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How to use it?
- Canonical form:
- 15x*x=0
- 4x^2+24xy+11y^2+64x+42y+51=0
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9x^2-16y^2+18x+64y-199=0
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xy=2400
- Plot:
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x^2+(y-cbrt(x^2))^2=1
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-1-(36*(x^2-y^2+5*x+6)-(36*x+132)*(2*x+5))/((x^2-y^2+5*x+6)^2+4*x^2*y+25*y^2+20*x*y^2)=0
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sqrt(x)+sqrt(y)=1
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tan(y)^2=tan(x)
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Identical expressions
- seventeen ^ two +1 two xy+8y^2-46x-28y+ seventeen = zero
- 17 squared plus 12xy plus 8y squared minus 46x minus 28y plus 17 equally 0
- seventeen to the power of two plus 1 two xy plus 8y squared minus 46x minus 28y plus seventeen equally zero
- 172+12xy+8y2-46x-28y+17=0
- 17²+12xy+8y²-46x-28y+17=0
- 17 to the power of 2+12xy+8y to the power of 2-46x-28y+17=0
- 17^2+12xy+8y^2-46x-28y+17=O
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Similar expressions
- 17^2+12xy+8y^2-46x-28y-17=0
- 17^2+12xy-8y^2-46x-28y+17=0
- 17^2+12xy+8y^2-46x+28y+17=0
- 17^2+12xy+8y^2+46x-28y+17=0
- 17^2-12xy+8y^2-46x-28y+17=0
17^2+12xy+8y^2-46x-28y+17=0 canonical form
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2 306 - 46*x - 28*y + 8*y + 12*x*y = 0
$$12 x y - 46 x + 8 y^{2} - 28 y + 306 = 0$$
12*x*y - 46*x + 8*y^2 - 28*y + 306 = 0
Detail solution
Given line equation of 2-order:
$$12 x y - 46 x + 8 y^{2} - 28 y + 306 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 6$$
$$a_{13} = -23$$
$$a_{22} = 8$$
$$a_{23} = -14$$
$$a_{33} = 306$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 6\\6 & 8\end{matrix}\right|$$
$$\Delta = -36$$
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
$$6 y_{0} - 23 = 0$$
$$6 x_{0} + 8 y_{0} - 14 = 0$$
then
$$x_{0} = - \frac{25}{9}$$
$$y_{0} = \frac{23}{6}$$
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
$$a'_{33} = - 23 x_{0} - 14 y_{0} + 306$$
$$a'_{33} = \frac{2846}{9}$$
then equation turns into
$$12 x' y' + 8 y'^{2} + \frac{2846}{9} = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{2}{3}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{2}{3} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{3 \sqrt{13}}{13}$$
$$\cos{\left(2 \phi \right)} = \frac{2 \sqrt{13}}{13}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}}$$
$$y' = - \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}} + \tilde y \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}}$$
then the equation turns from
$$12 x' y' + 8 y'^{2} + \frac{2846}{9} = 0$$
to
$$8 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}} + \tilde y \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}}\right)^{2} + 12 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}} + \tilde y \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}}\right) + \frac{2846}{9} = 0$$
simplify
$$- 12 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}} \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}} - \frac{8 \sqrt{13} \tilde x^{2}}{13} + 4 \tilde x^{2} - 16 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}} \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}} + \frac{24 \sqrt{13} \tilde x \tilde y}{13} + \frac{8 \sqrt{13} \tilde y^{2}}{13} + 4 \tilde y^{2} + 12 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}} \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}} + \frac{2846}{9} = 0$$
$$- 2 \sqrt{13} \tilde x^{2} + 4 \tilde x^{2} + 4 \tilde y^{2} + 2 \sqrt{13} \tilde y^{2} + \frac{2846}{9} = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{\frac{2846}{9} \frac{1}{-4 + 2 \sqrt{13}}} - \frac{\tilde y^{2}}{\frac{2846}{9} \frac{1}{4 + 2 \sqrt{13}}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}}\right)$$
$$\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}}, \ \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}}\right)$$
$$12 x y - 46 x + 8 y^{2} - 28 y + 306 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 6$$
$$a_{13} = -23$$
$$a_{22} = 8$$
$$a_{23} = -14$$
$$a_{33} = 306$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 6\\6 & 8\end{matrix}\right|$$
$$\Delta = -36$$
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
