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- 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:
- z=x^2-xy+y^2
- 7x^2+16xy-23y^2-14x-16y-218=0
- -4x^2+16y^2-64=0
- x1^2+6x1x2-2x1x3+5x2^2-18x2x3-10x3^2
- Plot:
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e^(y/x)=x*y+1
- |x+2/2-x|
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|3*y-x|+2=|y-4|
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|x-y+2|=1
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Identical expressions
- y^ two +6y+2x- one = zero
- y squared plus 6y plus 2x minus 1 equally 0
- y to the power of two plus 6y plus 2x minus one equally zero
- y2+6y+2x-1=0
- y²+6y+2x-1=0
- y to the power of 2+6y+2x-1=0
- y^2+6y+2x-1=O
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Similar expressions
- y^2+6y-2x-1=0
- y^2-6y+2x-1=0
- y^2+6y+2x+1=0
y^2+6y+2x-1=0 canonical form
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2 -1 + y + 2*x + 6*y = 0
$$2 x + y^{2} + 6 y - 1 = 0$$
2*x + y^2 + 6*y - 1 = 0
Detail solution
Given line equation of 2-order:
$$2 x + y^{2} + 6 y - 1 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 1$$
$$a_{22} = 1$$
$$a_{23} = 3$$
$$a_{33} = -1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 1\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\left(\tilde y + 3\right)^{2} = 10 - 2 \tilde x$$
$$\tilde y'^{2} = 10 - 2 \tilde x$$
Given equation is by parabola
- 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
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$2 x + y^{2} + 6 y - 1 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 1$$
$$a_{22} = 1$$
$$a_{23} = 3$$
$$a_{33} = -1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 1\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\left(\tilde y + 3\right)^{2} = 10 - 2 \tilde x$$
$$\tilde y'^{2} = 10 - 2 \tilde x$$
Given equation is by parabola
- 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:
$$2 x + y^{2} + 6 y - 1 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 1$$
$$a_{22} = 1$$
$$a_{23} = 3$$
$$a_{33} = -1$$
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
$$I_{1} = 1$$
$$I_{3} = \left|\begin{matrix}0 & 0 & 1\\0 & 1 & 3\\1 & 3 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = -1$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda$$
$$K_{2} = -11$$
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
$$2 \tilde x + \tilde y^{2} = 0$$
$$\tilde y^{2} = 2 \tilde x$$
- reduced to canonical form
$$2 x + y^{2} + 6 y - 1 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 1$$
$$a_{22} = 1$$
$$a_{23} = 3$$
$$a_{33} = -1$$
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
$$I_{1} = 1$$
|0 0|
I2 = | |
|0 1|$$I_{3} = \left|\begin{matrix}0 & 0 & 1\\0 & 1 & 3\\1 & 3 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|$$
|0 1 | |1 3 |
K2 = | | + | |
|1 -1| |3 -1|$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = -1$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda$$
$$K_{2} = -11$$
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
$$2 \tilde x + \tilde y^{2} = 0$$
$$\tilde y^{2} = 2 \tilde x$$
- reduced to canonical form