Given line equation of 2-order:
$$5 x + 3 y^{2} + 6 y + 13 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{5}{2}$$
$$a_{22} = 3$$
$$a_{23} = 3$$
$$a_{33} = 13$$
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
$$I_{1} = 3$$
|0 0|
I2 = | |
|0 3|
$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{5}{2}\\0 & 3 & 3\\\frac{5}{2} & 3 & 13\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 3 - \lambda\end{matrix}\right|$$
| 0 5/2| |3 3 |
K2 = | | + | |
|5/2 13 | |3 13|
$$I_{1} = 3$$
$$I_{2} = 0$$
$$I_{3} = - \frac{75}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 3 \lambda$$
$$K_{2} = \frac{95}{4}$$
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
$$5 \tilde x + 3 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{5 \tilde x}{3}$$
- reduced to canonical form