$$6 y_{0} - 23 = 0$$
$$6 x_{0} + 8 y_{0} - 14 = 0$$
then
$$x_{0} = - \frac{25}{9}$$
$$y_{0} = \frac{23}{6}$$
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
$$a'_{33} = - 23 x_{0} - 14 y_{0} + 306$$
$$a'_{33} = \frac{2846}{9}$$
then equation turns into
$$12 x' y' + 8 y'^{2} + \frac{2846}{9} = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{2}{3}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{2}{3} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{3 \sqrt{13}}{13}$$
$$\cos{\left(2 \phi \right)} = \frac{2 \sqrt{13}}{13}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}}$$
$$y' = - \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}} + \tilde y \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}}$$
then the equation turns from
$$12 x' y' + 8 y'^{2} + \frac{2846}{9} = 0$$
to
$$8 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}} + \tilde y \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}}\right)^{2} + 12 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}} + \tilde y \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}}\right) + \frac{2846}{9} = 0$$
simplify
$$- 12 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}} \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}} - \frac{8 \sqrt{13} \tilde x^{2}}{13} + 4 \tilde x^{2} - 16 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}} \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}} + \frac{24 \sqrt{13} \tilde x \tilde y}{13} + \frac{8 \sqrt{13} \tilde y^{2}}{13} + 4 \tilde y^{2} + 12 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}} \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}} + \frac{2846}{9} = 0$$
$$- 2 \sqrt{13} \tilde x^{2} + 4 \tilde x^{2} + 4 \tilde y^{2} + 2 \sqrt{13} \tilde y^{2} + \frac{2846}{9} = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{\frac{2846}{9} \frac{1}{-4 + 2 \sqrt{13}}} - \frac{\tilde y^{2}}{\frac{2846}{9} \frac{1}{4 + 2 \sqrt{13}}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(-25/9, 23/6)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}}\right)$$
$$\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{\sqrt{13}}{13}}, \ \sqrt{\frac{\sqrt{13}}{13} + \frac{1}{2}}\right)$$
Invariants method
Given line equation of 2-order:
$$12 x y - 46 x + 8 y^{2} - 28 y + 306 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 6$$
$$a_{13} = -23$$
$$a_{22} = 8$$
$$a_{23} = -14$$
$$a_{33} = 306$$
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
$$I_{1} = 8$$
$$I_{3} = \left|\begin{matrix}0 & 6 & -23\\6 & 8 & -14\\-23 & -14 & 306\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 6\\6 & 8 - \lambda\end{matrix}\right|$$
$$I_{1} = 8$$
$$I_{2} = -36$$
$$I_{3} = -11384$$
$$I{\left(\lambda \right)} = \lambda^{2} - 8 \lambda - 36$$
$$K_{2} = 1723$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 8 \lambda - 36 = 0$$
$$\lambda_{1} = 4 - 2 \sqrt{13}$$
$$\lambda_{2} = 4 + 2 \sqrt{13}$$
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
$$\tilde x^{2} \left(4 - 2 \sqrt{13}\right) + \tilde y^{2} \left(4 + 2 \sqrt{13}\right) + \frac{2846}{9} = 0$$
$$\frac{\tilde x^{2}}{\frac{2846}{9} \frac{1}{-4 + 2 \sqrt{13}}} - \frac{\tilde y^{2}}{\frac{2846}{9} \frac{1}{4 + 2 \sqrt{13}}} = 1$$
- reduced to canonical form
$$12 x y - 46 x + 8 y^{2} - 28 y + 306 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 6$$
$$a_{13} = -23$$
$$a_{22} = 8$$
$$a_{23} = -14$$
$$a_{33} = 306$$
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
$$I_{1} = 8$$
|0 6|
I2 = | |
|6 8|$$I_{3} = \left|\begin{matrix}0 & 6 & -23\\6 & 8 & -14\\-23 & -14 & 306\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 6\\6 & 8 - \lambda\end{matrix}\right|$$
| 0 -23| | 8 -14|
K2 = | | + | |
|-23 306| |-14 306|$$I_{1} = 8$$
$$I_{2} = -36$$
$$I_{3} = -11384$$
$$I{\left(\lambda \right)} = \lambda^{2} - 8 \lambda - 36$$
$$K_{2} = 1723$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 8 \lambda - 36 = 0$$
$$\lambda_{1} = 4 - 2 \sqrt{13}$$
$$\lambda_{2} = 4 + 2 \sqrt{13}$$
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
$$\tilde x^{2} \left(4 - 2 \sqrt{13}\right) + \tilde y^{2} \left(4 + 2 \sqrt{13}\right) + \frac{2846}{9} = 0$$
$$\frac{\tilde x^{2}}{\frac{2846}{9} \frac{1}{-4 + 2 \sqrt{13}}} - \frac{\tilde y^{2}}{\frac{2846}{9} \frac{1}{4 + 2 \sqrt{13}}} = 1$$
- reduced to canonical form