Mister Exam

9x2+121y2=1089 canonical form

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-1089 + 9*x2 + 121*y2 = 0
$$9 x_{2} + 121 y_{2} - 1089 = 0$$
9*x2 + 121*y2 - 1089 = 0
Detail solution
Given line equation of 2-order:
$$9 x_{2} + 121 y_{2} - 1089 = 0$$
This equation looks like:
$$a_{11} y_{2}^{2} + 2 a_{12} x_{2} y_{2} + 2 a_{13} y_{2} + a_{22} x_{2}^{2} + 2 a_{23} x_{2} + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{121}{2}$$
$$a_{22} = 0$$
$$a_{23} = \frac{9}{2}$$
$$a_{33} = -1089$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
Given equation is straight line
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$y_{20} = - \tilde x2 \sin{\left(\phi \right)} + \tilde y2 \cos{\left(\phi \right)}$$
$$x_{20} = \tilde x2 \cos{\left(\phi \right)} + \tilde y2 \sin{\left(\phi \right)}$$
$$y_{20} = 0 \cdot 0$$
$$x_{20} = 0 \cdot 0$$
$$y_{20} = 0$$
$$x_{20} = 0$$
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:
$$9 x_{2} + 121 y_{2} - 1089 = 0$$
This equation looks like:
$$a_{11} y_{2}^{2} + 2 a_{12} x_{2} y_{2} + 2 a_{13} y_{2} + a_{22} x_{2}^{2} + 2 a_{23} x_{2} + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{121}{2}$$
$$a_{22} = 0$$
$$a_{23} = \frac{9}{2}$$
$$a_{33} = -1089$$
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|

$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{121}{2}\\0 & 0 & \frac{9}{2}\\\frac{121}{2} & \frac{9}{2} & -1089\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
     |  0    121/2|   | 0    9/2 |
K2 = |            | + |          |
     |121/2  -1089|   |9/2  -1089|

$$I_{1} = 0$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2}$$
$$K_{2} = - \frac{7361}{2}$$
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
$$I_{1} \tilde x2^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
False

None

- reduced to canonical form