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:
- =4x^21+(−4b+8)x^22+(−4b−3)x^23+2x1x2−2x1x3+2(4b−1)x2x3
- (x-4)*(x+2)-(x-5)*(x-6)
- -3x1^2-3x2^2-2x1x2
- x^2+y^2-2*(1/4)*xy=0
- Plot:
- x^2+(y-((x^2)^(1/(3))))^2=1
- (x^2+y^2+2*z*x)^2-4*z^2*(x^2+y^2)=0
-
(x+1)^2+(y+1)^2=4
- z=x^2+y^2
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Identical expressions
- -3x1^ two -3x two ^2-2x1x2
- minus 3x1 squared minus 3x2 squared minus 2x1x2
- minus 3x1 to the power of two minus 3x two squared minus 2x1x2
- -3x12-3x22-2x1x2
- -3x1²-3x2²-2x1x2
- -3x1 to the power of 2-3x2 to the power of 2-2x1x2
-
Similar expressions
- -3x1^2-3x2^2+2x1x2
- -3x1^2+3x2^2-2x1x2
- 3x1^2-3x2^2-2x1x2
-3x1^2-3x2^2-2x1x2 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
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2 2 - 3*x1 - 3*x2 - 2*x1*x2 = 0
$$- 3 x_{1}^{2} - 2 x_{1} x_{2} - 3 x_{2}^{2} = 0$$
-3*x1^2 - 2*x1*x2 - 3*x2^2 = 0
Detail solution
Given line equation of 2-order:
$$- 3 x_{1}^{2} - 2 x_{1} x_{2} - 3 x_{2}^{2} = 0$$
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
$$a_{11} = -3$$
$$a_{12} = -1$$
$$a_{13} = 0$$
$$a_{22} = -3$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-3 & -1\\-1 & -3\end{matrix}\right|$$
$$\Delta = 8$$
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
$$- x_{10} - 3 x_{20} = 0$$
$$- 3 x_{10} - x_{20} = 0$$
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
$$- 3 x1'^{2} - 2 x1' x2' - 3 x2'^{2} = 0$$
Rotate the resulting coordinate system by an angle φ
$$x2' = - \tilde x1 \sin{\left(\phi \right)} + \tilde x2 \cos{\left(\phi \right)}$$
$$x1' = \tilde x1 \cos{\left(\phi \right)} + \tilde x2 \sin{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = 0$$
then
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
substitute coefficients
$$x2' = - \frac{\sqrt{2} \tilde x1}{2} + \frac{\sqrt{2} \tilde x2}{2}$$
$$x1' = \frac{\sqrt{2} \tilde x1}{2} + \frac{\sqrt{2} \tilde x2}{2}$$
then the equation turns from
$$- 3 x1'^{2} - 2 x1' x2' - 3 x2'^{2} = 0$$
to
$$- 3 \left(- \frac{\sqrt{2} \tilde x1}{2} + \frac{\sqrt{2} \tilde x2}{2}\right)^{2} - 2 \left(- \frac{\sqrt{2} \tilde x1}{2} + \frac{\sqrt{2} \tilde x2}{2}\right) \left(\frac{\sqrt{2} \tilde x1}{2} + \frac{\sqrt{2} \tilde x2}{2}\right) - 3 \left(\frac{\sqrt{2} \tilde x1}{2} + \frac{\sqrt{2} \tilde x2}{2}\right)^{2} = 0$$
simplify
$$- 2 \tilde x1^{2} - 4 \tilde x2^{2} = 0$$
$$2 \tilde x1^{2} + 4 \tilde x2^{2} = 0$$
Given equation is degenerate ellipse
$$\frac{\tilde x1^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}} + \frac{\tilde x2^{2}}{\left(\frac{1}{2}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$- 3 x_{1}^{2} - 2 x_{1} x_{2} - 3 x_{2}^{2} = 0$$
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
$$a_{11} = -3$$
$$a_{12} = -1$$
$$a_{13} = 0$$
$$a_{22} = -3$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-3 & -1\\-1 & -3\end{matrix}\right|$$
$$\Delta = 8$$
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
$$- x_{10} - 3 x_{20} = 0$$
$$- 3 x_{10} - x_{20} = 0$$
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
