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:
- 0.20x1+0.50x2=0.40*(x1+x2)
- y^2+z=0
- x^2+y^2+2z^2+6xy-4xz-12yz
- x^2+2y^2-4x+2y=20
- Plot:
- x^2+(y-((x^2)^(1/(3))))^2=1
-
x^2+(y+(x)^(2/3))^2=1
-
tg(x)^arctg(x)=e^((y(x))/x)
- x^2+(y-x^2/3)^2=1
- Sum of series:
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20
- Integral of d{x}:
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20
- Limit of the function:
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20
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Identical expressions
- x^ two + two y^2-4x+2y= twenty
- x squared plus 2y squared minus 4x plus 2y equally 20
- x to the power of two plus two y squared minus 4x plus 2y equally twenty
- x2+2y2-4x+2y=20
- x²+2y²-4x+2y=20
- x to the power of 2+2y to the power of 2-4x+2y=20
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Similar expressions
- x^2+2y^2+4x+2y=20
- x^2-2y^2-4x+2y=20
- x^2+2y^2-4x-2y=20
x^2+2y^2-4x+2y=20 canonical form
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The solution
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2 2 -20 + x - 4*x + 2*y + 2*y = 0
$$x^{2} - 4 x + 2 y^{2} + 2 y - 20 = 0$$
x^2 - 4*x + 2*y^2 + 2*y - 20 = 0
Detail solution
Given line equation of 2-order:
$$x^{2} - 4 x + 2 y^{2} + 2 y - 20 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = -2$$
$$a_{22} = 2$$
$$a_{23} = 1$$
$$a_{33} = -20$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & 2\end{matrix}\right|$$
$$\Delta = 2$$
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
$$x_{0} - 2 = 0$$
$$2 y_{0} + 1 = 0$$
then
$$x_{0} = 2$$
$$y_{0} = - \frac{1}{2}$$
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
$$a'_{33} = - 2 x_{0} + y_{0} - 20$$
$$a'_{33} = - \frac{49}{2}$$
then equation turns into
$$x'^{2} + 2 y'^{2} - \frac{49}{2} = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{1}{7} \sqrt{2}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{7} \sqrt{2}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$x^{2} - 4 x + 2 y^{2} + 2 y - 20 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = -2$$
$$a_{22} = 2$$
$$a_{23} = 1$$
$$a_{33} = -20$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & 2\end{matrix}\right|$$
$$\Delta = 2$$
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
$$x_{0} - 2 = 0$$
$$2 y_{0} + 1 = 0$$
then
$$x_{0} = 2$$
$$y_{0} = - \frac{1}{2}$$
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
$$a'_{33} = - 2 x_{0} + y_{0} - 20$$
$$a'_{33} = - \frac{49}{2}$$
then equation turns into
$$x'^{2} + 2 y'^{2} - \frac{49}{2} = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{1}{7} \sqrt{2}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{7} \sqrt{2}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(2, -1/2)
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:
$$x^{2} - 4 x + 2 y^{2} + 2 y - 20 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = -2$$
$$a_{22} = 2$$
$$a_{23} = 1$$
$$a_{33} = -20$$
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
$$I_{1} = 3$$
$$I_{3} = \left|\begin{matrix}1 & 0 & -2\\0 & 2 & 1\\-2 & 1 & -20\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & 2 - \lambda\end{matrix}\right|$$
$$I_{1} = 3$$
$$I_{2} = 2$$
$$I_{3} = -49$$
$$I{\left(\lambda \right)} = \lambda^{2} - 3 \lambda + 2$$
$$K_{2} = -65$$
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
$$\lambda^{2} - 3 \lambda + 2 = 0$$
$$\lambda_{1} = 2$$
$$\lambda_{2} = 1$$
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
$$2 \tilde x^{2} + \tilde y^{2} - \frac{49}{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{7} \sqrt{2}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{1}{7} \sqrt{2}}\right)^{2}} = 1$$
- reduced to canonical form
$$x^{2} - 4 x + 2 y^{2} + 2 y - 20 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = -2$$
$$a_{22} = 2$$
$$a_{23} = 1$$
$$a_{33} = -20$$
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
$$I_{1} = 3$$
|1 0|
I2 = | |
|0 2|$$I_{3} = \left|\begin{matrix}1 & 0 & -2\\0 & 2 & 1\\-2 & 1 & -20\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & 2 - \lambda\end{matrix}\right|$$
|1 -2 | |2 1 |
K2 = | | + | |
|-2 -20| |1 -20|$$I_{1} = 3$$
$$I_{2} = 2$$
$$I_{3} = -49$$
$$I{\left(\lambda \right)} = \lambda^{2} - 3 \lambda + 2$$
$$K_{2} = -65$$
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
$$\lambda^{2} - 3 \lambda + 2 = 0$$
$$\lambda_{1} = 2$$
$$\lambda_{2} = 1$$
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
$$2 \tilde x^{2} + \tilde y^{2} - \frac{49}{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{7} \sqrt{2}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{1}{7} \sqrt{2}}\right)^{2}} = 1$$
- reduced to canonical form