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:
- 2*x^2-y^2+5*z^2-10=0
- (y-2)^2=-6(x+1)
- 8y^2+4xy+2xz+4yz+4x+8y-9=0
- 2x^2+2y^2+2z^2+4xy-4xz+4yz
- Plot:
-
x^2+(y+(x)^(2/3))^2=1
-
tg(x)^arctg(x)=e^((y(x))/x)
- x^2+(y-x^2/3)^2=1
-
-x^2*y^3+(x^2+y^2-1)^3=0
- Graphing y =:
- (y-2)^2
- (y-2)^2
-
Identical expressions
- (y- two)^ two =- six (x+ one)
- (y minus 2) squared equally minus 6(x plus 1)
- (y minus two) to the power of two equally minus six (x plus one)
- (y-2)2=-6(x+1)
- y-22=-6x+1
- (y-2)²=-6(x+1)
- (y-2) to the power of 2=-6(x+1)
- y-2^2=-6x+1
-
Similar expressions
- x^3-6*x+1
- (y-2)^2=+6(x+1)
- (y-2)^2=-6(x-1)
- y^2=-6x+12
- 5x^2+2xy+16xz+2zy+13z^2-8z+2y^2-6x+1=0
- (y+2)^2=-6(x+1)
- 2x^2+2y^2-6x+10y+7=0
(y-2)^2=-6(x+1) canonical form
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The solution
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2 6 + (-2 + y) + 6*x = 0
$$6 x + \left(y - 2\right)^{2} + 6 = 0$$
6*x + (y - 2)^2 + 6 = 0
Detail solution
Given line equation of 2-order:
$$6 x + \left(y - 2\right)^{2} + 6 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 3$$
$$a_{22} = 1$$
$$a_{23} = -2$$
$$a_{33} = 10$$
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
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
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$6 x + \left(y - 2\right)^{2} + 6 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 3$$
$$a_{22} = 1$$
$$a_{23} = -2$$
$$a_{33} = 10$$
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
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:
$$6 x + \left(y - 2\right)^{2} + 6 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 3$$
$$a_{22} = 1$$
$$a_{23} = -2$$
$$a_{33} = 10$$
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
$$I_{1} = 1$$
$$I_{3} = \left|\begin{matrix}0 & 0 & 3\\0 & 1 & -2\\3 & -2 & 10\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = -9$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda$$
$$K_{2} = -3$$
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
$$6 \tilde x + \tilde y^{2} = 0$$
$$\tilde y^{2} = 6 \tilde x$$
- reduced to canonical form
$$6 x + \left(y - 2\right)^{2} + 6 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 3$$
$$a_{22} = 1$$
$$a_{23} = -2$$
$$a_{33} = 10$$
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|$$I_{3} = \left|\begin{matrix}0 & 0 & 3\\0 & 1 & -2\\3 & -2 & 10\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|$$
|0 3 | |1 -2|
K2 = | | + | |
|3 10| |-2 10|$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = -9$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda$$
$$K_{2} = -3$$
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
$$6 \tilde x + \tilde y^{2} = 0$$
$$\tilde y^{2} = 6 \tilde x$$
- reduced to canonical form