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How to use it?
- Canonical form:
- 4x^2+9y^2=25
- xy+4=0
- 8x^2+20y^2-24+y=7
- x1^2+x2^2+3x3^2+4x1x2+2x1x3+2x2x3=0
- Plot:
-
x²+(y-x⅔)²=1
- x^2+(y-(x^2)^(1/3))^2=1
-
3x^2+5y^2=25
- y=cbrt(x)
-
Identical expressions
- 8x^ two + two 0y^2- twenty-four +y= seven
- 8x squared plus 20y squared minus 24 plus y equally 7
- 8x to the power of two plus two 0y squared minus twenty minus four plus y equally seven
- 8x2+20y2-24+y=7
- 8x²+20y²-24+y=7
- 8x to the power of 2+20y to the power of 2-24+y=7
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Similar expressions
- 8x^2+20y^2+24+y=7
- 8x^2-20y^2-24+y=7
- 8x^2+20y^2-24-y=7
8x^2+20y^2-24+y=7 canonical form
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2 2 -31 + y + 8*x + 20*y = 0
$$8 x^{2} + 20 y^{2} + y - 31 = 0$$
8*x^2 + 20*y^2 + y - 31 = 0
Detail solution
Given line equation of 2-order:
$$8 x^{2} + 20 y^{2} + y - 31 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 20$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = -31$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}8 & 0\\0 & 20\end{matrix}\right|$$
$$\Delta = 160$$
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
$$8 x_{0} = 0$$
$$20 y_{0} + \frac{1}{2} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = - \frac{1}{40}$$
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
$$a'_{33} = \frac{y_{0}}{2} - 31$$
$$a'_{33} = - \frac{2481}{80}$$
then equation turns into
$$8 x'^{2} + 20 y'^{2} - \frac{2481}{80} = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{4} \sqrt{2}}{\frac{4}{2481} \sqrt{12405}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{10} \sqrt{5}}{\frac{4}{2481} \sqrt{12405}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$8 x^{2} + 20 y^{2} + y - 31 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 20$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = -31$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}8 & 0\\0 & 20\end{matrix}\right|$$
$$\Delta = 160$$
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
$$8 x_{0} = 0$$
$$20 y_{0} + \frac{1}{2} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = - \frac{1}{40}$$
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
$$a'_{33} = \frac{y_{0}}{2} - 31$$
$$a'_{33} = - \frac{2481}{80}$$
then equation turns into
$$8 x'^{2} + 20 y'^{2} - \frac{2481}{80} = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{4} \sqrt{2}}{\frac{4}{2481} \sqrt{12405}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{10} \sqrt{5}}{\frac{4}{2481} \sqrt{12405}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, -1/40)
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
Invariants method
Given line equation of 2-order:
$$8 x^{2} + 20 y^{2} + y - 31 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 20$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = -31$$
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
$$I_{1} = 28$$
$$I_{3} = \left|\begin{matrix}8 & 0 & 0\\0 & 20 & \frac{1}{2}\\0 & \frac{1}{2} & -31\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}8 - \lambda & 0\\0 & 20 - \lambda\end{matrix}\right|$$
$$I_{1} = 28$$
$$I_{2} = 160$$
$$I_{3} = -4962$$
$$I{\left(\lambda \right)} = \lambda^{2} - 28 \lambda + 160$$
$$K_{2} = - \frac{3473}{4}$$
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
$$\lambda^{2} - 28 \lambda + 160 = 0$$
$$\lambda_{1} = 20$$
$$\lambda_{2} = 8$$
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
$$20 \tilde x^{2} + 8 \tilde y^{2} - \frac{2481}{80} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{10} \sqrt{5}}{\frac{4}{2481} \sqrt{12405}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{4} \sqrt{2}}{\frac{4}{2481} \sqrt{12405}}\right)^{2}} = 1$$
- reduced to canonical form
$$8 x^{2} + 20 y^{2} + y - 31 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 20$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = -31$$
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
$$I_{1} = 28$$
|8 0 |
I2 = | |
|0 20|$$I_{3} = \left|\begin{matrix}8 & 0 & 0\\0 & 20 & \frac{1}{2}\\0 & \frac{1}{2} & -31\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}8 - \lambda & 0\\0 & 20 - \lambda\end{matrix}\right|$$
|8 0 | |20 1/2|
K2 = | | + | |
|0 -31| |1/2 -31|$$I_{1} = 28$$
$$I_{2} = 160$$
$$I_{3} = -4962$$
$$I{\left(\lambda \right)} = \lambda^{2} - 28 \lambda + 160$$
$$K_{2} = - \frac{3473}{4}$$
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
$$\lambda^{2} - 28 \lambda + 160 = 0$$
$$\lambda_{1} = 20$$
$$\lambda_{2} = 8$$
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
$$20 \tilde x^{2} + 8 \tilde y^{2} - \frac{2481}{80} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{10} \sqrt{5}}{\frac{4}{2481} \sqrt{12405}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{4} \sqrt{2}}{\frac{4}{2481} \sqrt{12405}}\right)^{2}} = 1$$
- reduced to canonical form