1/4 canonical form

Given line equation of 2-order:

False

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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = \frac{1}{4}$$
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
$$I_{1} = 0$$

     |0  0|
I2 = |    |
     |0  0|

$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & \frac{1}{4}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & - \lambda\end{matrix}\right|$$

     |0   0 |   |0   0 |
K2 = |      | + |      |
     |0  1/4|   |0  1/4|

$$I_{1} = 0$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2}$$
$$K_{2} = 0$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge \left(I_{1} = 0 \vee K_{2} = 0\right)$$
then by line type:
this equation is of type : two coincident straight lines
$$I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0$$
or

False

None

- reduced to canonical form