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
-
How to use it?
- Canonical form:
- y^2+z+1=0
- 2x^2+y^2-4xy-4yz+12x-12y+18=0
- sqrt(x)/x
- x+1/2
- Plot:
- 5*y^2-3*x=0
- z=sqrt((x+1)*(y-1))
- y=sin(x^(cos(y)))
-
sin(x-2.2*y)-y*x+1=0
- Derivative of:
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x+1/2
-
Identical expressions
- x+ one / two
- x plus 1 divide by 2
- x plus one divide by two
- x+1 divide by 2
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Similar expressions
- xy-2xz+yz+x+(1/2)y=0
- x-1/2
- sqrt(15-x^2+2x)+1/2-1=0
- ((17/20)+(1/10)*x+(1/20)*x^3)^3
x+1/2 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$x + \frac{1}{2} = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{1}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = \frac{1}{2}$$
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
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 1 + 0 \cdot 0$$
$$y_{0} = 0 \cdot 0 + 0 \cdot 1$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$x + \frac{1}{2} = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{1}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = \frac{1}{2}$$
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
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 1 + 0 \cdot 0$$
$$y_{0} = 0 \cdot 0 + 0 \cdot 1$$
$$x_{0} = 0$$
$$y_{0} = 0$$
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:
$$x + \frac{1}{2} = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{1}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = \frac{1}{2}$$
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
$$I_{1} = 0$$
$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{1}{2}\\0 & 0 & 0\\\frac{1}{2} & 0 & \frac{1}{2}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 0$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2}$$
$$K_{2} = - \frac{1}{4}$$
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
$$I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
- reduced to canonical form
$$x + \frac{1}{2} = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{1}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = \frac{1}{2}$$
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|$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{1}{2}\\0 & 0 & 0\\\frac{1}{2} & 0 & \frac{1}{2}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
| 0 1/2| |0 0 |
K2 = | | + | |
|1/2 1/2| |0 1/2|$$I_{1} = 0$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2}$$
$$K_{2} = - \frac{1}{4}$$
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
$$I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
False
None
- reduced to canonical form