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How to use it?
- Canonical form:
- y^2-z^2+2*x+(z-y)^2=0
- -z=2x^2+18y^2
- 5x^2+10y^2-50=0
- 25y^2+42x+144y=0
- Plot:
- x^2+y^2=z^2
-
x^2+(y-|x|^(2/3))^2=1
- u=1,2v
-
2*x^3-3*x^2*y+4*y^3=1
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Identical expressions
- 5x^ two +1 zero y^ two - fifty =0
- 5x squared plus 10y squared minus 50 equally 0
- 5x to the power of two plus 1 zero y to the power of two minus fifty equally 0
- 5x2+10y2-50=0
- 5x²+10y²-50=0
- 5x to the power of 2+10y to the power of 2-50=0
- 5x^2+10y^2-50=O
-
Similar expressions
- 5x^2-10y^2-50=0
- 5x^2+10y^2+50=0
5x^2+10y^2-50=0 canonical form
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The solution
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2 2 -50 + 5*x + 10*y = 0
$$5 x^{2} + 10 y^{2} - 50 = 0$$
5*x^2 + 10*y^2 - 50 = 0
Detail solution
Given line equation of 2-order:
$$5 x^{2} + 10 y^{2} - 50 = 0$$
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
$$a_{11} = 5$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 10$$
$$a_{23} = 0$$
$$a_{33} = -50$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}5 & 0\\0 & 10\end{matrix}\right|$$
$$\Delta = 50$$
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
$$5 x_{0} = 0$$
$$10 y_{0} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = 0$$
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
$$a'_{33} = -50$$
$$a'_{33} = -50$$
then equation turns into
$$5 x'^{2} + 10 y'^{2} - 50 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{10} \sqrt{2}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{10} \sqrt{10}}{\frac{1}{10} \sqrt{2}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$5 x^{2} + 10 y^{2} - 50 = 0$$
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
$$a_{11} = 5$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 10$$
$$a_{23} = 0$$
$$a_{33} = -50$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}5 & 0\\0 & 10\end{matrix}\right|$$
$$\Delta = 50$$
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
$$5 x_{0} = 0$$
$$10 y_{0} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = 0$$
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
$$a'_{33} = -50$$
$$a'_{33} = -50$$
then equation turns into
$$5 x'^{2} + 10 y'^{2} - 50 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{10} \sqrt{2}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{10} \sqrt{10}}{\frac{1}{10} \sqrt{2}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)
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:
$$5 x^{2} + 10 y^{2} - 50 = 0$$
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
$$a_{11} = 5$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 10$$
$$a_{23} = 0$$
$$a_{33} = -50$$
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
$$I_{1} = 15$$
$$I_{3} = \left|\begin{matrix}5 & 0 & 0\\0 & 10 & 0\\0 & 0 & -50\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}5 - \lambda & 0\\0 & 10 - \lambda\end{matrix}\right|$$
$$I_{1} = 15$$
$$I_{2} = 50$$
$$I_{3} = -2500$$
$$I{\left(\lambda \right)} = \lambda^{2} - 15 \lambda + 50$$
$$K_{2} = -750$$
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
$$\lambda^{2} - 15 \lambda + 50 = 0$$
$$\lambda_{1} = 10$$
$$\lambda_{2} = 5$$
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
$$10 \tilde x^{2} + 5 \tilde y^{2} - 50 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{10} \sqrt{10}}{\frac{1}{10} \sqrt{2}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{10} \sqrt{2}}\right)^{2}} = 1$$
- reduced to canonical form
$$5 x^{2} + 10 y^{2} - 50 = 0$$
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
$$a_{11} = 5$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 10$$
$$a_{23} = 0$$
$$a_{33} = -50$$
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
$$I_{1} = 15$$
|5 0 |
I2 = | |
|0 10|$$I_{3} = \left|\begin{matrix}5 & 0 & 0\\0 & 10 & 0\\0 & 0 & -50\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}5 - \lambda & 0\\0 & 10 - \lambda\end{matrix}\right|$$
|5 0 | |10 0 |
K2 = | | + | |
|0 -50| |0 -50|$$I_{1} = 15$$
$$I_{2} = 50$$
$$I_{3} = -2500$$
$$I{\left(\lambda \right)} = \lambda^{2} - 15 \lambda + 50$$
$$K_{2} = -750$$
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
$$\lambda^{2} - 15 \lambda + 50 = 0$$
$$\lambda_{1} = 10$$
$$\lambda_{2} = 5$$
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
$$10 \tilde x^{2} + 5 \tilde y^{2} - 50 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{10} \sqrt{10}}{\frac{1}{10} \sqrt{2}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{10} \sqrt{2}}\right)^{2}} = 1$$
- reduced to canonical form