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- Canonical form of a degenerate ellipse
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How to use it?
- Canonical form:
-
25x^2+50x+36y^2-216y-551=0
-
25x^2-16y^2+150x+128y-431=0
- x^2-4xy+4y^2-20x+10y-50=0
- x^2=2y^2+3z^2
- Plot:
-
4x^2+4y^2=25
-
xy=4
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(x^2+y^2-1)^3-(x^2)*y^3=0
- xyz
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Identical expressions
- two 5x^ two +5 zero x+36y^2-216y- five hundred and fifty-one =0
- 25x squared plus 50x plus 36y squared minus 216y minus 551 equally 0
- two 5x to the power of two plus 5 zero x plus 36y squared minus 216y minus five hundred and fifty minus one equally 0
- 25x2+50x+36y2-216y-551=0
- 25x²+50x+36y²-216y-551=0
- 25x to the power of 2+50x+36y to the power of 2-216y-551=0
- 25x^2+50x+36y^2-216y-551=O
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Similar expressions
- 25x^2+50x-36y^2-216y-551=0
- 25x^2+50x+36y^2-216y+551=0
- 25x^2-50x+36y^2-216y-551=0
- 25x^2+50x+36y^2+216y-551=0
25x^2+50x+36y^2-216y-551=0 canonical form
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The solution
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2 2 -551 - 216*y + 25*x + 36*y + 50*x = 0
$$25 x^{2} + 50 x + 36 y^{2} - 216 y - 551 = 0$$
25*x^2 + 50*x + 36*y^2 - 216*y - 551 = 0
Detail solution
Given line equation of 2-order:
$$25 x^{2} + 50 x + 36 y^{2} - 216 y - 551 = 0$$
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
$$a_{11} = 25$$
$$a_{12} = 0$$
$$a_{13} = 25$$
$$a_{22} = 36$$
$$a_{23} = -108$$
$$a_{33} = -551$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}25 & 0\\0 & 36\end{matrix}\right|$$
$$\Delta = 900$$
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
$$25 x_{0} + 25 = 0$$
$$36 y_{0} - 108 = 0$$
then
$$x_{0} = -1$$
$$y_{0} = 3$$
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
$$a'_{33} = 25 x_{0} - 108 y_{0} - 551$$
$$a'_{33} = -900$$
then equation turns into
$$25 x'^{2} + 36 y'^{2} - 900 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{6^{2}} + \frac{\tilde y^{2}}{5^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$25 x^{2} + 50 x + 36 y^{2} - 216 y - 551 = 0$$
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
$$a_{11} = 25$$
$$a_{12} = 0$$
$$a_{13} = 25$$
$$a_{22} = 36$$
$$a_{23} = -108$$
$$a_{33} = -551$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}25 & 0\\0 & 36\end{matrix}\right|$$
$$\Delta = 900$$
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
$$25 x_{0} + 25 = 0$$
$$36 y_{0} - 108 = 0$$
then
$$x_{0} = -1$$
$$y_{0} = 3$$
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
$$a'_{33} = 25 x_{0} - 108 y_{0} - 551$$
$$a'_{33} = -900$$
then equation turns into
$$25 x'^{2} + 36 y'^{2} - 900 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{6^{2}} + \frac{\tilde y^{2}}{5^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(-1, 3)
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:
$$25 x^{2} + 50 x + 36 y^{2} - 216 y - 551 = 0$$
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
$$a_{11} = 25$$
$$a_{12} = 0$$
$$a_{13} = 25$$
$$a_{22} = 36$$
$$a_{23} = -108$$
$$a_{33} = -551$$
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
$$I_{1} = 61$$
$$I_{3} = \left|\begin{matrix}25 & 0 & 25\\0 & 36 & -108\\25 & -108 & -551\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}25 - \lambda & 0\\0 & 36 - \lambda\end{matrix}\right|$$
$$I_{1} = 61$$
$$I_{2} = 900$$
$$I_{3} = -810000$$
$$I{\left(\lambda \right)} = \lambda^{2} - 61 \lambda + 900$$
$$K_{2} = -45900$$
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
$$\lambda^{2} - 61 \lambda + 900 = 0$$
Solve this equation
$$\lambda_{1} = 36$$
$$\lambda_{2} = 25$$
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
$$36 \tilde x^{2} + 25 \tilde y^{2} - 900 = 0$$
$$\frac{\tilde x^{2}}{5^{2}} + \frac{\tilde y^{2}}{6^{2}} = 1$$
- reduced to canonical form
$$25 x^{2} + 50 x + 36 y^{2} - 216 y - 551 = 0$$
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
$$a_{11} = 25$$
$$a_{12} = 0$$
$$a_{13} = 25$$
$$a_{22} = 36$$
$$a_{23} = -108$$
$$a_{33} = -551$$
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
$$I_{1} = 61$$
|25 0 |
I2 = | |
|0 36|$$I_{3} = \left|\begin{matrix}25 & 0 & 25\\0 & 36 & -108\\25 & -108 & -551\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}25 - \lambda & 0\\0 & 36 - \lambda\end{matrix}\right|$$
|25 25 | | 36 -108|
K2 = | | + | |
|25 -551| |-108 -551|$$I_{1} = 61$$
$$I_{2} = 900$$
$$I_{3} = -810000$$
$$I{\left(\lambda \right)} = \lambda^{2} - 61 \lambda + 900$$
$$K_{2} = -45900$$
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
$$\lambda^{2} - 61 \lambda + 900 = 0$$
Solve this equation
$$\lambda_{1} = 36$$
$$\lambda_{2} = 25$$
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
$$36 \tilde x^{2} + 25 \tilde y^{2} - 900 = 0$$
$$\frac{\tilde x^{2}}{5^{2}} + \frac{\tilde y^{2}}{6^{2}} = 1$$
- reduced to canonical form
The graph