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:
- 2x^2-4xy+5y^2+8x-2y+9=0
- 2x^2+4y=6
- 7x^2+(sqrt3)xy+6y^2=1
- 9x^2+-4xy+6y^2+16x-8y-2=0
- Plot:
- tan(x)=-x/10
- sqrt(36*x^2)/sqrt(49)
-
sqrt(2-x^2)=sqrt(2-y^2)
- sqrt(2*x+1)*x-y-1=0
-
Identical expressions
- two x^2+4y= six
- 2x squared plus 4y equally 6
- two x squared plus 4y equally six
- 2x2+4y=6
- 2x²+4y=6
- 2x to the power of 2+4y=6
-
Similar expressions
- 2x^2+4y^2+4x-8y-10=0
- 2x^2-4y=6
2x^2+4y=6 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$2 x^{2} + 4 y - 6 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = 2$$
$$a_{33} = -6$$
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
$$2 \tilde x^{2} = 6 - 4 \tilde y$$
$$\tilde x^{2} = 3 - 2 \tilde y$$
$$\tilde x'^{2} = 3 - 2 \tilde y$$
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
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$2 x^{2} + 4 y - 6 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = 2$$
$$a_{33} = -6$$
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
$$2 \tilde x^{2} = 6 - 4 \tilde y$$
$$\tilde x^{2} = 3 - 2 \tilde y$$
$$\tilde x'^{2} = 3 - 2 \tilde y$$
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:
$$2 x^{2} + 4 y - 6 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = 2$$
$$a_{33} = -6$$
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
$$I_{1} = 2$$
$$I_{3} = \left|\begin{matrix}2 & 0 & 0\\0 & 0 & 2\\0 & 2 & -6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 2$$
$$I_{2} = 0$$
$$I_{3} = -8$$
$$I{\left(\lambda \right)} = \lambda^{2} - 2 \lambda$$
$$K_{2} = -16$$
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
$$4 \tilde x + 2 \tilde y^{2} = 0$$
$$\tilde y^{2} = 2 \tilde x$$
- reduced to canonical form
$$2 x^{2} + 4 y - 6 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = 2$$
$$a_{33} = -6$$
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|$$I_{3} = \left|\begin{matrix}2 & 0 & 0\\0 & 0 & 2\\0 & 2 & -6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
|2 0 | |0 2 |
K2 = | | + | |
|0 -6| |2 -6|$$I_{1} = 2$$
$$I_{2} = 0$$
$$I_{3} = -8$$
$$I{\left(\lambda \right)} = \lambda^{2} - 2 \lambda$$
$$K_{2} = -16$$
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
$$4 \tilde x + 2 \tilde y^{2} = 0$$
$$\tilde y^{2} = 2 \tilde x$$
- reduced to canonical form