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How to use it?
- Canonical form:
- x^2-4xy+4y^2-20x+10y-50=0
- y=-7+2/5*sqrt(16+6x-x^2)
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25x^2-16y^2+150x+128y-431=0
- x^2=2y^2+3z^2
- Plot:
- xyz
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(x^2+y^2-1)^3-(x^2)*y^3=0
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x^2+(y+cbrt(x^2))^2=1
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Identical expressions
- x^ two -4xy+4y^ two -2 zero x+10y- fifty =0
- x squared minus 4xy plus 4y squared minus 20x plus 10y minus 50 equally 0
- x to the power of two minus 4xy plus 4y to the power of two minus 2 zero x plus 10y minus fifty equally 0
- x2-4xy+4y2-20x+10y-50=0
- x²-4xy+4y²-20x+10y-50=0
- x to the power of 2-4xy+4y to the power of 2-20x+10y-50=0
- x^2-4xy+4y^2-20x+10y-50=O
-
Similar expressions
- x^2-4xy+4y^2-20x+10y+50=0
- x^2-4xy-4y^2-20x+10y-50=0
- x^2+4xy+4y^2-20x+10y-50=0
- x^2-4xy+4y^2+20x+10y-50=0
- x^2-4xy+4y^2-20x-10y-50=0
x^2-4xy+4y^2-20x+10y-50=0 canonical form
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2 2 -50 + x - 20*x + 4*y + 10*y - 4*x*y = 0
$$x^{2} - 4 x y - 20 x + 4 y^{2} + 10 y - 50 = 0$$
x^2 - 4*x*y - 20*x + 4*y^2 + 10*y - 50 = 0
Detail solution
Given line equation of 2-order:
$$x^{2} - 4 x y - 20 x + 4 y^{2} + 10 y - 50 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -2$$
$$a_{13} = -10$$
$$a_{22} = 4$$
$$a_{23} = 5$$
$$a_{33} = -50$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & -2\\-2 & 4\end{matrix}\right|$$
$$\Delta = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{3}{4}$$
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
$$x'^{2} - 4 x' y' - 20 x' + 4 y'^{2} + 10 y' - 50 = 0$$
to
$$4 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} - 4 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) + 10 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) + \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 20 \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) - 50 = 0$$
simplify
$$- 6 \sqrt{5} \tilde x + 5 \tilde y^{2} + 8 \sqrt{5} \tilde y - 50 = 0$$
$$6 \sqrt{5} \tilde x - 5 \tilde y^{2} - 8 \sqrt{5} \tilde y + 50 = 0$$
$$\left(\sqrt{5} \tilde y - 4\right)^{2} = 6 \sqrt{5} \tilde x + 66$$
$$\left(\tilde y - \frac{4 \sqrt{5}}{5}\right)^{2} = \frac{6 \sqrt{5} \left(\tilde x + \frac{11 \sqrt{5}}{5}\right)}{5}$$
$$\tilde y'^{2} = \frac{6 \sqrt{5} \tilde x'}{5}$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \frac{2 \sqrt{5}}{5} + 0 \frac{\sqrt{5}}{5}$$
$$y_{0} = 0 \frac{\sqrt{5}}{5} + 0 \frac{2 \sqrt{5}}{5}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
$$x^{2} - 4 x y - 20 x + 4 y^{2} + 10 y - 50 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -2$$
$$a_{13} = -10$$
$$a_{22} = 4$$
$$a_{23} = 5$$
$$a_{33} = -50$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & -2\\-2 & 4\end{matrix}\right|$$
$$\Delta = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{3}{4}$$
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
$$x'^{2} - 4 x' y' - 20 x' + 4 y'^{2} + 10 y' - 50 = 0$$
to
$$4 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} - 4 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) + 10 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) + \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 20 \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) - 50 = 0$$
simplify
$$- 6 \sqrt{5} \tilde x + 5 \tilde y^{2} + 8 \sqrt{5} \tilde y - 50 = 0$$
$$6 \sqrt{5} \tilde x - 5 \tilde y^{2} - 8 \sqrt{5} \tilde y + 50 = 0$$
$$\left(\sqrt{5} \tilde y - 4\right)^{2} = 6 \sqrt{5} \tilde x + 66$$
$$\left(\tilde y - \frac{4 \sqrt{5}}{5}\right)^{2} = \frac{6 \sqrt{5} \left(\tilde x + \frac{11 \sqrt{5}}{5}\right)}{5}$$
$$\tilde y'^{2} = \frac{6 \sqrt{5} \tilde x'}{5}$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \frac{2 \sqrt{5}}{5} + 0 \frac{\sqrt{5}}{5}$$
$$y_{0} = 0 \frac{\sqrt{5}}{5} + 0 \frac{2 \sqrt{5}}{5}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
Invariants method
Given line equation of 2-order:
$$x^{2} - 4 x y - 20 x + 4 y^{2} + 10 y - 50 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -2$$
$$a_{13} = -10$$
$$a_{22} = 4$$
$$a_{23} = 5$$
$$a_{33} = -50$$
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
$$I_{1} = 5$$
$$I_{3} = \left|\begin{matrix}1 & -2 & -10\\-2 & 4 & 5\\-10 & 5 & -50\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & -2\\-2 & 4 - \lambda\end{matrix}\right|$$
$$I_{1} = 5$$
$$I_{2} = 0$$
$$I_{3} = -225$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda$$
$$K_{2} = -375$$
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
$$6 \sqrt{5} \tilde x + 5 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{6 \sqrt{5}}{5} \tilde x$$
- reduced to canonical form
$$x^{2} - 4 x y - 20 x + 4 y^{2} + 10 y - 50 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -2$$
$$a_{13} = -10$$
$$a_{22} = 4$$
$$a_{23} = 5$$
$$a_{33} = -50$$
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 -2|
I2 = | |
|-2 4 |$$I_{3} = \left|\begin{matrix}1 & -2 & -10\\-2 & 4 & 5\\-10 & 5 & -50\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & -2\\-2 & 4 - \lambda\end{matrix}\right|$$
| 1 -10| |4 5 |
K2 = | | + | |
|-10 -50| |5 -50|$$I_{1} = 5$$
$$I_{2} = 0$$
$$I_{3} = -225$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda$$
$$K_{2} = -375$$
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
$$6 \sqrt{5} \tilde x + 5 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{6 \sqrt{5}}{5} \tilde x$$
- reduced to canonical form