Mister Exam
5x^2+4xy+8y^2+8x+z+14y+5=0 canonical form
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2 2 5 + z + 5*x + 8*x + 8*y + 14*y + 4*x*y = 0
$$5 x^{2} + 4 x y + 8 x + 8 y^{2} + 14 y + z + 5 = 0$$
5*x^2 + 4*x*y + 8*x + 8*y^2 + 14*y + z + 5 = 0
Invariants method
Given equation of the surface of 2-order:
$$5 x^{2} + 4 x y + 8 x + 8 y^{2} + 14 y + z + 5 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0$$
where
$$a_{11} = 5$$
$$a_{12} = 2$$
$$a_{13} = 0$$
$$a_{14} = 4$$
$$a_{22} = 8$$
$$a_{23} = 0$$
$$a_{24} = 7$$
$$a_{33} = 0$$
$$a_{34} = \frac{1}{2}$$
$$a_{44} = 5$$
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
$$I_{1} = 13$$
$$I_{3} = \left|\begin{matrix}5 & 2 & 0\\2 & 8 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}5 & 2 & 0 & 4\\2 & 8 & 0 & 7\\0 & 0 & 0 & \frac{1}{2}\\4 & 7 & \frac{1}{2} & 5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}5 - \lambda & 2 & 0\\2 & 8 - \lambda & 0\\0 & 0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 13$$
$$I_{2} = 36$$
$$I_{3} = 0$$
$$I_{4} = -9$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 13 \lambda^{2} - 36 \lambda$$
$$K_{2} = - \frac{1}{4}$$
$$K_{3} = - \frac{337}{4}$$
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
$$\lambda^{3} - 13 \lambda^{2} + 36 \lambda = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 4$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
and
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$9 \tilde x^{2} + 4 \tilde y^{2} + \tilde z = 0$$
and
$$9 \tilde x^{2} + 4 \tilde y^{2} - \tilde z = 0$$
and
this equation is fora type elliptical paraboloid
- reduced to canonical form
$$5 x^{2} + 4 x y + 8 x + 8 y^{2} + 14 y + z + 5 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0$$
where
$$a_{11} = 5$$
$$a_{12} = 2$$
$$a_{13} = 0$$
$$a_{14} = 4$$
$$a_{22} = 8$$
$$a_{23} = 0$$
$$a_{24} = 7$$
$$a_{33} = 0$$
$$a_{34} = \frac{1}{2}$$
$$a_{44} = 5$$
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
$$I_{1} = 13$$
|5 2| |8 0| |5 0|
I2 = | | + | | + | |
|2 8| |0 0| |0 0|$$I_{3} = \left|\begin{matrix}5 & 2 & 0\\2 & 8 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}5 & 2 & 0 & 4\\2 & 8 & 0 & 7\\0 & 0 & 0 & \frac{1}{2}\\4 & 7 & \frac{1}{2} & 5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}5 - \lambda & 2 & 0\\2 & 8 - \lambda & 0\\0 & 0 & - \lambda\end{matrix}\right|$$
|5 4| |8 7| | 0 1/2|
K2 = | | + | | + | |
|4 5| |7 5| |1/2 5 | |5 2 4| |8 0 7 | |5 0 4 |
| | | | | |
K3 = |2 8 7| + |0 0 1/2| + |0 0 1/2|
| | | | | |
|4 7 5| |7 1/2 5 | |4 1/2 5 |$$I_{1} = 13$$
$$I_{2} = 36$$
$$I_{3} = 0$$
$$I_{4} = -9$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 13 \lambda^{2} - 36 \lambda$$
$$K_{2} = - \frac{1}{4}$$
$$K_{3} = - \frac{337}{4}$$
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
$$\lambda^{3} - 13 \lambda^{2} + 36 \lambda = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 4$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
and
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$9 \tilde x^{2} + 4 \tilde y^{2} + \tilde z = 0$$
and
$$9 \tilde x^{2} + 4 \tilde y^{2} - \tilde z = 0$$
2 2 \tilde x \tilde y --------- + --------- + 2*\tilde z = 0 1/18 1/8
and
2 2 \tilde x \tilde y --------- + --------- - 2*\tilde z = 0 1/18 1/8
this equation is fora type elliptical paraboloid
- reduced to canonical form