Given line equation of 2-order:
$$14 x^{2} + 50 x - 18 y + 44 = 0$$
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
$$a_{11} = 14$$
$$a_{12} = 0$$
$$a_{13} = 25$$
$$a_{22} = 0$$
$$a_{23} = -9$$
$$a_{33} = 44$$
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
$$I_{1} = 14$$
|14 0|
I2 = | |
|0 0|
$$I_{3} = \left|\begin{matrix}14 & 0 & 25\\0 & 0 & -9\\25 & -9 & 44\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}14 - \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
|14 25| |0 -9|
K2 = | | + | |
|25 44| |-9 44|
$$I_{1} = 14$$
$$I_{2} = 0$$
$$I_{3} = -1134$$
$$I{\left(\lambda \right)} = \lambda^{2} - 14 \lambda$$
$$K_{2} = -90$$
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
$$18 \tilde x + 14 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{9 \tilde x}{7}$$
- reduced to canonical form