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:
- 4x^2-9y^2+8x-32=0
- 4*x1^2+2*x2^2+14*x3^2-4*x1x2-6*x2x3
- -2x^2+8xy+y^2-2xy-8z+x^2+y^2=0
- 9x^2+24xy+16y^2-24x+110y-475=0
- Plot:
- (x^2+y^2-1)^3-x^2*y^3=0
- (x-3)*(3*x-9-y)=(|x-3|)^3
-
(x-2)y^2=1
- (x-1)^2+y^2=5
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Identical expressions
- y^ two +z- one = zero
- y squared plus z minus 1 equally 0
- y to the power of two plus z minus one equally zero
- y2+z-1=0
- y²+z-1=0
- y to the power of 2+z-1=0
- y^2+z-1=O
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Similar expressions
- y^2-z-1=0
- y^2+z+1=0
y^2+z-1=0 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:
$$y^{2} + z - 1 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = \frac{1}{2}$$
$$a_{44} = -1$$
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
$$I_{1} = 1$$
$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 1 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 0 & \frac{1}{2}\\0 & 0 & \frac{1}{2} & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & 1 - \lambda & 0\\0 & 0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \lambda^{2}$$
$$K_{2} = - \frac{5}{4}$$
$$K_{3} = - \frac{1}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge I_{4} = 0 \wedge I_{1} \neq 0 \wedge K_{3} \neq 0$$
then by type of surface:
you need to
then the canonical form of the equation will be
$$I_{1} \tilde x^{2} + \tilde y 2 \sqrt{\frac{\left(-1\right) K_{3}}{I_{1}}} = 0$$
and
$$I_{1} \tilde x^{2} - \tilde y 2 \sqrt{\frac{\left(-1\right) K_{3}}{I_{1}}} = 0$$
$$\tilde x^{2} + \tilde y = 0$$
and
$$\tilde x^{2} - \tilde y = 0$$
$$\tilde x^{2} = \tilde y$$
and
$$\tilde x^{2} = - \tilde y$$
this equation is fora type parabolic cylinder
- reduced to canonical form
$$y^{2} + z - 1 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = \frac{1}{2}$$
$$a_{44} = -1$$
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
$$I_{1} = 1$$
|0 0| |1 0| |0 0|
I2 = | | + | | + | |
|0 1| |0 0| |0 0|$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 1 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 0 & \frac{1}{2}\\0 & 0 & \frac{1}{2} & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & 1 - \lambda & 0\\0 & 0 & - \lambda\end{matrix}\right|$$
|0 0 | |1 0 | | 0 1/2|
K2 = | | + | | + | |
|0 -1| |0 -1| |1/2 -1 | |0 0 0 | |1 0 0 | |0 0 0 |
| | | | | |
K3 = |0 1 0 | + |0 0 1/2| + |0 0 1/2|
| | | | | |
|0 0 -1| |0 1/2 -1 | |0 1/2 -1 |$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \lambda^{2}$$
$$K_{2} = - \frac{5}{4}$$
$$K_{3} = - \frac{1}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge I_{4} = 0 \wedge I_{1} \neq 0 \wedge K_{3} \neq 0$$
then by type of surface:
you need to
then the canonical form of the equation will be
$$I_{1} \tilde x^{2} + \tilde y 2 \sqrt{\frac{\left(-1\right) K_{3}}{I_{1}}} = 0$$
and
$$I_{1} \tilde x^{2} - \tilde y 2 \sqrt{\frac{\left(-1\right) K_{3}}{I_{1}}} = 0$$
$$\tilde x^{2} + \tilde y = 0$$
and
$$\tilde x^{2} - \tilde y = 0$$
$$\tilde x^{2} = \tilde y$$
and
$$\tilde x^{2} = - \tilde y$$
this equation is fora type parabolic cylinder
- reduced to canonical form