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How to use it?
- Canonical form:
- x^2+y^2-14x-8y+40=0
- 4x^2-9y^2-18y+16z^2-72z=-27
- 2x^2+2y^2+14x-10y+35=0
- 5*x^2+y^2+4*z^2+6*x*y+8*x*z+4*y*z-4*x-4*y-4*z=0
- Plot:
-
sqrt((x-4)^2+y^2)+sqrt(x^2+(y+4)^2)=8*sqrt(2)
- sin(999*x)+y=1-y
- 3*x-3*y-3=6
- 3*x+2*y=12
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Identical expressions
- -4x^ two + two xy-4y^2+ one zero x-10y+1=0
- minus 4x squared plus 2xy minus 4y squared plus 10x minus 10y plus 1 equally 0
- minus 4x to the power of two plus two xy minus 4y squared plus one zero x minus 10y plus 1 equally 0
- -4x2+2xy-4y2+10x-10y+1=0
- -4x²+2xy-4y²+10x-10y+1=0
- -4x to the power of 2+2xy-4y to the power of 2+10x-10y+1=0
- -4x^2+2xy-4y^2+10x-10y+1=O
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Similar expressions
- -4x^2-2xy-4y^2+10x-10y+1=0
- -4x^2+2xy-4y^2-10x-10y+1=0
- -4x^2+2xy-4y^2+10x-10y-1=0
- -4x^2+2xy+4y^2+10x-10y+1=0
- 4x^2+2xy-4y^2+10x-10y+1=0
- -4x^2+2xy-4y^2+10x+10y+1=0
-4x^2+2xy-4y^2+10x-10y+1=0 canonical form
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The solution
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2 2 1 - 10*y - 4*x - 4*y + 10*x + 2*x*y = 0
$$- 4 x^{2} + 2 x y + 10 x - 4 y^{2} - 10 y + 1 = 0$$
-4*x^2 + 2*x*y + 10*x - 4*y^2 - 10*y + 1 = 0
Detail solution
Given line equation of 2-order:
$$- 4 x^{2} + 2 x y + 10 x - 4 y^{2} - 10 y + 1 = 0$$
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
$$a_{11} = -4$$
$$a_{12} = 1$$
$$a_{13} = 5$$
$$a_{22} = -4$$
$$a_{23} = -5$$
$$a_{33} = 1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-4 & 1\\1 & -4\end{matrix}\right|$$
$$\Delta = 15$$
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
$$- 4 x_{0} + y_{0} + 5 = 0$$
$$x_{0} - 4 y_{0} - 5 = 0$$
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
$$a'_{33} = 5 x_{0} - 5 y_{0} + 1$$
$$a'_{33} = 11$$
then equation turns into
$$- 4 x'^{2} + 2 x' y' - 4 y'^{2} + 11 = 0$$
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
$$\cot{\left(2 \phi \right)} = 0$$
then
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
substitute coefficients
$$x' = \frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}$$
$$y' = \frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}$$
then the equation turns from
$$- 4 x'^{2} + 2 x' y' - 4 y'^{2} + 11 = 0$$
to
$$- 4 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right)^{2} + 2 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) - 4 \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right)^{2} + 11 = 0$$
simplify
$$- 3 \tilde x^{2} - 5 \tilde y^{2} + 11 = 0$$
$$3 \tilde x^{2} + 5 \tilde y^{2} - 11 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{11} \sqrt{11}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{11} \sqrt{11}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$- 4 x^{2} + 2 x y + 10 x - 4 y^{2} - 10 y + 1 = 0$$
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
$$a_{11} = -4$$
$$a_{12} = 1$$
$$a_{13} = 5$$
$$a_{22} = -4$$
$$a_{23} = -5$$
$$a_{33} = 1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-4 & 1\\1 & -4\end{matrix}\right|$$
$$\Delta = 15$$
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
$$- 4 x_{0} + y_{0} + 5 = 0$$
$$x_{0} - 4 y_{0} - 5 = 0$$
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
$$a'_{33} = 5 x_{0} - 5 y_{0} + 1$$
$$a'_{33} = 11$$
then equation turns into
$$- 4 x'^{2} + 2 x' y' - 4 y'^{2} + 11 = 0$$
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
$$\cot{\left(2 \phi \right)} = 0$$
then
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
substitute coefficients
$$x' = \frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}$$
$$y' = \frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}$$
then the equation turns from
$$- 4 x'^{2} + 2 x' y' - 4 y'^{2} + 11 = 0$$
to
$$- 4 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right)^{2} + 2 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) - 4 \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right)^{2} + 11 = 0$$
simplify
$$- 3 \tilde x^{2} - 5 \tilde y^{2} + 11 = 0$$
$$3 \tilde x^{2} + 5 \tilde y^{2} - 11 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{11} \sqrt{11}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{11} \sqrt{11}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(1, -1)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
Invariants method
Given line equation of 2-order:
$$- 4 x^{2} + 2 x y + 10 x - 4 y^{2} - 10 y + 1 = 0$$
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
$$a_{11} = -4$$
$$a_{12} = 1$$
$$a_{13} = 5$$
$$a_{22} = -4$$
$$a_{23} = -5$$
$$a_{33} = 1$$
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
$$I_{1} = -8$$
$$I_{3} = \left|\begin{matrix}-4 & 1 & 5\\1 & -4 & -5\\5 & -5 & 1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 4 & 1\\1 & - \lambda - 4\end{matrix}\right|$$
$$I_{1} = -8$$
$$I_{2} = 15$$
$$I_{3} = 165$$
$$I{\left(\lambda \right)} = \lambda^{2} + 8 \lambda + 15$$
$$K_{2} = -58$$
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
$$\lambda^{2} + 8 \lambda + 15 = 0$$
$$\lambda_{1} = -3$$
$$\lambda_{2} = -5$$
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
$$- 3 \tilde x^{2} - 5 \tilde y^{2} + 11 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{11} \sqrt{11}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{11} \sqrt{11}}\right)^{2}} = 1$$
- reduced to canonical form
$$- 4 x^{2} + 2 x y + 10 x - 4 y^{2} - 10 y + 1 = 0$$
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
$$a_{11} = -4$$
$$a_{12} = 1$$
$$a_{13} = 5$$
$$a_{22} = -4$$
$$a_{23} = -5$$
$$a_{33} = 1$$
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
$$I_{1} = -8$$
|-4 1 |
I2 = | |
|1 -4|$$I_{3} = \left|\begin{matrix}-4 & 1 & 5\\1 & -4 & -5\\5 & -5 & 1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 4 & 1\\1 & - \lambda - 4\end{matrix}\right|$$
|-4 5| |-4 -5|
K2 = | | + | |
|5 1| |-5 1 |$$I_{1} = -8$$
$$I_{2} = 15$$
$$I_{3} = 165$$
$$I{\left(\lambda \right)} = \lambda^{2} + 8 \lambda + 15$$
$$K_{2} = -58$$
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
$$\lambda^{2} + 8 \lambda + 15 = 0$$
$$\lambda_{1} = -3$$
$$\lambda_{2} = -5$$
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
$$- 3 \tilde x^{2} - 5 \tilde y^{2} + 11 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{11} \sqrt{11}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{11} \sqrt{11}}\right)^{2}} = 1$$
- reduced to canonical form