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How to use it?
- Canonical form:
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5x^2 + 5y^2 + 6xy – 6x – 10y – 3 = 0
- 9x2+12xy+4y2-24x-16y+3=0
- x^2+y^2+4z^2−2xy+4xz−4yz−2x+2y+2z=0
- 9x^2+16y^2-24xy-20x+110y-50=0
- Plot:
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xy=4
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xy=6
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4x^2+4y^2=25
- xyz
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Identical expressions
- 5x^ two + 5y^ two + 6xy – 6x – 1 zero y – three = 0
- 5x squared plus 5y squared plus 6xy – 6x – 10y – 3 equally 0
- 5x to the power of two plus 5y to the power of two plus 6xy – 6x – 1 zero y – three equally 0
- 5x2 + 5y2 + 6xy – 6x – 10y – 3 = 0
- 5x² + 5y² + 6xy – 6x – 10y – 3 = 0
- 5x to the power of 2 + 5y to the power of 2 + 6xy – 6x – 10y – 3 = 0
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Similar expressions
- 5x^2 - 5y^2 + 6xy – 6x – 10y – 3 = 0
- 5x^2 + 5y^2 - 6xy – 6x – 10y – 3 = 0
5x^2 + 5y^2 + 6xy – 6x – 10y – 3 = 0 canonical form
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The solution
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2 2 -3 - 10*y - 6*x + 5*x + 5*y + 6*x*y = 0
$$5 x^{2} + 6 x y + 5 y^{2} - 6 x - 10 y - 3 = 0$$
5*x^2 + 6*x*y - 6*x + 5*y^2 - 10*y - 3 = 0
Detail solution
Given line equation of 2-order:
$$5 x^{2} + 6 x y + 5 y^{2} - 6 x - 10 y - 3 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 5$$
$$a_{12} = 3$$
$$a_{13} = -3$$
$$a_{22} = 5$$
$$a_{23} = -5$$
$$a_{33} = -3$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}5 & 3\\3 & 5\end{matrix}\right|$$
$$\Delta = 16$$
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
$$5 x_{0} + 3 y_{0} - 3 = 0$$
$$3 x_{0} + 5 y_{0} - 5 = 0$$
then
$$x_{0} = 0$$
$$y_{0} = 1$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - 3 x_{0} - 5 y_{0} - 3$$
$$a'_{33} = -8$$
then The equation is transformed to
$$5 x'^{2} + 6 x' y' + 5 y'^{2} - 8 = 0$$
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
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{- \cos^{2}{\left(\phi \right)} + 1}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
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
$$5 x'^{2} + 6 x' y' + 5 y'^{2} - 8 = 0$$
to
$$5 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right)^{2} + 6 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) + 5 \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 8 = 0$$
simplify
$$8 \tilde x^{2} + 2 \tilde y^{2} - 8 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{1^{2}} + \frac{\tilde y^{2}}{2^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$\vec e_{2} = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$5 x^{2} + 6 x y + 5 y^{2} - 6 x - 10 y - 3 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 5$$
$$a_{12} = 3$$
$$a_{13} = -3$$
$$a_{22} = 5$$
$$a_{23} = -5$$
$$a_{33} = -3$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}5 & 3\\3 & 5\end{matrix}\right|$$
$$\Delta = 16$$
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
$$5 x_{0} + 3 y_{0} - 3 = 0$$
$$3 x_{0} + 5 y_{0} - 5 = 0$$
then
$$x_{0} = 0$$
$$y_{0} = 1$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - 3 x_{0} - 5 y_{0} - 3$$
$$a'_{33} = -8$$
then The equation is transformed to
$$5 x'^{2} + 6 x' y' + 5 y'^{2} - 8 = 0$$
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
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{- \cos^{2}{\left(\phi \right)} + 1}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
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
$$5 x'^{2} + 6 x' y' + 5 y'^{2} - 8 = 0$$
to
$$5 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right)^{2} + 6 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) + 5 \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 8 = 0$$
simplify
$$8 \tilde x^{2} + 2 \tilde y^{2} - 8 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{1^{2}} + \frac{\tilde y^{2}}{2^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 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:
$$5 x^{2} + 6 x y + 5 y^{2} - 6 x - 10 y - 3 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 5$$
$$a_{12} = 3$$
$$a_{13} = -3$$
$$a_{22} = 5$$
$$a_{23} = -5$$
$$a_{33} = -3$$
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
$$I_{1} = 10$$
$$I_{3} = \left|\begin{matrix}5 & 3 & -3\\3 & 5 & -5\\-3 & -5 & -3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 5 & 3\\3 & - \lambda + 5\end{matrix}\right|$$
$$I_{1} = 10$$
$$I_{2} = 16$$
$$I_{3} = -128$$
$$I{\left(\lambda \right)} = \lambda^{2} - 10 \lambda + 16$$
$$K_{2} = -64$$
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:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - 10 \lambda + 16 = 0$$
Solve this equation
$$\lambda_{1} = 8$$
$$\lambda_{2} = 2$$
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
$$8 \tilde x^{2} + 2 \tilde y^{2} - 8 = 0$$
$$\frac{\tilde x^{2}}{1^{2}} + \frac{\tilde y^{2}}{2^{2}} = 1$$
- reduced to canonical form
$$5 x^{2} + 6 x y + 5 y^{2} - 6 x - 10 y - 3 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 5$$
$$a_{12} = 3$$
$$a_{13} = -3$$
$$a_{22} = 5$$
$$a_{23} = -5$$
$$a_{33} = -3$$
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
$$I_{1} = 10$$
|5 3|
I2 = | |
|3 5|$$I_{3} = \left|\begin{matrix}5 & 3 & -3\\3 & 5 & -5\\-3 & -5 & -3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 5 & 3\\3 & - \lambda + 5\end{matrix}\right|$$
|5 -3| |5 -5|
K2 = | | + | |
|-3 -3| |-5 -3|$$I_{1} = 10$$
$$I_{2} = 16$$
$$I_{3} = -128$$
$$I{\left(\lambda \right)} = \lambda^{2} - 10 \lambda + 16$$
$$K_{2} = -64$$
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:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - 10 \lambda + 16 = 0$$
Solve this equation
$$\lambda_{1} = 8$$
$$\lambda_{2} = 2$$
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
$$8 \tilde x^{2} + 2 \tilde y^{2} - 8 = 0$$
$$\frac{\tilde x^{2}}{1^{2}} + \frac{\tilde y^{2}}{2^{2}} = 1$$
- reduced to canonical form
The graph