Mister Exam
2y^2+x-y-1=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$x + 2 y^{2} - y - 1 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{1}{2}$$
$$a_{22} = 2$$
$$a_{23} = - \frac{1}{2}$$
$$a_{33} = -1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 2\end{matrix}\right|$$
$$\Delta = 0$$
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
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$x + 2 y^{2} - y - 1 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{1}{2}$$
$$a_{22} = 2$$
$$a_{23} = - \frac{1}{2}$$
$$a_{33} = -1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 2\end{matrix}\right|$$
$$\Delta = 0$$
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:
$$x + 2 y^{2} - y - 1 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{1}{2}$$
$$a_{22} = 2$$
$$a_{23} = - \frac{1}{2}$$
$$a_{33} = -1$$
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
$$I_{1} = 2$$
$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{1}{2}\\0 & 2 & - \frac{1}{2}\\\frac{1}{2} & - \frac{1}{2} & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 2 - \lambda\end{matrix}\right|$$
$$I_{1} = 2$$
$$I_{2} = 0$$
$$I_{3} = - \frac{1}{2}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 2 \lambda$$
$$K_{2} = - \frac{5}{2}$$
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
$$x + 2 y^{2} - y - 1 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{1}{2}$$
$$a_{22} = 2$$
$$a_{23} = - \frac{1}{2}$$
$$a_{33} = -1$$
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
$$I_{1} = 2$$
|0 0|
I2 = | |
|0 2|$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{1}{2}\\0 & 2 & - \frac{1}{2}\\\frac{1}{2} & - \frac{1}{2} & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 2 - \lambda\end{matrix}\right|$$
| 0 1/2| | 2 -1/2|
K2 = | | + | |
|1/2 -1 | |-1/2 -1 |$$I_{1} = 2$$
$$I_{2} = 0$$
$$I_{3} = - \frac{1}{2}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 2 \lambda$$
$$K_{2} = - \frac{5}{2}$$
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