Given line equation of 2-order:
$$\frac{29 x^{2}}{100} - \frac{y^{2}}{16} - 1 = 0$$
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
$$a_{11} = \frac{29}{100}$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = - \frac{1}{16}$$
$$a_{23} = 0$$
$$a_{33} = -1$$
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
$$I_{1} = \frac{91}{400}$$
| 29 |
|--- 0 |
I2 = |100 |
| |
| 0 -1/16|
$$I_{3} = \left|\begin{matrix}\frac{29}{100} & 0 & 0\\0 & - \frac{1}{16} & 0\\0 & 0 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}\frac{29}{100} - \lambda & 0\\0 & - \lambda - \frac{1}{16}\end{matrix}\right|$$
| 29 |
|--- 0 | |-1/16 0 |
K2 = |100 | + | |
| | | 0 -1|
| 0 -1|
$$I_{1} = \frac{91}{400}$$
$$I_{2} = - \frac{29}{1600}$$
$$I_{3} = \frac{29}{1600}$$
$$I{\left(\lambda \right)} = \lambda^{2} - \frac{91 \lambda}{400} - \frac{29}{1600}$$
$$K_{2} = - \frac{91}{400}$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - \frac{91 \lambda}{400} - \frac{29}{1600} = 0$$
$$\lambda_{1} = - \frac{1}{16}$$
$$\lambda_{2} = \frac{29}{100}$$
then the canonical form of the equation will be
$$\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0$$
or
$$- \frac{\tilde x^{2}}{16} + \frac{29 \tilde y^{2}}{100} - 1 = 0$$
$$\frac{\tilde x^{2}}{16} - \frac{\tilde y^{2}}{\frac{100}{29}} = -1$$
- reduced to canonical form