Mister Exam
8x^2-2xy+2y^2-6x-4y+5=0 canonical form
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2 2 5 - 6*x - 4*y + 2*y + 8*x - 2*x*y = 0
$$8 x^{2} - 2 x y - 6 x + 2 y^{2} - 4 y + 5 = 0$$
8*x^2 - 2*x*y - 6*x + 2*y^2 - 4*y + 5 = 0
Detail solution
Given line equation of 2-order:
$$8 x^{2} - 2 x y - 6 x + 2 y^{2} - 4 y + 5 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = -1$$
$$a_{13} = -3$$
$$a_{22} = 2$$
$$a_{23} = -2$$
$$a_{33} = 5$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}8 & -1\\-1 & 2\end{matrix}\right|$$
$$\Delta = 15$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
substitute coefficients
$$8 x_{0} - y_{0} - 3 = 0$$
$$- x_{0} + 2 y_{0} - 2 = 0$$
then
$$x_{0} = \frac{8}{15}$$
$$y_{0} = \frac{19}{15}$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - 3 x_{0} - 2 y_{0} + 5$$
$$a'_{33} = \frac{13}{15}$$
then equation turns into
$$8 x'^{2} - 2 x' y' + 2 y'^{2} + \frac{13}{15} = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = -3$$
then
$$\phi = - \frac{\operatorname{acot}{\left(3 \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{\sqrt{10}}{10}$$
$$\cos{\left(2 \phi \right)} = \frac{3 \sqrt{10}}{10}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}}$$
$$y' = - \tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}} + \tilde y \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}}$$
then the equation turns from
$$8 x'^{2} - 2 x' y' + 2 y'^{2} + \frac{13}{15} = 0$$
to
$$2 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}} + \tilde y \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}}\right)^{2} - 2 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}} + \tilde y \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}}\right) + 8 \left(\tilde x \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}}\right)^{2} + \frac{13}{15} = 0$$
simplify
$$2 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}} \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}} + \frac{9 \sqrt{10} \tilde x^{2}}{10} + 5 \tilde x^{2} - \frac{3 \sqrt{10} \tilde x \tilde y}{5} + 12 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}} \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}} - \frac{9 \sqrt{10} \tilde y^{2}}{10} - 2 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}} \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}} + 5 \tilde y^{2} + \frac{13}{15} = 0$$
$$\sqrt{10} \tilde x^{2} + 5 \tilde x^{2} - \sqrt{10} \tilde y^{2} + 5 \tilde y^{2} + \frac{13}{15} = 0$$
Given equation is imaginary ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{\sqrt{195}}{13} \sqrt{\sqrt{10} + 5}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{\sqrt{195}}{13} \sqrt{5 - \sqrt{10}}}\right)^{2}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}}\right)$$
$$\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}}, \ \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}}\right)$$
$$8 x^{2} - 2 x y - 6 x + 2 y^{2} - 4 y + 5 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = -1$$
$$a_{13} = -3$$
$$a_{22} = 2$$
$$a_{23} = -2$$
$$a_{33} = 5$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}8 & -1\\-1 & 2\end{matrix}\right|$$
$$\Delta = 15$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
substitute coefficients
$$8 x_{0} - y_{0} - 3 = 0$$
$$- x_{0} + 2 y_{0} - 2 = 0$$
then
$$x_{0} = \frac{8}{15}$$
$$y_{0} = \frac{19}{15}$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - 3 x_{0} - 2 y_{0} + 5$$
$$a'_{33} = \frac{13}{15}$$
then equation turns into
$$8 x'^{2} - 2 x' y' + 2 y'^{2} + \frac{13}{15} = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = -3$$
then
$$\phi = - \frac{\operatorname{acot}{\left(3 \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{\sqrt{10}}{10}$$
$$\cos{\left(2 \phi \right)} = \frac{3 \sqrt{10}}{10}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}}$$
$$y' = - \tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}} + \tilde y \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}}$$
then the equation turns from
$$8 x'^{2} - 2 x' y' + 2 y'^{2} + \frac{13}{15} = 0$$
