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- y*y-6*x-6*y-15=0
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- x^2+y^2-z^2-2*x-4*y+6=0
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x=449.4*(y/10000-1)*(5/(1.4*(6*y/100000+1)))^(1/2)
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Identical expressions
- x^ two -4xy+3y^ two
- x squared minus 4xy plus 3y squared
- x to the power of two minus 4xy plus 3y to the power of two
- x2-4xy+3y2
- x²-4xy+3y²
- x to the power of 2-4xy+3y to the power of 2
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Similar expressions
- x^2+4xy+3y^2
- x^2-4xy-3y^2
x^2-4xy+3y^2 canonical form
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2 2 x + 3*y - 4*x*y = 0
$$x^{2} - 4 x y + 3 y^{2} = 0$$
x^2 - 4*x*y + 3*y^2 = 0
Detail solution
Given line equation of 2-order:
$$x^{2} - 4 x y + 3 y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -2$$
$$a_{13} = 0$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & -2\\-2 & 3\end{matrix}\right|$$
$$\Delta = -1$$
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
$$x_{0} - 2 y_{0} = 0$$
$$- 2 x_{0} + 3 y_{0} = 0$$
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
$$x'^{2} - 4 x' y' + 3 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{1}{2}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{1}{2} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{2 \sqrt{5}}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{\sqrt{5}}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}} + \tilde y \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}}$$
then the equation turns from
$$x'^{2} - 4 x' y' + 3 y'^{2} = 0$$
to
$$3 \left(\tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}} + \tilde y \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}}\right)^{2} - 4 \left(\tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}} + \tilde y \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}}\right) + \left(\tilde x \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}}\right)^{2} = 0$$
simplify
$$- 4 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}} \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}} - \frac{\sqrt{5} \tilde x^{2}}{5} + 2 \tilde x^{2} - \frac{4 \sqrt{5} \tilde x \tilde y}{5} + 4 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}} \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}} + \frac{\sqrt{5} \tilde y^{2}}{5} + 4 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}} \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}} + 2 \tilde y^{2} = 0$$
$$- \sqrt{5} \tilde x^{2} + 2 \tilde x^{2} + 2 \tilde y^{2} + \sqrt{5} \tilde y^{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{-2 + \sqrt{5}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{2 + \sqrt{5}}}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}}, \ \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}}\right)$$
$$x^{2} - 4 x y + 3 y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -2$$
$$a_{13} = 0$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & -2\\-2 & 3\end{matrix}\right|$$
$$\Delta = -1$$
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
$$x_{0} - 2 y_{0} = 0$$
$$- 2 x_{0} + 3 y_{0} = 0$$
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
$$x'^{2} - 4 x' y' + 3 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{1}{2}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{1}{2} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{2 \sqrt{5}}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{\sqrt{5}}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}} + \tilde y \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}}$$
then the equation turns from
$$x'^{2} - 4 x' y' + 3 y'^{2} = 0$$
to
$$3 \left(\tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}} + \tilde y \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}}\right)^{2} - 4 \left(\tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}} + \tilde y \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}}\right) + \left(\tilde x \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}}\right)^{2} = 0$$
simplify
$$- 4 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}} \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}} - \frac{\sqrt{5} \tilde x^{2}}{5} + 2 \tilde x^{2} - \frac{4 \sqrt{5} \tilde x \tilde y}{5} + 4 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}} \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}} + \frac{\sqrt{5} \tilde y^{2}}{5} + 4 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}} \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}} + 2 \tilde y^{2} = 0$$
$$- \sqrt{5} \tilde x^{2} + 2 \tilde x^{2} + 2 \tilde y^{2} + \sqrt{5} \tilde y^{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{-2 + \sqrt{5}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{2 + \sqrt{5}}}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}}, \ \sqrt{\frac{\sqrt{5}}{10} + \frac{1}{2}}\right)$$
Invariants method
Given line equation of 2-order:
$$x^{2} - 4 x y + 3 y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -2$$
$$a_{13} = 0$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 4$$
$$I_{3} = \left|\begin{matrix}1 & -2 & 0\\-2 & 3 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & -2\\-2 & 3 - \lambda\end{matrix}\right|$$
$$I_{1} = 4$$
$$I_{2} = -1$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 4 \lambda - 1$$
$$K_{2} = 0$$
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
$$\lambda^{2} - 4 \lambda - 1 = 0$$
$$\lambda_{1} = 2 - \sqrt{5}$$
$$\lambda_{2} = 2 + \sqrt{5}$$
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
$$\tilde x^{2} \left(2 - \sqrt{5}\right) + \tilde y^{2} \left(2 + \sqrt{5}\right) = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{-2 + \sqrt{5}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{2 + \sqrt{5}}}\right)^{2}} = 0$$
- reduced to canonical form
$$x^{2} - 4 x y + 3 y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -2$$
$$a_{13} = 0$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 4$$
|1 -2|
I2 = | |
|-2 3 |$$I_{3} = \left|\begin{matrix}1 & -2 & 0\\-2 & 3 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & -2\\-2 & 3 - \lambda\end{matrix}\right|$$
|1 0| |3 0|
K2 = | | + | |
|0 0| |0 0|$$I_{1} = 4$$
$$I_{2} = -1$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 4 \lambda - 1$$
$$K_{2} = 0$$
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
$$\lambda^{2} - 4 \lambda - 1 = 0$$
$$\lambda_{1} = 2 - \sqrt{5}$$
$$\lambda_{2} = 2 + \sqrt{5}$$
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
$$\tilde x^{2} \left(2 - \sqrt{5}\right) + \tilde y^{2} \left(2 + \sqrt{5}\right) = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{-2 + \sqrt{5}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{2 + \sqrt{5}}}\right)^{2}} = 0$$
- reduced to canonical form