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:
- 25x^2-120xy+144y^2-242x-298y+491=0
- 11x^2+24xy+4y^2+42x+64y+51=0
-
13x^2-8y^2+20xy-46x-4y=26
- 3y^2+3x^2=z
- Plot:
-
x^2*y=x^2-y^2
- y=x*(x+1)*(x+2)*(x+3)
-
y=x-sqrt(abs(y))
- x+y=-2
-
Identical expressions
- 3y^ two +3x^ two =z
- 3y squared plus 3x squared equally z
- 3y to the power of two plus 3x to the power of two equally z
- 3y2+3x2=z
- 3y²+3x²=z
- 3y to the power of 2+3x to the power of 2=z
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Similar expressions
- 3y^2-3x^2=z
3y^2+3x^2=z 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:
$$3 x^{2} + 3 y^{2} - z = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = - \frac{1}{2}$$
$$a_{44} = 0$$
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
$$I_{1} = 6$$
$$I_{3} = \left|\begin{matrix}3 & 0 & 0\\0 & 3 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}3 & 0 & 0 & 0\\0 & 3 & 0 & 0\\0 & 0 & 0 & - \frac{1}{2}\\0 & 0 & - \frac{1}{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0 & 0\\0 & 3 - \lambda & 0\\0 & 0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 6$$
$$I_{2} = 9$$
$$I_{3} = 0$$
$$I_{4} = - \frac{9}{4}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 6 \lambda^{2} - 9 \lambda$$
$$K_{2} = - \frac{1}{4}$$
$$K_{3} = - \frac{3}{2}$$
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
$$\lambda^{3} - 6 \lambda^{2} + 9 \lambda = 0$$
Solve this equation
$$\lambda_{1} = 3$$
$$\lambda_{2} = 3$$
$$\lambda_{3} = 0$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$3 \tilde x^{2} + 3 \tilde y^{2} + \tilde z = 0$$
and
$$3 \tilde x^{2} + 3 \tilde y^{2} - \tilde z = 0$$
$$6 \tilde x^{2} + 6 \tilde y^{2} + 2 \tilde z = 0$$
and
$$6 \tilde x^{2} + 6 \tilde y^{2} - 2 \tilde z = 0$$
this equation is fora type elliptical paraboloid
- reduced to canonical form
$$3 x^{2} + 3 y^{2} - z = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = - \frac{1}{2}$$
$$a_{44} = 0$$
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
$$I_{1} = 6$$
|3 0| |3 0| |3 0|
I2 = | | + | | + | |
|0 3| |0 0| |0 0|$$I_{3} = \left|\begin{matrix}3 & 0 & 0\\0 & 3 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}3 & 0 & 0 & 0\\0 & 3 & 0 & 0\\0 & 0 & 0 & - \frac{1}{2}\\0 & 0 & - \frac{1}{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0 & 0\\0 & 3 - \lambda & 0\\0 & 0 & - \lambda\end{matrix}\right|$$
|3 0| |3 0| | 0 -1/2|
K2 = | | + | | + | |
|0 0| |0 0| |-1/2 0 | |3 0 0| |3 0 0 | |3 0 0 |
| | | | | |
K3 = |0 3 0| + |0 0 -1/2| + |0 0 -1/2|
| | | | | |
|0 0 0| |0 -1/2 0 | |0 -1/2 0 |$$I_{1} = 6$$
$$I_{2} = 9$$
$$I_{3} = 0$$
$$I_{4} = - \frac{9}{4}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 6 \lambda^{2} - 9 \lambda$$
$$K_{2} = - \frac{1}{4}$$
$$K_{3} = - \frac{3}{2}$$
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
$$\lambda^{3} - 6 \lambda^{2} + 9 \lambda = 0$$
Solve this equation
$$\lambda_{1} = 3$$
$$\lambda_{2} = 3$$
$$\lambda_{3} = 0$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$3 \tilde x^{2} + 3 \tilde y^{2} + \tilde z = 0$$
and
$$3 \tilde x^{2} + 3 \tilde y^{2} - \tilde z = 0$$
$$6 \tilde x^{2} + 6 \tilde y^{2} + 2 \tilde z = 0$$
and
$$6 \tilde x^{2} + 6 \tilde y^{2} - 2 \tilde z = 0$$
this equation is fora type elliptical paraboloid
- reduced to canonical form