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How to use it?
- Canonical form:
- x^2-8xy-8y^2=0
- 2x^2+2y^2+2yz+2z^2-1=0
- 11x^2+24xy+4y^2+42x+64y+51=0
- y^2+8*z^2=64
- Plot:
- (x^2+y^2-1)^3-x^2*y^3=0
- x^2*e^-x=y
- 3x^2+5y^2+2z^2=25
-
x*y-x+y+(-x+y)^3/3-3/10=0
-
Identical expressions
- x^ two -8xy-8y^ two = zero
- x squared minus 8xy minus 8y squared equally 0
- x to the power of two minus 8xy minus 8y to the power of two equally zero
- x2-8xy-8y2=0
- x²-8xy-8y²=0
- x to the power of 2-8xy-8y to the power of 2=0
- x^2-8xy-8y^2=O
-
Similar expressions
- x^2-8xy+8y^2=0
- x^2+8xy-8y^2=0
x^2-8xy-8y^2=0 canonical form
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2 2 x - 8*y - 8*x*y = 0
$$x^{2} - 8 x y - 8 y^{2} = 0$$
x^2 - 8*x*y - 8*y^2 = 0
Detail solution
Given line equation of 2-order:
$$x^{2} - 8 x y - 8 y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -4$$
$$a_{13} = 0$$
$$a_{22} = -8$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & -4\\-4 & -8\end{matrix}\right|$$
$$\Delta = -24$$
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
$$x_{0} - 4 y_{0} = 0$$
$$- 4 x_{0} - 8 y_{0} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = 0$$
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
$$a'_{33} = 0$$
$$a'_{33} = 0$$
then equation turns into
$$x'^{2} - 8 x' y' - 8 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{9}{8}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{9}{8} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{8 \sqrt{145}}{145}$$
$$\cos{\left(2 \phi \right)} = \frac{9 \sqrt{145}}{145}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}}$$
$$y' = - \tilde x \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}} + \tilde y \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}}$$
then the equation turns from
$$x'^{2} - 8 x' y' - 8 y'^{2} = 0$$
to
$$- 8 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}} + \tilde y \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}}\right)^{2} - 8 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}} + \tilde y \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}}\right) + \left(\tilde x \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}}\right)^{2} = 0$$
simplify
$$- \frac{7 \tilde x^{2}}{2} + 8 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}} \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}} + \frac{81 \sqrt{145} \tilde x^{2}}{290} - \frac{72 \sqrt{145} \tilde x \tilde y}{145} + 18 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}} \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}} - \frac{7 \tilde y^{2}}{2} - \frac{81 \sqrt{145} \tilde y^{2}}{290} - 8 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}} \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}} = 0$$
$$- \frac{\sqrt{145} \tilde x^{2}}{2} + \frac{7 \tilde x^{2}}{2} + \frac{7 \tilde y^{2}}{2} + \frac{\sqrt{145} \tilde y^{2}}{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{- \frac{7}{2} + \frac{\sqrt{145}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\frac{7}{2} + \frac{\sqrt{145}}{2}}}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}}\right)$$
$$\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}}, \ \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}}\right)$$
$$x^{2} - 8 x y - 8 y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -4$$
$$a_{13} = 0$$
$$a_{22} = -8$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & -4\\-4 & -8\end{matrix}\right|$$
$$\Delta = -24$$
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
$$x_{0} - 4 y_{0} = 0$$
$$- 4 x_{0} - 8 y_{0} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = 0$$
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
$$a'_{33} = 0$$
$$a'_{33} = 0$$
then equation turns into
$$x'^{2} - 8 x' y' - 8 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{9}{8}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{9}{8} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{8 \sqrt{145}}{145}$$
$$\cos{\left(2 \phi \right)} = \frac{9 \sqrt{145}}{145}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}}$$
$$y' = - \tilde x \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}} + \tilde y \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}}$$
then the equation turns from
$$x'^{2} - 8 x' y' - 8 y'^{2} = 0$$
to
$$- 8 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}} + \tilde y \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}}\right)^{2} - 8 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}} + \tilde y \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}}\right) + \left(\tilde x \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}}\right)^{2} = 0$$
simplify
$$- \frac{7 \tilde x^{2}}{2} + 8 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}} \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}} + \frac{81 \sqrt{145} \tilde x^{2}}{290} - \frac{72 \sqrt{145} \tilde x \tilde y}{145} + 18 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}} \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}} - \frac{7 \tilde y^{2}}{2} - \frac{81 \sqrt{145} \tilde y^{2}}{290} - 8 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}} \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}} = 0$$
$$- \frac{\sqrt{145} \tilde x^{2}}{2} + \frac{7 \tilde x^{2}}{2} + \frac{7 \tilde y^{2}}{2} + \frac{\sqrt{145} \tilde y^{2}}{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{- \frac{7}{2} + \frac{\sqrt{145}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\frac{7}{2} + \frac{\sqrt{145}}{2}}}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}}\right)$$
$$\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{9 \sqrt{145}}{290}}, \ \sqrt{\frac{9 \sqrt{145}}{290} + \frac{1}{2}}\right)$$
Invariants method
Given line equation of 2-order:
$$x^{2} - 8 x y - 8 y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -4$$
$$a_{13} = 0$$
$$a_{22} = -8$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = -7$$
$$I_{3} = \left|\begin{matrix}1 & -4 & 0\\-4 & -8 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & -4\\-4 & - \lambda - 8\end{matrix}\right|$$
$$I_{1} = -7$$
$$I_{2} = -24$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + 7 \lambda - 24$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} < 0$$
then by line type:
this equation is of type : degenerate hyperbole
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} + 7 \lambda - 24 = 0$$
$$\lambda_{1} = - \frac{\sqrt{145}}{2} - \frac{7}{2}$$
$$\lambda_{2} = - \frac{7}{2} + \frac{\sqrt{145}}{2}$$
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
$$\tilde x^{2} \left(- \frac{\sqrt{145}}{2} - \frac{7}{2}\right) + \tilde y^{2} \left(- \frac{7}{2} + \frac{\sqrt{145}}{2}\right) = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{\frac{7}{2} + \frac{\sqrt{145}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{- \frac{7}{2} + \frac{\sqrt{145}}{2}}}\right)^{2}} = 0$$
- reduced to canonical form
$$x^{2} - 8 x y - 8 y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -4$$
$$a_{13} = 0$$
$$a_{22} = -8$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = -7$$
|1 -4|
I2 = | |
|-4 -8|$$I_{3} = \left|\begin{matrix}1 & -4 & 0\\-4 & -8 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & -4\\-4 & - \lambda - 8\end{matrix}\right|$$
|1 0| |-8 0|
K2 = | | + | |
|0 0| |0 0|$$I_{1} = -7$$
$$I_{2} = -24$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + 7 \lambda - 24$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} < 0$$
then by line type:
this equation is of type : degenerate hyperbole
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} + 7 \lambda - 24 = 0$$
$$\lambda_{1} = - \frac{\sqrt{145}}{2} - \frac{7}{2}$$
$$\lambda_{2} = - \frac{7}{2} + \frac{\sqrt{145}}{2}$$
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
$$\tilde x^{2} \left(- \frac{\sqrt{145}}{2} - \frac{7}{2}\right) + \tilde y^{2} \left(- \frac{7}{2} + \frac{\sqrt{145}}{2}\right) = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{\frac{7}{2} + \frac{\sqrt{145}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{- \frac{7}{2} + \frac{\sqrt{145}}{2}}}\right)^{2}} = 0$$
- reduced to canonical form