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y^2+x-4y+5=0 canonical form

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5 + x + y  - 4*y = 0
$$x + y^{2} - 4 y + 5 = 0$$
x + y^2 - 4*y + 5 = 0
Detail solution
Given line equation of 2-order:
$$x + y^{2} - 4 y + 5 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{1}{2}$$
$$a_{22} = 1$$
$$a_{23} = -2$$
$$a_{33} = 5$$
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
$$\left(\tilde y - 2\right)^{2} = - \tilde x - 1$$
$$\tilde y'^{2} = - \tilde x'$$
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:
$$x + y^{2} - 4 y + 5 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{1}{2}$$
$$a_{22} = 1$$
$$a_{23} = -2$$
$$a_{33} = 5$$
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|

$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{1}{2}\\0 & 1 & -2\\\frac{1}{2} & -2 & 5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|$$
     | 0   1/2|   |1   -2|
K2 = |        | + |      |
     |1/2   5 |   |-2  5 |

$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = - \frac{1}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda$$
$$K_{2} = \frac{3}{4}$$
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
$$\tilde x + \tilde y^{2} = 0$$
$$\tilde y^{2} = \tilde x$$
- reduced to canonical form