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:
- -x^2+2y^2-2xy-2x-2y+2=0
- x^2+4*y^2+8*z^2-16
- 3x^2+6x-15y+15=0
- -17x^2+18xy+7y^2-38x-34y+11=0
- Plot:
-
x^2-4xy+3y^2=6
- x^2+y^2=26*y
-
x^2*y^2+y=1
- x=y^2+z^2
- Factor polynomial:
- x1^2+x2^2
-
Identical expressions
- x1^ two +x two ^2
- x1 squared plus x2 squared
- x1 to the power of two plus x two squared
- x12+x22
- x1²+x2²
- x1 to the power of 2+x2 to the power of 2
-
Similar expressions
- x1^2-x2^2
x1^2+x2^2 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$x_{1}^{2} + x_{2}^{2} = 0$$
This equation looks like:
$$a_{11} x_{2}^{2} + 2 a_{12} x_{1} x_{2} + 2 a_{13} x_{2} + a_{22} x_{1}^{2} + 2 a_{23} x_{1} + a_{33} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & 1\end{matrix}\right|$$
$$\Delta = 1$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{20} + a_{12} x_{10} + a_{13} = 0$$
$$a_{12} x_{20} + a_{22} x_{10} + a_{23} = 0$$
substitute coefficients
$$x_{20} = 0$$
$$x_{10} = 0$$
then
$$x_{20} = 0$$
$$x_{10} = 0$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x2'^{2} + 2 a_{12} x1' x2' + a_{22} x1'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{20} + a_{23} x_{10} + a_{33}$$
or
$$a'_{33} = 0$$
$$a'_{33} = 0$$
then equation turns into
$$x1'^{2} + x2'^{2} = 0$$
Given equation is degenerate ellipse
$$\frac{\tilde x1^{2}}{1^{2}} + \frac{\tilde x2^{2}}{1^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$x_{1}^{2} + x_{2}^{2} = 0$$
This equation looks like:
$$a_{11} x_{2}^{2} + 2 a_{12} x_{1} x_{2} + 2 a_{13} x_{2} + a_{22} x_{1}^{2} + 2 a_{23} x_{1} + a_{33} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & 1\end{matrix}\right|$$
$$\Delta = 1$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{20} + a_{12} x_{10} + a_{13} = 0$$
$$a_{12} x_{20} + a_{22} x_{10} + a_{23} = 0$$
substitute coefficients
$$x_{20} = 0$$
$$x_{10} = 0$$
then
$$x_{20} = 0$$
$$x_{10} = 0$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x2'^{2} + 2 a_{12} x1' x2' + a_{22} x1'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{20} + a_{23} x_{10} + a_{33}$$
or
$$a'_{33} = 0$$
$$a'_{33} = 0$$
then equation turns into
$$x1'^{2} + x2'^{2} = 0$$
Given equation is degenerate ellipse
$$\frac{\tilde x1^{2}}{1^{2}} + \frac{\tilde x2^{2}}{1^{2}} = 0$$
- 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:
$$x_{1}^{2} + x_{2}^{2} = 0$$
This equation looks like:
$$a_{11} x_{2}^{2} + 2 a_{12} x_{1} x_{2} + 2 a_{13} x_{2} + a_{22} x_{1}^{2} + 2 a_{23} x_{1} + a_{33} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 2$$
$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 2$$
$$I_{2} = 1$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 2 \lambda + 1$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} > 0$$
then by line type:
this equation is of type : degenerate ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 2 \lambda + 1 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = 1$$
then the canonical form of the equation will be
$$\tilde x1^{2} \lambda_{2} + \tilde x2^{2} \lambda_{1} + \frac{I_{3}}{I_{2}} = 0$$
or
$$\tilde x1^{2} + \tilde x2^{2} = 0$$
$$\frac{\tilde x1^{2}}{1^{2}} + \frac{\tilde x2^{2}}{1^{2}} = 0$$
- reduced to canonical form
$$x_{1}^{2} + x_{2}^{2} = 0$$
This equation looks like:
$$a_{11} x_{2}^{2} + 2 a_{12} x_{1} x_{2} + 2 a_{13} x_{2} + a_{22} x_{1}^{2} + 2 a_{23} x_{1} + a_{33} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 2$$
|1 0|
I2 = | |
|0 1|$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|$$
|1 0| |1 0|
K2 = | | + | |
|0 0| |0 0|$$I_{1} = 2$$
$$I_{2} = 1$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 2 \lambda + 1$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} > 0$$
then by line type:
this equation is of type : degenerate ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 2 \lambda + 1 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = 1$$
then the canonical form of the equation will be
$$\tilde x1^{2} \lambda_{2} + \tilde x2^{2} \lambda_{1} + \frac{I_{3}}{I_{2}} = 0$$
or
$$\tilde x1^{2} + \tilde x2^{2} = 0$$
$$\frac{\tilde x1^{2}}{1^{2}} + \frac{\tilde x2^{2}}{1^{2}} = 0$$
- reduced to canonical form