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How to use it?
- Canonical form:
- uxy+uxy-2uxy-3uy-6y=0
- 2x^2+12xy-7y^2+16x-2y-3=0
- 9x^2-36y^2+4z^2-18x+72y+16z-47=0
- x^2+y^2+z^2+8x-6y+2z+17=0
- Plot:
- tan(x*factorial(y))-1=0
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sqrtx+sqrty=4
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|y|=|x|
- (x^2+y^3-1)^3*x^2*y^3=0
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Identical expressions
- two x^ two +12xy-7y^2+16x-2y- three = zero
- 2x squared plus 12xy minus 7y squared plus 16x minus 2y minus 3 equally 0
- two x to the power of two plus 12xy minus 7y squared plus 16x minus 2y minus three equally zero
- 2x2+12xy-7y2+16x-2y-3=0
- 2x²+12xy-7y²+16x-2y-3=0
- 2x to the power of 2+12xy-7y to the power of 2+16x-2y-3=0
- 2x^2+12xy-7y^2+16x-2y-3=O
-
Similar expressions
- 2x^2-12xy-7y^2+16x-2y-3=0
- 2x^2+12xy-7y^2+16x+2y-3=0
- 2x^2+12xy-7y^2+16x-2y+3=0
- 2x^2+12xy+7y^2+16x-2y-3=0
- 2x^2+12xy-7y^2-16x-2y-3=0
2x^2+12xy-7y^2+16x-2y-3=0 canonical form
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The solution
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2 2 -3 - 7*y - 2*y + 2*x + 16*x + 12*x*y = 0
$$2 x^{2} + 12 x y + 16 x - 7 y^{2} - 2 y - 3 = 0$$
2*x^2 + 12*x*y + 16*x - 7*y^2 - 2*y - 3 = 0
Detail solution
Given line equation of 2-order:
$$2 x^{2} + 12 x y + 16 x - 7 y^{2} - 2 y - 3 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 6$$
$$a_{13} = 8$$
$$a_{22} = -7$$
$$a_{23} = -1$$
$$a_{33} = -3$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}2 & 6\\6 & -7\end{matrix}\right|$$
$$\Delta = -50$$
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
$$2 x_{0} + 6 y_{0} + 8 = 0$$
$$6 x_{0} - 7 y_{0} - 1 = 0$$
then
$$x_{0} = -1$$
$$y_{0} = -1$$
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
$$a'_{33} = 8 x_{0} - y_{0} - 3$$
$$a'_{33} = -10$$
then equation turns into
$$2 x'^{2} + 12 x' y' - 7 y'^{2} - 10 = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{3}{4}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{3}{4} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{4}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{2 \sqrt{5}}{5}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{5}}{5}$$
substitute coefficients
$$x' = \frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}$$
$$y' = \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}$$
then the equation turns from
$$2 x'^{2} + 12 x' y' - 7 y'^{2} - 10 = 0$$
to
$$- 7 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} + 12 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) + 2 \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 10 = 0$$
simplify
$$5 \tilde x^{2} - 10 \tilde y^{2} - 10 = 0$$
$$- 5 \tilde x^{2} + 10 \tilde y^{2} + 10 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{2} - \frac{\tilde y^{2}}{1} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
$$2 x^{2} + 12 x y + 16 x - 7 y^{2} - 2 y - 3 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 6$$
$$a_{13} = 8$$
$$a_{22} = -7$$
$$a_{23} = -1$$
$$a_{33} = -3$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}2 & 6\\6 & -7\end{matrix}\right|$$
$$\Delta = -50$$
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
$$2 x_{0} + 6 y_{0} + 8 = 0$$
$$6 x_{0} - 7 y_{0} - 1 = 0$$
then
$$x_{0} = -1$$
$$y_{0} = -1$$
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
$$a'_{33} = 8 x_{0} - y_{0} - 3$$
$$a'_{33} = -10$$
then equation turns into
$$2 x'^{2} + 12 x' y' - 7 y'^{2} - 10 = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{3}{4}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{3}{4} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{4}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{2 \sqrt{5}}{5}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{5}}{5}$$
substitute coefficients
$$x' = \frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}$$
$$y' = \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}$$
then the equation turns from
$$2 x'^{2} + 12 x' y' - 7 y'^{2} - 10 = 0$$
to
$$- 7 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} + 12 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) + 2 \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 10 = 0$$
simplify
$$5 \tilde x^{2} - 10 \tilde y^{2} - 10 = 0$$
$$- 5 \tilde x^{2} + 10 \tilde y^{2} + 10 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{2} - \frac{\tilde y^{2}}{1} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(-1, -1)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
Invariants method
Given line equation of 2-order:
$$2 x^{2} + 12 x y + 16 x - 7 y^{2} - 2 y - 3 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 6$$
$$a_{13} = 8$$
$$a_{22} = -7$$
$$a_{23} = -1$$
$$a_{33} = -3$$
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
$$I_{1} = -5$$
$$I_{3} = \left|\begin{matrix}2 & 6 & 8\\6 & -7 & -1\\8 & -1 & -3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 6\\6 & - \lambda - 7\end{matrix}\right|$$
$$I_{1} = -5$$
$$I_{2} = -50$$
$$I_{3} = 500$$
$$I{\left(\lambda \right)} = \lambda^{2} + 5 \lambda - 50$$
$$K_{2} = -50$$
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
$$\lambda^{2} + 5 \lambda - 50 = 0$$
$$\lambda_{1} = 5$$
$$\lambda_{2} = -10$$
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
$$5 \tilde x^{2} - 10 \tilde y^{2} - 10 = 0$$
$$\frac{\tilde x^{2}}{2} - \frac{\tilde y^{2}}{1} = 1$$
- reduced to canonical form
$$2 x^{2} + 12 x y + 16 x - 7 y^{2} - 2 y - 3 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 6$$
$$a_{13} = 8$$
$$a_{22} = -7$$
$$a_{23} = -1$$
$$a_{33} = -3$$
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
$$I_{1} = -5$$
|2 6 |
I2 = | |
|6 -7|$$I_{3} = \left|\begin{matrix}2 & 6 & 8\\6 & -7 & -1\\8 & -1 & -3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 6\\6 & - \lambda - 7\end{matrix}\right|$$
|2 8 | |-7 -1|
K2 = | | + | |
|8 -3| |-1 -3|$$I_{1} = -5$$
$$I_{2} = -50$$
$$I_{3} = 500$$
$$I{\left(\lambda \right)} = \lambda^{2} + 5 \lambda - 50$$
$$K_{2} = -50$$
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
$$\lambda^{2} + 5 \lambda - 50 = 0$$
$$\lambda_{1} = 5$$
$$\lambda_{2} = -10$$
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
$$5 \tilde x^{2} - 10 \tilde y^{2} - 10 = 0$$
$$\frac{\tilde x^{2}}{2} - \frac{\tilde y^{2}}{1} = 1$$
- reduced to canonical form