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:
- 5*x^2+4*y^2-4*z^2-2*x+2*y-4*z-3*y*z-3*x*z-x*y=0
- x^2-2x+y^2+6y-2=0
- y^2-5*x-8*y-14=0
- 2x^2=y^2+1
- Plot:
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23*x-4*y+37=-x-sqrt(x^2-4*y)
-
2*x-4*y+3=-x-sqrt(x^2-4*y)
-
10+15x-18y+2x^2-5xy+3y^2
- (|x|)*(x-1)-5*x*y=(|x|)*(x-1)-5*x
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Identical expressions
- f=5x+2x-3x
- f equally 5x plus 2x minus 3x
-
Similar expressions
- f=5x-2x-3x
- f=5x+2x+3x
f=5x+2x-3x canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$f - 4 x = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} f x + 2 a_{13} x + a_{22} f^{2} + 2 a_{23} f + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = -2$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
Given equation is straight line
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = - \tilde f \sin{\left(\phi \right)} + \tilde x \cos{\left(\phi \right)}$$
$$f_{0} = \tilde f \cos{\left(\phi \right)} + \tilde x \sin{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$f_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$f_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$f - 4 x = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} f x + 2 a_{13} x + a_{22} f^{2} + 2 a_{23} f + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = -2$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
Given equation is straight line
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = - \tilde f \sin{\left(\phi \right)} + \tilde x \cos{\left(\phi \right)}$$
$$f_{0} = \tilde f \cos{\left(\phi \right)} + \tilde x \sin{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$f_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$f_{0} = 0$$
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
Invariants method
Given line equation of 2-order:
$$f - 4 x = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} f x + 2 a_{13} x + a_{22} f^{2} + 2 a_{23} f + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = -2$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 0$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
substitute coefficients
$$I_{1} = 0$$
$$I_{3} = \left|\begin{matrix}0 & 0 & -2\\0 & 0 & \frac{1}{2}\\-2 & \frac{1}{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 0$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2}$$
$$K_{2} = - \frac{17}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge \left(I_{1} = 0 \vee K_{2} = 0\right)$$
then by line type:
this equation is of type : two coincident straight lines
$$I_{1} \tilde f^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
- reduced to canonical form
$$f - 4 x = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} f x + 2 a_{13} x + a_{22} f^{2} + 2 a_{23} f + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = -2$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 0$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
|a11 a12|
I2 = | |
|a12 a22|$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
|a11 a13| |a22 a23|
K2 = | | + | |
|a13 a33| |a23 a33|substitute coefficients
$$I_{1} = 0$$
|0 0|
I2 = | |
|0 0|$$I_{3} = \left|\begin{matrix}0 & 0 & -2\\0 & 0 & \frac{1}{2}\\-2 & \frac{1}{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
|0 -2| | 0 1/2|
K2 = | | + | |
|-2 0 | |1/2 0 |$$I_{1} = 0$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2}$$
$$K_{2} = - \frac{17}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge \left(I_{1} = 0 \vee K_{2} = 0\right)$$
then by line type:
this equation is of type : two coincident straight lines
$$I_{1} \tilde f^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
False
None
- reduced to canonical form