3 equally 0
- (x2 plus y2– one) three –x2y3 equally zero
- x2+y2–13–x2y3=0
- (x2+y2–1)3–x2y3=O
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Similar expressions
- (x2-y2–1)3–x2y3=0
(x2+y2–1)3–x2y3=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
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-3 + 3*x2 + 3*y2 - x2*y3 = 0
$$- x_{2} y_{3} + 3 x_{2} + 3 y_{2} - 3 = 0$$
-x2*y3 + 3*x2 + 3*y2 - 3 = 0
Invariants method
Given equation of the surface of 2-order:
$$- x_{2} y_{3} + 3 x_{2} + 3 y_{2} - 3 = 0$$
This equation looks like:
$$a_{11} y_{3}^{2} + 2 a_{12} y_{2} y_{3} + 2 a_{13} x_{2} y_{3} + 2 a_{14} y_{3} + a_{22} y_{2}^{2} + 2 a_{23} x_{2} y_{2} + 2 a_{24} y_{2} + a_{33} x_{2}^{2} + 2 a_{34} x_{2} + a_{44} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = - \frac{1}{2}$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = \frac{3}{2}$$
$$a_{33} = 0$$
$$a_{34} = \frac{3}{2}$$
$$a_{44} = -3$$
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
$$I_{1} = 0$$
$$I_{3} = \left|\begin{matrix}0 & 0 & - \frac{1}{2}\\0 & 0 & 0\\- \frac{1}{2} & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & - \frac{1}{2} & 0\\0 & 0 & 0 & \frac{3}{2}\\- \frac{1}{2} & 0 & 0 & \frac{3}{2}\\0 & \frac{3}{2} & \frac{3}{2} & -3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & - \frac{1}{2}\\0 & - \lambda & 0\\- \frac{1}{2} & 0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 0$$
$$I_{2} = - \frac{1}{4}$$
$$I_{3} = 0$$
$$I_{4} = \frac{9}{16}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{\lambda}{4}$$
$$K_{2} = - \frac{9}{2}$$
$$K_{3} = \frac{3}{4}$$
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
$$\lambda^{3} - \frac{\lambda}{4} = 0$$
Solve this equation
$$\lambda_{1} = - \frac{1}{2}$$
$$\lambda_{2} = \frac{1}{2}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\tilde x2 \cdot 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde y2^{2} \lambda_{2} + \tilde y3^{2} \lambda_{1}\right) = 0$$
and
$$- \tilde x2 \cdot 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde y2^{2} \lambda_{2} + \tilde y3^{2} \lambda_{1}\right) = 0$$
$$3 \tilde x2 + \frac{\tilde y2^{2}}{2} - \frac{\tilde y3^{2}}{2} = 0$$
and
$$- 3 \tilde x2 + \frac{\tilde y2^{2}}{2} - \frac{\tilde y3^{2}}{2} = 0$$
$$- 2 \tilde x2 - \left(\frac{\tilde y2^{2}}{3} - \frac{\tilde y3^{2}}{3}\right) = 0$$
and
$$2 \tilde x2 - \left(\frac{\tilde y2^{2}}{3} - \frac{\tilde y3^{2}}{3}\right) = 0$$
this equation is fora type hyperbolic paraboloid
- reduced to canonical form
$$- x_{2} y_{3} + 3 x_{2} + 3 y_{2} - 3 = 0$$
This equation looks like:
$$a_{11} y_{3}^{2} + 2 a_{12} y_{2} y_{3} + 2 a_{13} x_{2} y_{3} + 2 a_{14} y_{3} + a_{22} y_{2}^{2} + 2 a_{23} x_{2} y_{2} + 2 a_{24} y_{2} + a_{33} x_{2}^{2} + 2 a_{34} x_{2} + a_{44} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = - \frac{1}{2}$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = \frac{3}{2}$$
$$a_{33} = 0$$
$$a_{34} = \frac{3}{2}$$
$$a_{44} = -3$$
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 0| |0 0| | 0 -1/2|
I2 = | | + | | + | |
|0 0| |0 0| |-1/2 0 |$$I_{3} = \left|\begin{matrix}0 & 0 & - \frac{1}{2}\\0 & 0 & 0\\- \frac{1}{2} & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & - \frac{1}{2} & 0\\0 & 0 & 0 & \frac{3}{2}\\- \frac{1}{2} & 0 & 0 & \frac{3}{2}\\0 & \frac{3}{2} & \frac{3}{2} & -3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & - \frac{1}{2}\\0 & - \lambda & 0\\- \frac{1}{2} & 0 & - \lambda\end{matrix}\right|$$
|0 0 | | 0 3/2| | 0 3/2|
K2 = | | + | | + | |
|0 -3| |3/2 -3 | |3/2 -3 | |0 0 0 | | 0 0 3/2| | 0 -1/2 0 |
| | | | | |
K3 = |0 0 3/2| + | 0 0 3/2| + |-1/2 0 3/2|
| | | | | |
|0 3/2 -3 | |3/2 3/2 -3 | | 0 3/2 -3 |$$I_{1} = 0$$
$$I_{2} = - \frac{1}{4}$$
$$I_{3} = 0$$
$$I_{4} = \frac{9}{16}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{\lambda}{4}$$
$$K_{2} = - \frac{9}{2}$$
$$K_{3} = \frac{3}{4}$$
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
$$\lambda^{3} - \frac{\lambda}{4} = 0$$
Solve this equation
$$\lambda_{1} = - \frac{1}{2}$$
$$\lambda_{2} = \frac{1}{2}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\tilde x2 \cdot 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde y2^{2} \lambda_{2} + \tilde y3^{2} \lambda_{1}\right) = 0$$
and
$$- \tilde x2 \cdot 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde y2^{2} \lambda_{2} + \tilde y3^{2} \lambda_{1}\right) = 0$$
$$3 \tilde x2 + \frac{\tilde y2^{2}}{2} - \frac{\tilde y3^{2}}{2} = 0$$
and
$$- 3 \tilde x2 + \frac{\tilde y2^{2}}{2} - \frac{\tilde y3^{2}}{2} = 0$$
$$- 2 \tilde x2 - \left(\frac{\tilde y2^{2}}{3} - \frac{\tilde y3^{2}}{3}\right) = 0$$
and
$$2 \tilde x2 - \left(\frac{\tilde y2^{2}}{3} - \frac{\tilde y3^{2}}{3}\right) = 0$$
this equation is fora type hyperbolic paraboloid
- reduced to canonical form