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-
How to use it?
- Canonical form:
- x^2-4x+y+1=0
- x^2+6xy+y^2=0
- 4x^2+36y^2=144
- 4x^2-16x-y=0
- Plot:
- (x^(2)+y^(2)-1)^(3)-x^(2)*y^(3)=0
- 3*x*y^2+9*x*y+y^2+6*x+3*y=2020
- 230+14050∙e^(-51∙t)+e^(-25∙t)∙0.042∙cos0.568∙t+1636∙sin0.568∙t
- (|y|)=(|2*x-5|)
- Derivative of:
- 144
- Sum of series:
-
144
-
Identical expressions
- 4x^ two +36y^ two = one hundred and forty-four
- 4x squared plus 36y squared equally 144
- 4x to the power of two plus 36y to the power of two equally one hundred and forty minus four
- 4x2+36y2=144
- 4x²+36y²=144
- 4x to the power of 2+36y to the power of 2=144
-
Similar expressions
- 4x^2-36y^2=144
4x^2+36y^2=144 canonical form
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The solution
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2 2 -144 + 4*x + 36*y = 0
$$4 x^{2} + 36 y^{2} - 144 = 0$$
4*x^2 + 36*y^2 - 144 = 0
Detail solution
Given line equation of 2-order:
$$4 x^{2} + 36 y^{2} - 144 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 36$$
$$a_{23} = 0$$
$$a_{33} = -144$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & 36\end{matrix}\right|$$
$$\Delta = 144$$
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
$$4 x_{0} = 0$$
$$36 y_{0} = 0$$
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
$$a'_{33} = -144$$
$$a'_{33} = -144$$
then equation turns into
$$4 x'^{2} + 36 y'^{2} - 144 = 0$$
Given equation is ellipse
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$4 x^{2} + 36 y^{2} - 144 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 36$$
$$a_{23} = 0$$
$$a_{33} = -144$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & 36\end{matrix}\right|$$
$$\Delta = 144$$
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
$$4 x_{0} = 0$$
$$36 y_{0} = 0$$
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
$$a'_{33} = -144$$
$$a'_{33} = -144$$
then equation turns into
$$4 x'^{2} + 36 y'^{2} - 144 = 0$$
Given equation is ellipse
2 2
\tilde x \tilde y
--------- + --------- = 1
2 2
/ 1 \ / 1 \
|------| |------|
\2*1/12/ \6*1/12/ - 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:
$$4 x^{2} + 36 y^{2} - 144 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 36$$
$$a_{23} = 0$$
$$a_{33} = -144$$
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
$$I_{1} = 40$$
$$I_{3} = \left|\begin{matrix}4 & 0 & 0\\0 & 36 & 0\\0 & 0 & -144\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0\\0 & 36 - \lambda\end{matrix}\right|$$
$$I_{1} = 40$$
$$I_{2} = 144$$
$$I_{3} = -20736$$
$$I{\left(\lambda \right)} = \lambda^{2} - 40 \lambda + 144$$
$$K_{2} = -5760$$
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
$$\lambda^{2} - 40 \lambda + 144 = 0$$
$$\lambda_{1} = 36$$
$$\lambda_{2} = 4$$
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
$$36 \tilde x^{2} + 4 \tilde y^{2} - 144 = 0$$
- reduced to canonical form
$$4 x^{2} + 36 y^{2} - 144 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 36$$
$$a_{23} = 0$$
$$a_{33} = -144$$
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
$$I_{1} = 40$$
|4 0 |
I2 = | |
|0 36|$$I_{3} = \left|\begin{matrix}4 & 0 & 0\\0 & 36 & 0\\0 & 0 & -144\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0\\0 & 36 - \lambda\end{matrix}\right|$$
|4 0 | |36 0 |
K2 = | | + | |
|0 -144| |0 -144|$$I_{1} = 40$$
$$I_{2} = 144$$
$$I_{3} = -20736$$
$$I{\left(\lambda \right)} = \lambda^{2} - 40 \lambda + 144$$
$$K_{2} = -5760$$
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
$$\lambda^{2} - 40 \lambda + 144 = 0$$
$$\lambda_{1} = 36$$
$$\lambda_{2} = 4$$
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
$$36 \tilde x^{2} + 4 \tilde y^{2} - 144 = 0$$
2 2
\tilde x \tilde y
--------- + --------- = 1
2 2
/ 1 \ / 1 \
|------| |------|
\6*1/12/ \2*1/12/ - reduced to canonical form