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How to use it?
- Canonical form:
- 2x^2+y^2-z^2-6y=2z+80
- x^2+y^2–z^2–4x–6y+4z+10=0
-
7y^2+3x-28y
- 4x^2-12xy+9y^2+z^2+4xz-6yz+4x-6y+2z-5=0
- Plot:
- x^2+(y-3*sqrt(x)^2)^2=1
-
4*y^2=x^5+4*x^2*y
- z=4*(x-y)-x^2-y^2
-
x^2*y^2=(x-1)*(x-2)
-
Identical expressions
- three x^ two -3.4 six xy+5y^ two =6
- 3x squared minus 3.46xy plus 5y squared equally 6
- three x to the power of two minus 3.4 six xy plus 5y to the power of two equally 6
- 3x2-3.46xy+5y2=6
- 3x²-3.46xy+5y²=6
- 3x to the power of 2-3.46xy+5y to the power of 2=6
-
Similar expressions
- 3x^2+3.46xy+5y^2=6
- 3x^2-3.46xy-5y^2=6
3x^2-3.46xy+5y^2=6 canonical form
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The solution
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2 2 173*x*y
-6 + 3*x + 5*y - ------- = 0
50
$$3 x^{2} - \frac{173 x y}{50} + 5 y^{2} - 6 = 0$$
3*x^2 - 173*x*y/50 + 5*y^2 - 6 = 0
Detail solution
Given line equation of 2-order:
$$3 x^{2} - \frac{173 x y}{50} + 5 y^{2} - 6 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = - \frac{173}{100}$$
$$a_{13} = 0$$
$$a_{22} = 5$$
$$a_{23} = 0$$
$$a_{33} = -6$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & - \frac{173}{100}\\- \frac{173}{100} & 5\end{matrix}\right|$$
$$\Delta = \frac{120071}{10000}$$
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
$$3 x_{0} - \frac{173 y_{0}}{100} = 0$$
$$- \frac{173 x_{0}}{100} + 5 y_{0} = 0$$
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
$$a'_{33} = -6$$
$$a'_{33} = -6$$
then equation turns into
$$3 x'^{2} - \frac{173 x' y'}{50} + 5 y'^{2} - 6 = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{100}{173}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{100}{173} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{173 \sqrt{39929}}{39929}$$
$$\cos{\left(2 \phi \right)} = \frac{100 \sqrt{39929}}{39929}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}} + \tilde y \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}}$$
then the equation turns from
$$3 x'^{2} - \frac{173 x' y'}{50} + 5 y'^{2} - 6 = 0$$
to
$$5 \left(\tilde x \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}} + \tilde y \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}}\right)^{2} - \frac{173 \left(\tilde x \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}} + \tilde y \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}}\right)}{50} + 3 \left(\tilde x \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}}\right)^{2} - 6 = 0$$
simplify
$$- \frac{173 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}} \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}}}{50} - \frac{100 \sqrt{39929} \tilde x^{2}}{39929} + 4 \tilde x^{2} - \frac{346 \sqrt{39929} \tilde x \tilde y}{39929} + 4 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}} \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}} + \frac{100 \sqrt{39929} \tilde y^{2}}{39929} + \frac{173 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}} \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}}}{50} + 4 \tilde y^{2} - 6 = 0$$
$$- \frac{\sqrt{39929} \tilde x^{2}}{100} + 4 \tilde x^{2} + \frac{\sqrt{39929} \tilde y^{2}}{100} + 4 \tilde y^{2} - 6 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{\sqrt{6}}{6} \sqrt{4 - \frac{\sqrt{39929}}{100}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{\sqrt{6}}{6} \sqrt{\frac{\sqrt{39929}}{100} + 4}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}}, \ \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}}\right)$$
$$3 x^{2} - \frac{173 x y}{50} + 5 y^{2} - 6 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = - \frac{173}{100}$$
$$a_{13} = 0$$
$$a_{22} = 5$$
$$a_{23} = 0$$
$$a_{33} = -6$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & - \frac{173}{100}\\- \frac{173}{100} & 5\end{matrix}\right|$$
$$\Delta = \frac{120071}{10000}$$
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
$$3 x_{0} - \frac{173 y_{0}}{100} = 0$$
$$- \frac{173 x_{0}}{100} + 5 y_{0} = 0$$
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
$$a'_{33} = -6$$
$$a'_{33} = -6$$
then equation turns into
$$3 x'^{2} - \frac{173 x' y'}{50} + 5 y'^{2} - 6 = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{100}{173}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{100}{173} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{173 \sqrt{39929}}{39929}$$
$$\cos{\left(2 \phi \right)} = \frac{100 \sqrt{39929}}{39929}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}} + \tilde y \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}}$$
