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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:
- x2+y2−4y+3=0
- 2x^2+2y^2-8sqrt(2)y
- 9*x^2+18*x+18xy+22y+13y^2=11
- 3x^2+5y^2+3z^2+2*x*z-2*x*y-2y*z=0
- Plot:
- (|y|)=(|x^2-x+1|)
- x^2+y^2=4
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3*cos(x)*sin(y)-sin(x)*cos(y)-4*cos(y)=3*sqrt(3)
- x=x-sqrt(abs(y))
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Identical expressions
- four *x-y^ two + two *y+ seven = zero
- 4 multiply by x minus y squared plus 2 multiply by y plus 7 equally 0
- four multiply by x minus y to the power of two plus two multiply by y plus seven equally zero
- 4*x-y2+2*y+7=0
- 4*x-y²+2*y+7=0
- 4*x-y to the power of 2+2*y+7=0
- 4x-y^2+2y+7=0
- 4x-y2+2y+7=0
- 4*x-y^2+2*y+7=O
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Similar expressions
- 4*x-y^2-2*y+7=0
- 4*x-y^2+2*y-7=0
- 4*x+y^2+2*y+7=0
4*x-y^2+2*y+7=0 canonical form
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The solution
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2 7 - y + 2*y + 4*x = 0
$$4 x - y^{2} + 2 y + 7 = 0$$
4*x - y^2 + 2*y + 7 = 0
Detail solution
Given line equation of 2-order:
$$4 x - y^{2} + 2 y + 7 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 2$$
$$a_{22} = -1$$
$$a_{23} = 1$$
$$a_{33} = 7$$
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
$$\left(\tilde y + 1\right)^{2} = 4 \tilde x + 8$$
$$\tilde y'^{2} = 4 \tilde x + 8$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 1 + 0 \cdot 0$$
$$y_{0} = 0 \cdot 0 + 0 \cdot 1$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$4 x - y^{2} + 2 y + 7 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 2$$
$$a_{22} = -1$$
$$a_{23} = 1$$
$$a_{33} = 7$$
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
$$\left(\tilde y + 1\right)^{2} = 4 \tilde x + 8$$
$$\tilde y'^{2} = 4 \tilde x + 8$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 1 + 0 \cdot 0$$
$$y_{0} = 0 \cdot 0 + 0 \cdot 1$$
$$x_{0} = 0$$
$$y_{0} = 0$$
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:
$$4 x - y^{2} + 2 y + 7 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 2$$
$$a_{22} = -1$$
$$a_{23} = 1$$
$$a_{33} = 7$$
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
$$I_{1} = -1$$
$$I_{3} = \left|\begin{matrix}0 & 0 & 2\\0 & -1 & 1\\2 & 1 & 7\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & - \lambda - 1\end{matrix}\right|$$
$$I_{1} = -1$$
$$I_{2} = 0$$
$$I_{3} = 4$$
$$I{\left(\lambda \right)} = \lambda^{2} + \lambda$$
$$K_{2} = -12$$
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
$$4 \tilde x - \tilde y^{2} = 0$$
$$\tilde y^{2} = - 4 \tilde x$$
- reduced to canonical form
$$4 x - y^{2} + 2 y + 7 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 2$$
$$a_{22} = -1$$
$$a_{23} = 1$$
$$a_{33} = 7$$
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|$$I_{3} = \left|\begin{matrix}0 & 0 & 2\\0 & -1 & 1\\2 & 1 & 7\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & - \lambda - 1\end{matrix}\right|$$
|0 2| |-1 1|
K2 = | | + | |
|2 7| |1 7|$$I_{1} = -1$$
$$I_{2} = 0$$
$$I_{3} = 4$$
$$I{\left(\lambda \right)} = \lambda^{2} + \lambda$$
$$K_{2} = -12$$
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
$$4 \tilde x - \tilde y^{2} = 0$$
$$\tilde y^{2} = - 4 \tilde x$$
- reduced to canonical form
The graph