$$- 3 x1'^{2} - 2 x1' x2' - 3 x2'^{2} = 0$$
Rotate the resulting coordinate system by an angle φ
$$x2' = - \tilde x1 \sin{\left(\phi \right)} + \tilde x2 \cos{\left(\phi \right)}$$
$$x1' = \tilde x1 \cos{\left(\phi \right)} + \tilde x2 \sin{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = 0$$
then
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
substitute coefficients
$$x2' = - \frac{\sqrt{2} \tilde x1}{2} + \frac{\sqrt{2} \tilde x2}{2}$$
$$x1' = \frac{\sqrt{2} \tilde x1}{2} + \frac{\sqrt{2} \tilde x2}{2}$$
then the equation turns from
$$- 3 x1'^{2} - 2 x1' x2' - 3 x2'^{2} = 0$$
to
$$- 3 \left(- \frac{\sqrt{2} \tilde x1}{2} + \frac{\sqrt{2} \tilde x2}{2}\right)^{2} - 2 \left(- \frac{\sqrt{2} \tilde x1}{2} + \frac{\sqrt{2} \tilde x2}{2}\right) \left(\frac{\sqrt{2} \tilde x1}{2} + \frac{\sqrt{2} \tilde x2}{2}\right) - 3 \left(\frac{\sqrt{2} \tilde x1}{2} + \frac{\sqrt{2} \tilde x2}{2}\right)^{2} = 0$$
simplify
$$- 2 \tilde x1^{2} - 4 \tilde x2^{2} = 0$$
$$2 \tilde x1^{2} + 4 \tilde x2^{2} = 0$$
Given equation is degenerate ellipse
$$\frac{\tilde x1^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}} + \frac{\tilde x2^{2}}{\left(\frac{1}{2}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
Invariants method
Given line equation of 2-order:
$$- 3 x_{1}^{2} - 2 x_{1} x_{2} - 3 x_{2}^{2} = 0$$
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
$$a_{11} = -3$$
$$a_{12} = -1$$
$$a_{13} = 0$$
$$a_{22} = -3$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = -6$$
$$I_{3} = \left|\begin{matrix}-3 & -1 & 0\\-1 & -3 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 3 & -1\\-1 & - \lambda - 3\end{matrix}\right|$$
$$I_{1} = -6$$
$$I_{2} = 8$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + 6 \lambda + 8$$
$$K_{2} = 0$$
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
$$\lambda^{2} + 6 \lambda + 8 = 0$$
$$\lambda_{1} = -2$$
$$\lambda_{2} = -4$$
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
$$- 4 \tilde x1^{2} - 2 \tilde x2^{2} = 0$$
$$\frac{\tilde x1^{2}}{\left(\frac{1}{2}\right)^{2}} + \frac{\tilde x2^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}} = 0$$
- reduced to canonical form
$$- 3 x_{1}^{2} - 2 x_{1} x_{2} - 3 x_{2}^{2} = 0$$
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
$$a_{11} = -3$$
$$a_{12} = -1$$
$$a_{13} = 0$$
$$a_{22} = -3$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = -6$$
|-3 -1|
I2 = | |
|-1 -3|$$I_{3} = \left|\begin{matrix}-3 & -1 & 0\\-1 & -3 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 3 & -1\\-1 & - \lambda - 3\end{matrix}\right|$$
|-3 0| |-3 0|
K2 = | | + | |
|0 0| |0 0|$$I_{1} = -6$$
$$I_{2} = 8$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + 6 \lambda + 8$$
$$K_{2} = 0$$
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
$$\lambda^{2} + 6 \lambda + 8 = 0$$
$$\lambda_{1} = -2$$
$$\lambda_{2} = -4$$
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
$$- 4 \tilde x1^{2} - 2 \tilde x2^{2} = 0$$
$$\frac{\tilde x1^{2}}{\left(\frac{1}{2}\right)^{2}} + \frac{\tilde x2^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}} = 0$$
- reduced to canonical form