Mister Exam
f(x)=176x2 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Invariants method
Given equation of the surface of 2-order:
$$f x - 176 x_{2} = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x x_{2} + 2 a_{13} f x + 2 a_{14} x + a_{22} x_{2}^{2} + 2 a_{23} f x_{2} + 2 a_{24} x_{2} + a_{33} f^{2} + 2 a_{34} f + a_{44} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{1}{2}$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = -88$$
$$a_{33} = 0$$
$$a_{34} = 0$$
$$a_{44} = 0$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$
substitute coefficients
$$I_{1} = 0$$
$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{1}{2}\\0 & 0 & 0\\\frac{1}{2} & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & \frac{1}{2} & 0\\0 & 0 & 0 & -88\\\frac{1}{2} & 0 & 0 & 0\\0 & -88 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & \frac{1}{2}\\0 & - \lambda & 0\\\frac{1}{2} & 0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 0$$
$$I_{2} = - \frac{1}{4}$$
$$I_{3} = 0$$
$$I_{4} = 1936$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{\lambda}{4}$$
$$K_{2} = -7744$$
$$K_{3} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0$$
or
$$\lambda^{3} - \frac{\lambda}{4} = 0$$
$$\lambda_{1} = - \frac{1}{2}$$
$$\lambda_{2} = \frac{1}{2}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\tilde f 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde x2^{2} \lambda_{2}\right) = 0$$
and
$$- \tilde f 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde x2^{2} \lambda_{2}\right) = 0$$
$$176 \tilde f - \frac{\tilde x^{2}}{2} + \frac{\tilde x2^{2}}{2} = 0$$
and
$$- 176 \tilde f - \frac{\tilde x^{2}}{2} + \frac{\tilde x2^{2}}{2} = 0$$
$$- 2 \tilde f + \left(\frac{\tilde x^{2}}{176} - \frac{\tilde x2^{2}}{176}\right) = 0$$
and
$$2 \tilde f + \left(\frac{\tilde x^{2}}{176} - \frac{\tilde x2^{2}}{176}\right) = 0$$
this equation is fora type hyperbolic paraboloid
- reduced to canonical form
$$f x - 176 x_{2} = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x x_{2} + 2 a_{13} f x + 2 a_{14} x + a_{22} x_{2}^{2} + 2 a_{23} f x_{2} + 2 a_{24} x_{2} + a_{33} f^{2} + 2 a_{34} f + a_{44} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{1}{2}$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = -88$$
$$a_{33} = 0$$
$$a_{34} = 0$$
$$a_{44} = 0$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
|a11 a12| |a22 a23| |a11 a13|
I2 = | | + | | + | |
|a12 a22| |a23 a33| |a13 a33|$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$
|a11 a14| |a22 a24| |a33 a34|
K2 = | | + | | + | |
|a14 a44| |a24 a44| |a34 a44| |a11 a12 a14| |a22 a23 a24| |a11 a13 a14|
| | | | | |
K3 = |a12 a22 a24| + |a23 a33 a34| + |a13 a33 a34|
| | | | | |
|a14 a24 a44| |a24 a34 a44| |a14 a34 a44|substitute coefficients
$$I_{1} = 0$$
|0 0| |0 0| | 0 1/2|
I2 = | | + | | + | |
|0 0| |0 0| |1/2 0 |$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{1}{2}\\0 & 0 & 0\\\frac{1}{2} & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & \frac{1}{2} & 0\\0 & 0 & 0 & -88\\\frac{1}{2} & 0 & 0 & 0\\0 & -88 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & \frac{1}{2}\\0 & - \lambda & 0\\\frac{1}{2} & 0 & - \lambda\end{matrix}\right|$$
|0 0| | 0 -88| |0 0|
K2 = | | + | | + | |
|0 0| |-88 0 | |0 0| |0 0 0 | | 0 0 -88| | 0 1/2 0|
| | | | | |
K3 = |0 0 -88| + | 0 0 0 | + |1/2 0 0|
| | | | | |
|0 -88 0 | |-88 0 0 | | 0 0 0|$$I_{1} = 0$$
$$I_{2} = - \frac{1}{4}$$
$$I_{3} = 0$$
$$I_{4} = 1936$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{\lambda}{4}$$
$$K_{2} = -7744$$
$$K_{3} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0$$
or
$$\lambda^{3} - \frac{\lambda}{4} = 0$$
$$\lambda_{1} = - \frac{1}{2}$$
$$\lambda_{2} = \frac{1}{2}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\tilde f 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde x2^{2} \lambda_{2}\right) = 0$$
and
$$- \tilde f 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde x2^{2} \lambda_{2}\right) = 0$$
$$176 \tilde f - \frac{\tilde x^{2}}{2} + \frac{\tilde x2^{2}}{2} = 0$$
and
$$- 176 \tilde f - \frac{\tilde x^{2}}{2} + \frac{\tilde x2^{2}}{2} = 0$$
$$- 2 \tilde f + \left(\frac{\tilde x^{2}}{176} - \frac{\tilde x2^{2}}{176}\right) = 0$$
and
$$2 \tilde f + \left(\frac{\tilde x^{2}}{176} - \frac{\tilde x2^{2}}{176}\right) = 0$$
this equation is fora type hyperbolic paraboloid
- reduced to canonical form