Mister Exam
Other calculators
- Canonical form of a elliptical paraboloid
- Canonical form of a double hyperboloid
- Canonical form of a imaginary ellipsoid
- Canonical form of a degenerate ellipse
- Canonical form of a parabola
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How to use it?
- Canonical form:
- 13x^2+10xy+13y^2-72=0
-
x^2-9y^2-4x+36y-41=0
- 9x2+121y2=1089
-
sqrt(6)*x=2*sqrt(6)+sqrt(35-y)
- Plot:
-
sqrt(x)+sqrt(y)=1
-
sin(x+y)=1.2*x
- x^2+(y-(x^2)^(1/3))^2=1
- y=4^x
- Sum of series:
-
1089
-
Identical expressions
- 9x2+121y2= one thousand and eighty-nine
- 9x2 plus 121y2 equally 1089
- 9x2 plus 121y2 equally one thousand and eighty minus nine
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Similar expressions
- 9x2-121y2=1089
- 9x^2+121y^2=1089
9x2+121y2=1089 canonical form
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The solution
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-1089 + 9*x2 + 121*y2 = 0
$$9 x_{2} + 121 y_{2} - 1089 = 0$$
9*x2 + 121*y2 - 1089 = 0
Detail solution
Given line equation of 2-order:
$$9 x_{2} + 121 y_{2} - 1089 = 0$$
This equation looks like:
$$a_{11} y_{2}^{2} + 2 a_{12} x_{2} y_{2} + 2 a_{13} y_{2} + a_{22} x_{2}^{2} + 2 a_{23} x_{2} + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{121}{2}$$
$$a_{22} = 0$$
$$a_{23} = \frac{9}{2}$$
$$a_{33} = -1089$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
Given equation is straight line
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$y_{20} = - \tilde x2 \sin{\left(\phi \right)} + \tilde y2 \cos{\left(\phi \right)}$$
$$x_{20} = \tilde x2 \cos{\left(\phi \right)} + \tilde y2 \sin{\left(\phi \right)}$$
$$y_{20} = 0 \cdot 0$$
$$x_{20} = 0 \cdot 0$$
$$y_{20} = 0$$
$$x_{20} = 0$$
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:
$$a_{11} y_{2}^{2} + 2 a_{12} x_{2} y_{2} + 2 a_{13} y_{2} + a_{22} x_{2}^{2} + 2 a_{23} x_{2} + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{121}{2}$$
$$a_{22} = 0$$
$$a_{23} = \frac{9}{2}$$
$$a_{33} = -1089$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
Given equation is straight line
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$y_{20} = - \tilde x2 \sin{\left(\phi \right)} + \tilde y2 \cos{\left(\phi \right)}$$
$$x_{20} = \tilde x2 \cos{\left(\phi \right)} + \tilde y2 \sin{\left(\phi \right)}$$
$$y_{20} = 0 \cdot 0$$
$$x_{20} = 0 \cdot 0$$
$$y_{20} = 0$$
$$x_{20} = 0$$
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:
$$a_{11} y_{2}^{2} + 2 a_{12} x_{2} y_{2} + 2 a_{13} y_{2} + a_{22} x_{2}^{2} + 2 a_{23} x_{2} + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{121}{2}$$
$$a_{22} = 0$$
$$a_{23} = \frac{9}{2}$$
$$a_{33} = -1089$$
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
$$I_{1} = 0$$
$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{121}{2}\\0 & 0 & \frac{9}{2}\\\frac{121}{2} & \frac{9}{2} & -1089\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 0$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2}$$
$$K_{2} = - \frac{7361}{2}$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge \left(I_{1} = 0 \vee K_{2} = 0\right)$$
then by line type:
this equation is of type : two coincident straight lines
$$I_{1} \tilde x2^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
- reduced to canonical form
$$9 x_{2} + 121 y_{2} - 1089 = 0$$
This equation looks like:
$$a_{11} y_{2}^{2} + 2 a_{12} x_{2} y_{2} + 2 a_{13} y_{2} + a_{22} x_{2}^{2} + 2 a_{23} x_{2} + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{121}{2}$$
$$a_{22} = 0$$
$$a_{23} = \frac{9}{2}$$
$$a_{33} = -1089$$
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
$$I_{1} = 0$$
|0 0|
I2 = | |
|0 0|$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{121}{2}\\0 & 0 & \frac{9}{2}\\\frac{121}{2} & \frac{9}{2} & -1089\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
| 0 121/2| | 0 9/2 |
K2 = | | + | |
|121/2 -1089| |9/2 -1089|$$I_{1} = 0$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2}$$
$$K_{2} = - \frac{7361}{2}$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge \left(I_{1} = 0 \vee K_{2} = 0\right)$$
then by line type:
this equation is of type : two coincident straight lines
$$I_{1} \tilde x2^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
False
None
- reduced to canonical form