then the equation turns from
$$3 x'^{2} - \frac{173 x' y'}{50} + 5 y'^{2} - 6 = 0$$
to
$$5 \left(\tilde x \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}} + \tilde y \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}}\right)^{2} - \frac{173 \left(\tilde x \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}} + \tilde y \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}}\right)}{50} + 3 \left(\tilde x \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}}\right)^{2} - 6 = 0$$
simplify
$$- \frac{173 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}} \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}}}{50} - \frac{100 \sqrt{39929} \tilde x^{2}}{39929} + 4 \tilde x^{2} - \frac{346 \sqrt{39929} \tilde x \tilde y}{39929} + 4 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}} \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}} + \frac{100 \sqrt{39929} \tilde y^{2}}{39929} + \frac{173 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}} \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}}}{50} + 4 \tilde y^{2} - 6 = 0$$
$$- \frac{\sqrt{39929} \tilde x^{2}}{100} + 4 \tilde x^{2} + \frac{\sqrt{39929} \tilde y^{2}}{100} + 4 \tilde y^{2} - 6 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{\sqrt{6}}{6} \sqrt{4 - \frac{\sqrt{39929}}{100}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{\sqrt{6}}{6} \sqrt{\frac{\sqrt{39929}}{100} + 4}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{50 \sqrt{39929}}{39929}}, \ \sqrt{\frac{50 \sqrt{39929}}{39929} + \frac{1}{2}}\right)$$
Invariants method
Given line equation of 2-order:
$$3 x^{2} - \frac{173 x y}{50} + 5 y^{2} - 6 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = - \frac{173}{100}$$
$$a_{13} = 0$$
$$a_{22} = 5$$
$$a_{23} = 0$$
$$a_{33} = -6$$
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
$$I_{1} = 8$$
$$I_{3} = \left|\begin{matrix}3 & - \frac{173}{100} & 0\\- \frac{173}{100} & 5 & 0\\0 & 0 & -6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & - \frac{173}{100}\\- \frac{173}{100} & 5 - \lambda\end{matrix}\right|$$
$$I_{1} = 8$$
$$I_{2} = \frac{120071}{10000}$$
$$I_{3} = - \frac{360213}{5000}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 8 \lambda + \frac{120071}{10000}$$
$$K_{2} = -48$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} < 0$$
then by line type:
this equation is of type : ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 8 \lambda + \frac{120071}{10000} = 0$$
$$\lambda_{1} = 4 - \frac{\sqrt{39929}}{100}$$
$$\lambda_{2} = \frac{\sqrt{39929}}{100} + 4$$
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
$$\tilde x^{2} \left(4 - \frac{\sqrt{39929}}{100}\right) + \tilde y^{2} \left(\frac{\sqrt{39929}}{100} + 4\right) - 6 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{\sqrt{6}}{6} \sqrt{4 - \frac{\sqrt{39929}}{100}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{\sqrt{6}}{6} \sqrt{\frac{\sqrt{39929}}{100} + 4}}\right)^{2}} = 1$$
- reduced to canonical form
$$3 x^{2} - \frac{173 x y}{50} + 5 y^{2} - 6 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = - \frac{173}{100}$$
$$a_{13} = 0$$
$$a_{22} = 5$$
$$a_{23} = 0$$
$$a_{33} = -6$$
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
$$I_{1} = 8$$
| -173 |
| 3 -----|
| 100 |
I2 = | |
|-173 |
|----- 5 |
| 100 |$$I_{3} = \left|\begin{matrix}3 & - \frac{173}{100} & 0\\- \frac{173}{100} & 5 & 0\\0 & 0 & -6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & - \frac{173}{100}\\- \frac{173}{100} & 5 - \lambda\end{matrix}\right|$$
|3 0 | |5 0 |
K2 = | | + | |
|0 -6| |0 -6|$$I_{1} = 8$$
$$I_{2} = \frac{120071}{10000}$$
$$I_{3} = - \frac{360213}{5000}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 8 \lambda + \frac{120071}{10000}$$
$$K_{2} = -48$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} < 0$$
then by line type:
this equation is of type : ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 8 \lambda + \frac{120071}{10000} = 0$$
$$\lambda_{1} = 4 - \frac{\sqrt{39929}}{100}$$
$$\lambda_{2} = \frac{\sqrt{39929}}{100} + 4$$
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
$$\tilde x^{2} \left(4 - \frac{\sqrt{39929}}{100}\right) + \tilde y^{2} \left(\frac{\sqrt{39929}}{100} + 4\right) - 6 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{\sqrt{6}}{6} \sqrt{4 - \frac{\sqrt{39929}}{100}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{\sqrt{6}}{6} \sqrt{\frac{\sqrt{39929}}{100} + 4}}\right)^{2}} = 1$$
- reduced to canonical form