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How to use it?
- Canonical form:
- sqrt(x^2+y^2)=5
- 9x^2+25y^2+18x−100y−116=0
- x²+2y²-12y+10=0
- x=-1+2*sqrt(-5-6y-y^2)
- Plot:
- x^2+(y-((x^2)^(1/(3))))^2=1
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x^2+(y+(x)^(2/3))^2=1
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tg(x)^arctg(x)=e^((y(x))/x)
- x^2+(y-x^2/3)^2=1
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Identical expressions
- 1xx-6xy+9yy+4x+3y- seven
- 1xx minus 6xy plus 9yy plus 4x plus 3y minus 7
- 1xx minus 6xy plus 9yy plus 4x plus 3y minus seven
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Similar expressions
- 1xx-6xy-9yy+4x+3y-7
- 1xx-6xy+9yy-4x+3y-7
- 1xx-6xy+9yy+4x+3y+7
- 1xx-6xy+9yy+4x-3y-7
- 1xx+6xy+9yy+4x+3y-7
1xx-6xy+9yy+4x+3y-7 canonical form
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The solution
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2 2 -7 + x + 3*y + 4*x + 9*y - 6*x*y = 0
$$x^{2} - 6 x y + 4 x + 9 y^{2} + 3 y - 7 = 0$$
x^2 - 6*x*y + 4*x + 9*y^2 + 3*y - 7 = 0
Detail solution
Given line equation of 2-order:
$$x^{2} - 6 x y + 4 x + 9 y^{2} + 3 y - 7 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -3$$
$$a_{13} = 2$$
$$a_{22} = 9$$
$$a_{23} = \frac{3}{2}$$
$$a_{33} = -7$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & -3\\-3 & 9\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
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
$$\cot{\left(2 \phi \right)} = \frac{4}{3}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{4}{3} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{4}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{3 \sqrt{10}}{10}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{10}}{10}$$
substitute coefficients
$$x' = \frac{3 \sqrt{10} \tilde x}{10} - \frac{\sqrt{10} \tilde y}{10}$$
$$y' = \frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}$$
then the equation turns from
$$x'^{2} - 6 x' y' + 4 x' + 9 y'^{2} + 3 y' - 7 = 0$$
to
$$9 \left(\frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right)^{2} - 6 \left(\frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right) \left(\frac{3 \sqrt{10} \tilde x}{10} - \frac{\sqrt{10} \tilde y}{10}\right) + 3 \left(\frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right) + \left(\frac{3 \sqrt{10} \tilde x}{10} - \frac{\sqrt{10} \tilde y}{10}\right)^{2} + 4 \left(\frac{3 \sqrt{10} \tilde x}{10} - \frac{\sqrt{10} \tilde y}{10}\right) - 7 = 0$$
simplify
$$\frac{3 \sqrt{10} \tilde x}{2} + 10 \tilde y^{2} + \frac{\sqrt{10} \tilde y}{2} - 7 = 0$$
$$\left(\sqrt{10} \tilde y + \frac{1}{4}\right)^{2} = - \frac{3 \sqrt{10} \tilde x}{2} + \frac{113}{16}$$
$$\left(\tilde y + \frac{\sqrt{10}}{40}\right)^{2} = - \frac{3 \sqrt{10} \left(\tilde x - \frac{113 \sqrt{10}}{240}\right)}{20}$$
$$\tilde y'^{2} = - \frac{3 \sqrt{10} \tilde x'}{20}$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \frac{3 \sqrt{10}}{10} + 0 \frac{\sqrt{10}}{10}$$
$$y_{0} = 0 \frac{\sqrt{10}}{10} + 0 \frac{3 \sqrt{10}}{10}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{3 \sqrt{10}}{10}, \ \frac{\sqrt{10}}{10}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{10}}{10}, \ \frac{3 \sqrt{10}}{10}\right)$$
$$x^{2} - 6 x y + 4 x + 9 y^{2} + 3 y - 7 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -3$$
$$a_{13} = 2$$
$$a_{22} = 9$$
$$a_{23} = \frac{3}{2}$$
$$a_{33} = -7$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & -3\\-3 & 9\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
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