to
$$2 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}} + \tilde y \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}}\right)^{2} - 2 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}} + \tilde y \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}}\right) + 8 \left(\tilde x \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}}\right)^{2} + \frac{13}{15} = 0$$
simplify
$$2 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}} \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}} + \frac{9 \sqrt{10} \tilde x^{2}}{10} + 5 \tilde x^{2} - \frac{3 \sqrt{10} \tilde x \tilde y}{5} + 12 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}} \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}} - \frac{9 \sqrt{10} \tilde y^{2}}{10} - 2 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}} \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}} + 5 \tilde y^{2} + \frac{13}{15} = 0$$
$$\sqrt{10} \tilde x^{2} + 5 \tilde x^{2} - \sqrt{10} \tilde y^{2} + 5 \tilde y^{2} + \frac{13}{15} = 0$$
Given equation is imaginary ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{\sqrt{195}}{13} \sqrt{\sqrt{10} + 5}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{\sqrt{195}}{13} \sqrt{5 - \sqrt{10}}}\right)^{2}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
19
(8/15, --)
15 Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}}\right)$$
$$\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{3 \sqrt{10}}{20}}, \ \sqrt{\frac{3 \sqrt{10}}{20} + \frac{1}{2}}\right)$$
Invariants method
Given line equation of 2-order:
$$8 x^{2} - 2 x y - 6 x + 2 y^{2} - 4 y + 5 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = -1$$
$$a_{13} = -3$$
$$a_{22} = 2$$
$$a_{23} = -2$$
$$a_{33} = 5$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
substitute coefficients
$$I_{1} = 10$$
$$I_{3} = \left|\begin{matrix}8 & -1 & -3\\-1 & 2 & -2\\-3 & -2 & 5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}8 - \lambda & -1\\-1 & 2 - \lambda\end{matrix}\right|$$
$$I_{1} = 10$$
$$I_{2} = 15$$
$$I_{3} = 13$$
$$I{\left(\lambda \right)} = \lambda^{2} - 10 \lambda + 15$$
$$K_{2} = 37$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} > 0$$
then by line type:
this equation is of type : imaginary ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 10 \lambda + 15 = 0$$
$$\lambda_{1} = 5 - \sqrt{10}$$
$$\lambda_{2} = \sqrt{10} + 5$$
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
$$\tilde x^{2} \left(5 - \sqrt{10}\right) + \tilde y^{2} \left(\sqrt{10} + 5\right) + \frac{13}{15} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{\sqrt{195}}{13} \sqrt{5 - \sqrt{10}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{\sqrt{195}}{13} \sqrt{\sqrt{10} + 5}}\right)^{2}} = -1$$
- reduced to canonical form
$$8 x^{2} - 2 x y - 6 x + 2 y^{2} - 4 y + 5 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = -1$$
$$a_{13} = -3$$
$$a_{22} = 2$$
$$a_{23} = -2$$
$$a_{33} = 5$$
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
$$I_{1} = 10$$
|8 -1|
I2 = | |
|-1 2 |$$I_{3} = \left|\begin{matrix}8 & -1 & -3\\-1 & 2 & -2\\-3 & -2 & 5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}8 - \lambda & -1\\-1 & 2 - \lambda\end{matrix}\right|$$
|8 -3| |2 -2|
K2 = | | + | |
|-3 5 | |-2 5 |$$I_{1} = 10$$
$$I_{2} = 15$$
$$I_{3} = 13$$
$$I{\left(\lambda \right)} = \lambda^{2} - 10 \lambda + 15$$
$$K_{2} = 37$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} > 0$$
then by line type:
this equation is of type : imaginary ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 10 \lambda + 15 = 0$$
$$\lambda_{1} = 5 - \sqrt{10}$$
$$\lambda_{2} = \sqrt{10} + 5$$
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
$$\tilde x^{2} \left(5 - \sqrt{10}\right) + \tilde y^{2} \left(\sqrt{10} + 5\right) + \frac{13}{15} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{\sqrt{195}}{13} \sqrt{5 - \sqrt{10}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{\sqrt{195}}{13} \sqrt{\sqrt{10} + 5}}\right)^{2}} = -1$$
- reduced to canonical form