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- Canonical form of a double hyperboloid
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- Canonical form of a parabola
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How to use it?
- Canonical form:
- 0x^2+xy+0y^2+2x+y+2=0
-
x^2+y^2+6x-4y+12=0
- 2x^2+3y^2-4x+12y-12z+2=0
- y=-2x^2+8x-6
- Plot:
- (|y|)=(|x^2-x+1|)
-
arctg(x/y)=x*y
- tan(x*factorial(y))-1=0
-
sqrtx+sqrty=4
- Integral of d{x}:
- -2x^2+8x-6
-
Identical expressions
- y=- two x^2+8x- six
- y equally minus 2x squared plus 8x minus 6
- y equally minus two x squared plus 8x minus six
- y=-2x2+8x-6
- y=-2x²+8x-6
- y=-2x to the power of 2+8x-6
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Similar expressions
- y=+2x^2+8x-6
- y=-2x^2+8x+6
- y=-2x^2-8x-6
y=-2x^2+8x-6 canonical form
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The solution
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2 6 + y - 8*x + 2*x = 0
$$2 x^{2} - 8 x + y + 6 = 0$$
2*x^2 - 8*x + y + 6 = 0
Detail solution
Given line equation of 2-order:
$$2 x^{2} - 8 x + y + 6 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 6$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}2 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
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
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$y_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$2 x^{2} - 8 x + y + 6 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 6$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}2 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
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
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$y_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$y_{0} = 0$$
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:
$$2 x^{2} - 8 x + y + 6 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 6$$
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
$$I_{1} = 2$$
$$I_{3} = \left|\begin{matrix}2 & 0 & -4\\0 & 0 & \frac{1}{2}\\-4 & \frac{1}{2} & 6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 2$$
$$I_{2} = 0$$
$$I_{3} = - \frac{1}{2}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 2 \lambda$$
$$K_{2} = - \frac{17}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : parabola
$$I_{1} \tilde y^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0$$
or
$$\tilde x + 2 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{\tilde x}{2}$$
- reduced to canonical form
$$2 x^{2} - 8 x + y + 6 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 6$$
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
$$I_{1} = 2$$
|2 0|
I2 = | |
|0 0|$$I_{3} = \left|\begin{matrix}2 & 0 & -4\\0 & 0 & \frac{1}{2}\\-4 & \frac{1}{2} & 6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
|2 -4| | 0 1/2|
K2 = | | + | |
|-4 6 | |1/2 6 |$$I_{1} = 2$$
$$I_{2} = 0$$
$$I_{3} = - \frac{1}{2}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 2 \lambda$$
$$K_{2} = - \frac{17}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : parabola
$$I_{1} \tilde y^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0$$
or
$$\tilde x + 2 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{\tilde x}{2}$$
- reduced to canonical form