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How to use it?
- Canonical form:
- x^2+2y^2+4x-4y=0
- 5y^2+2x-20y+20=0
- x^2+4*y^2+2*x-32*y+64=0
- -3*x^2+(-3)*y^2+4*x*y+5*x+5*y+3=0
- Plot:
- (x^(2)+y^(2)-1)^(3)-x^(2)*y^(3)=0
- y^2-(8/5)*x+x^2=0
-
y^2=x*y-2
- y=x3+4*x^2+4*x
- Sum of series:
-
1089
-
Identical expressions
- 9x^ two +1 two 1y^2= one thousand and eighty-nine
- 9x squared plus 121y squared equally 1089
- 9x to the power of two plus 1 two 1y squared equally one thousand and eighty minus nine
- 9x2+121y2=1089
- 9x²+121y²=1089
- 9x to the power of 2+121y to the power of 2=1089
-
Similar expressions
- 9x^2-121y^2=1089
9x^2+121y^2=1089 canonical form
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The solution
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2 2 -1089 + 9*x + 121*y = 0
$$9 x^{2} + 121 y^{2} - 1089 = 0$$
9*x^2 + 121*y^2 - 1089 = 0
Detail solution
Given line equation of 2-order:
$$9 x^{2} + 121 y^{2} - 1089 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 121$$
$$a_{23} = 0$$
$$a_{33} = -1089$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 0\\0 & 121\end{matrix}\right|$$
$$\Delta = 1089$$
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
$$9 x_{0} = 0$$
$$121 y_{0} = 0$$
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
$$a'_{33} = -1089$$
$$a'_{33} = -1089$$
then equation turns into
$$9 x'^{2} + 121 y'^{2} - 1089 = 0$$
Given equation is ellipse
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$9 x^{2} + 121 y^{2} - 1089 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 121$$
$$a_{23} = 0$$
$$a_{33} = -1089$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 0\\0 & 121\end{matrix}\right|$$
$$\Delta = 1089$$
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
$$9 x_{0} = 0$$
$$121 y_{0} = 0$$
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
$$a'_{33} = -1089$$
$$a'_{33} = -1089$$
then equation turns into
$$9 x'^{2} + 121 y'^{2} - 1089 = 0$$
Given equation is ellipse
2 2
\tilde x \tilde y
--------- + ---------- = 1
2 2
/ 1 \ / 1 \
|------| |-------|
\3*1/33/ \11*1/33/ - 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:
$$9 x^{2} + 121 y^{2} - 1089 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 121$$
$$a_{23} = 0$$
$$a_{33} = -1089$$
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
$$I_{1} = 130$$
$$I_{3} = \left|\begin{matrix}9 & 0 & 0\\0 & 121 & 0\\0 & 0 & -1089\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 0\\0 & 121 - \lambda\end{matrix}\right|$$
$$I_{1} = 130$$
$$I_{2} = 1089$$
$$I_{3} = -1185921$$
$$I{\left(\lambda \right)} = \lambda^{2} - 130 \lambda + 1089$$
$$K_{2} = -141570$$
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
$$\lambda^{2} - 130 \lambda + 1089 = 0$$
$$\lambda_{1} = 121$$
$$\lambda_{2} = 9$$
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
$$121 \tilde x^{2} + 9 \tilde y^{2} - 1089 = 0$$
- reduced to canonical form
$$9 x^{2} + 121 y^{2} - 1089 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 121$$
$$a_{23} = 0$$
$$a_{33} = -1089$$
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
$$I_{1} = 130$$
|9 0 |
I2 = | |
|0 121|$$I_{3} = \left|\begin{matrix}9 & 0 & 0\\0 & 121 & 0\\0 & 0 & -1089\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 0\\0 & 121 - \lambda\end{matrix}\right|$$
|9 0 | |121 0 |
K2 = | | + | |
|0 -1089| | 0 -1089|$$I_{1} = 130$$
$$I_{2} = 1089$$
$$I_{3} = -1185921$$
$$I{\left(\lambda \right)} = \lambda^{2} - 130 \lambda + 1089$$
$$K_{2} = -141570$$
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
$$\lambda^{2} - 130 \lambda + 1089 = 0$$
$$\lambda_{1} = 121$$
$$\lambda_{2} = 9$$
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
$$121 \tilde x^{2} + 9 \tilde y^{2} - 1089 = 0$$
2 2
\tilde x \tilde y
---------- + --------- = 1
2 2
/ 1 \ / 1 \
|-------| |------|
\11*1/33/ \3*1/33/ - reduced to canonical form