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How to use it?
- Canonical form:
- sqrt[x]*4*sqrt[y]-2x-y
- 3x^2+6y^2+3z^2-4x-8y-4z=0
- 3a^2+2b^2+4ac+6bc-4bc+2bd-2cd
- 41x^2+24xy+9y^2+24x+18y-36=0
- Plot:
- y=x^3-3*x^2+4
- x^3*y^2+5*x*y+4=0
- x^2+((3/4))=(1-y)^2
-
5x^2+10y^2=0
-
Identical expressions
- x^ two -5xy+4y^ two +x+ two y-2= zero
- x squared minus 5xy plus 4y squared plus x plus 2y minus 2 equally 0
- x to the power of two minus 5xy plus 4y to the power of two plus x plus two y minus 2 equally zero
- x2-5xy+4y2+x+2y-2=0
- x²-5xy+4y²+x+2y-2=0
- x to the power of 2-5xy+4y to the power of 2+x+2y-2=0
- x^2-5xy+4y^2+x+2y-2=O
-
Similar expressions
- x^2+5xy+4y^2+x+2y-2=0
- x^2-5xy-4y^2+x+2y-2=0
- x^2-5xy+4y^2+x+2y+2=0
- x^2-5xy+4y^2-x+2y-2=0
- x^2-5xy+4y^2+x-2y-2=0
x^2-5xy+4y^2+x+2y-2=0 canonical form
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The solution
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2 2 -2 + x + x + 2*y + 4*y - 5*x*y = 0
$$x^{2} - 5 x y + x + 4 y^{2} + 2 y - 2 = 0$$
x^2 - 5*x*y + x + 4*y^2 + 2*y - 2 = 0
Detail solution
Given line equation of 2-order:
$$x^{2} - 5 x y + x + 4 y^{2} + 2 y - 2 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = - \frac{5}{2}$$
$$a_{13} = \frac{1}{2}$$
$$a_{22} = 4$$
$$a_{23} = 1$$
$$a_{33} = -2$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & - \frac{5}{2}\\- \frac{5}{2} & 4\end{matrix}\right|$$
$$\Delta = - \frac{9}{4}$$
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
$$x_{0} - \frac{5 y_{0}}{2} + \frac{1}{2} = 0$$
$$- \frac{5 x_{0}}{2} + 4 y_{0} + 1 = 0$$
then
$$x_{0} = 2$$
$$y_{0} = 1$$
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
$$a'_{33} = \frac{x_{0}}{2} + y_{0} - 2$$
$$a'_{33} = 0$$
then equation turns into
$$x'^{2} - 5 x' y' + 4 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{3}{5}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{3}{5} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{5 \sqrt{34}}{34}$$
$$\cos{\left(2 \phi \right)} = \frac{3 \sqrt{34}}{34}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}} + \tilde y \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}}$$
then the equation turns from
$$x'^{2} - 5 x' y' + 4 y'^{2} = 0$$
to
$$4 \left(\tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}} + \tilde y \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}}\right)^{2} - 5 \left(\tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}} + \tilde y \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}}\right) + \left(\tilde x \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}}\right)^{2} = 0$$
simplify
$$- 5 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}} \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}} - \frac{9 \sqrt{34} \tilde x^{2}}{68} + \frac{5 \tilde x^{2}}{2} - \frac{15 \sqrt{34} \tilde x \tilde y}{34} + 6 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}} \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}} + \frac{9 \sqrt{34} \tilde y^{2}}{68} + 5 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}} \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}} + \frac{5 \tilde y^{2}}{2} = 0$$
$$- \frac{\sqrt{34} \tilde x^{2}}{2} + \frac{5 \tilde x^{2}}{2} + \frac{5 \tilde y^{2}}{2} + \frac{\sqrt{34} \tilde y^{2}}{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{- \frac{5}{2} + \frac{\sqrt{34}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\frac{5}{2} + \frac{\sqrt{34}}{2}}}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}}, \ \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}}\right)$$
$$x^{2} - 5 x y + x + 4 y^{2} + 2 y - 2 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = - \frac{5}{2}$$
$$a_{13} = \frac{1}{2}$$
$$a_{22} = 4$$
$$a_{23} = 1$$
$$a_{33} = -2$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & - \frac{5}{2}\\- \frac{5}{2} & 4\end{matrix}\right|$$
$$\Delta = - \frac{9}{4}$$
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
$$x_{0} - \frac{5 y_{0}}{2} + \frac{1}{2} = 0$$
$$- \frac{5 x_{0}}{2} + 4 y_{0} + 1 = 0$$
then
$$x_{0} = 2$$
$$y_{0} = 1$$
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
$$a'_{33} = \frac{x_{0}}{2} + y_{0} - 2$$
$$a'_{33} = 0$$
then equation turns into
$$x'^{2} - 5 x' y' + 4 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{3}{5}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{3}{5} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{5 \sqrt{34}}{34}$$