$$\cot{\left(2 \phi \right)} = \frac{4}{3}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{4}{3} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{4}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{3 \sqrt{10}}{10}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{10}}{10}$$
substitute coefficients
$$x' = \frac{3 \sqrt{10} \tilde x}{10} - \frac{\sqrt{10} \tilde y}{10}$$
$$y' = \frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}$$
then the equation turns from
$$x'^{2} - 6 x' y' + 4 x' + 9 y'^{2} + 3 y' - 7 = 0$$
to
$$9 \left(\frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right)^{2} - 6 \left(\frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right) \left(\frac{3 \sqrt{10} \tilde x}{10} - \frac{\sqrt{10} \tilde y}{10}\right) + 3 \left(\frac{\sqrt{10} \tilde x}{10} + \frac{3 \sqrt{10} \tilde y}{10}\right) + \left(\frac{3 \sqrt{10} \tilde x}{10} - \frac{\sqrt{10} \tilde y}{10}\right)^{2} + 4 \left(\frac{3 \sqrt{10} \tilde x}{10} - \frac{\sqrt{10} \tilde y}{10}\right) - 7 = 0$$
simplify
$$\frac{3 \sqrt{10} \tilde x}{2} + 10 \tilde y^{2} + \frac{\sqrt{10} \tilde y}{2} - 7 = 0$$
$$\left(\sqrt{10} \tilde y + \frac{1}{4}\right)^{2} = - \frac{3 \sqrt{10} \tilde x}{2} + \frac{113}{16}$$
$$\left(\tilde y + \frac{\sqrt{10}}{40}\right)^{2} = - \frac{3 \sqrt{10} \left(\tilde x - \frac{113 \sqrt{10}}{240}\right)}{20}$$
$$\tilde y'^{2} = - \frac{3 \sqrt{10} \tilde x'}{20}$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \frac{3 \sqrt{10}}{10} + 0 \frac{\sqrt{10}}{10}$$
$$y_{0} = 0 \frac{\sqrt{10}}{10} + 0 \frac{3 \sqrt{10}}{10}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{3 \sqrt{10}}{10}, \ \frac{\sqrt{10}}{10}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{10}}{10}, \ \frac{3 \sqrt{10}}{10}\right)$$
Invariants method
Given line equation of 2-order:
$$x^{2} - 6 x y + 4 x + 9 y^{2} + 3 y - 7 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -3$$
$$a_{13} = 2$$
$$a_{22} = 9$$
$$a_{23} = \frac{3}{2}$$
$$a_{33} = -7$$
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
$$I_{1} = 10$$
$$I_{3} = \left|\begin{matrix}1 & -3 & 2\\-3 & 9 & \frac{3}{2}\\2 & \frac{3}{2} & -7\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & -3\\-3 & 9 - \lambda\end{matrix}\right|$$
$$I_{1} = 10$$
$$I_{2} = 0$$
$$I_{3} = - \frac{225}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 10 \lambda$$
$$K_{2} = - \frac{305}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : parabola
$$I_{1} \tilde y^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0$$
or
$$\frac{3 \sqrt{10} \tilde x}{2} + 10 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{3 \sqrt{10}}{20} \tilde x$$
- reduced to canonical form
$$x^{2} - 6 x y + 4 x + 9 y^{2} + 3 y - 7 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -3$$
$$a_{13} = 2$$
$$a_{22} = 9$$
$$a_{23} = \frac{3}{2}$$
$$a_{33} = -7$$
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
$$I_{1} = 10$$
|1 -3|
I2 = | |
|-3 9 |$$I_{3} = \left|\begin{matrix}1 & -3 & 2\\-3 & 9 & \frac{3}{2}\\2 & \frac{3}{2} & -7\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & -3\\-3 & 9 - \lambda\end{matrix}\right|$$
|1 2 | | 9 3/2|
K2 = | | + | |
|2 -7| |3/2 -7 |$$I_{1} = 10$$
$$I_{2} = 0$$
$$I_{3} = - \frac{225}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 10 \lambda$$
$$K_{2} = - \frac{305}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : parabola
$$I_{1} \tilde y^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0$$
or
$$\frac{3 \sqrt{10} \tilde x}{2} + 10 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{3 \sqrt{10}}{20} \tilde x$$
- reduced to canonical form