$$\cos{\left(2 \phi \right)} = \frac{3 \sqrt{34}}{34}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}} + \tilde y \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}}$$
then the equation turns from
$$x'^{2} - 5 x' y' + 4 y'^{2} = 0$$
to
$$4 \left(\tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}} + \tilde y \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}}\right)^{2} - 5 \left(\tilde x \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}} + \tilde y \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}}\right) + \left(\tilde x \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}}\right)^{2} = 0$$
simplify
$$- 5 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}} \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}} - \frac{9 \sqrt{34} \tilde x^{2}}{68} + \frac{5 \tilde x^{2}}{2} - \frac{15 \sqrt{34} \tilde x \tilde y}{34} + 6 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}} \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}} + \frac{9 \sqrt{34} \tilde y^{2}}{68} + 5 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}} \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}} + \frac{5 \tilde y^{2}}{2} = 0$$
$$- \frac{\sqrt{34} \tilde x^{2}}{2} + \frac{5 \tilde x^{2}}{2} + \frac{5 \tilde y^{2}}{2} + \frac{\sqrt{34} \tilde y^{2}}{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{- \frac{5}{2} + \frac{\sqrt{34}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\frac{5}{2} + \frac{\sqrt{34}}{2}}}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
(2, 1)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{3 \sqrt{34}}{68}}, \ \sqrt{\frac{3 \sqrt{34}}{68} + \frac{1}{2}}\right)$$
Invariants method
Given line equation of 2-order:
$$x^{2} - 5 x y + x + 4 y^{2} + 2 y - 2 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = - \frac{5}{2}$$
$$a_{13} = \frac{1}{2}$$
$$a_{22} = 4$$
$$a_{23} = 1$$
$$a_{33} = -2$$
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
$$I_{1} = 5$$
$$I_{3} = \left|\begin{matrix}1 & - \frac{5}{2} & \frac{1}{2}\\- \frac{5}{2} & 4 & 1\\\frac{1}{2} & 1 & -2\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & - \frac{5}{2}\\- \frac{5}{2} & 4 - \lambda\end{matrix}\right|$$
$$I_{1} = 5$$
$$I_{2} = - \frac{9}{4}$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda - \frac{9}{4}$$
$$K_{2} = - \frac{45}{4}$$
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
$$\lambda^{2} - 5 \lambda - \frac{9}{4} = 0$$
$$\lambda_{1} = \frac{5}{2} - \frac{\sqrt{34}}{2}$$
$$\lambda_{2} = \frac{5}{2} + \frac{\sqrt{34}}{2}$$
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
$$\tilde x^{2} \left(\frac{5}{2} - \frac{\sqrt{34}}{2}\right) + \tilde y^{2} \left(\frac{5}{2} + \frac{\sqrt{34}}{2}\right) = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{- \frac{5}{2} + \frac{\sqrt{34}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\frac{5}{2} + \frac{\sqrt{34}}{2}}}\right)^{2}} = 0$$
- reduced to canonical form
$$x^{2} - 5 x y + x + 4 y^{2} + 2 y - 2 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = - \frac{5}{2}$$
$$a_{13} = \frac{1}{2}$$
$$a_{22} = 4$$
$$a_{23} = 1$$
$$a_{33} = -2$$
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
$$I_{1} = 5$$
| 1 -5/2|
I2 = | |
|-5/2 4 |$$I_{3} = \left|\begin{matrix}1 & - \frac{5}{2} & \frac{1}{2}\\- \frac{5}{2} & 4 & 1\\\frac{1}{2} & 1 & -2\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & - \frac{5}{2}\\- \frac{5}{2} & 4 - \lambda\end{matrix}\right|$$
| 1 1/2| |4 1 |
K2 = | | + | |
|1/2 -2 | |1 -2|$$I_{1} = 5$$
$$I_{2} = - \frac{9}{4}$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda - \frac{9}{4}$$
$$K_{2} = - \frac{45}{4}$$
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
$$\lambda^{2} - 5 \lambda - \frac{9}{4} = 0$$
$$\lambda_{1} = \frac{5}{2} - \frac{\sqrt{34}}{2}$$
$$\lambda_{2} = \frac{5}{2} + \frac{\sqrt{34}}{2}$$
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
$$\tilde x^{2} \left(\frac{5}{2} - \frac{\sqrt{34}}{2}\right) + \tilde y^{2} \left(\frac{5}{2} + \frac{\sqrt{34}}{2}\right) = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{- \frac{5}{2} + \frac{\sqrt{34}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\frac{5}{2} + \frac{\sqrt{34}}{2}}}\right)^{2}} = 0$$
- reduced to canonical form