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
-
How to use it?
- Canonical form:
- x^2+4*y^2+2*x-32*y+64=0
- y^2-x-4y+5=0
- (y^2)+2y+x=0
- x^2+2xy+6xz+5y^2+2yz+a*z^2
- Plot:
-
y^2=(x^4+83*x^2-408+60*x+10*x^3)/(21+2*x^2+10*x-y^2)
- y=-0,8x;y-2,5x;y=0,4x
- y=-2*sqrt(|x|)+4
-
y=(x^2-1)^1/2
-
Identical expressions
- 4x^ two + two y^2-8x-4y+ ten = zero
- 4x squared plus 2y squared minus 8x minus 4y plus 10 equally 0
- 4x to the power of two plus two y squared minus 8x minus 4y plus ten equally zero
- 4x2+2y2-8x-4y+10=0
- 4x²+2y²-8x-4y+10=0
- 4x to the power of 2+2y to the power of 2-8x-4y+10=0
- 4x^2+2y^2-8x-4y+10=O
-
Similar expressions
- 4x^2+2y^2-8x+4y+10=0
- 4x^2+2y^2-8x-4y-10=0
- 4x^2-2y^2-8x-4y+10=0
- 4x^2+2y^2+8x-4y+10=0
4x^2+2y^2-8x-4y+10=0 canonical form
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The solution
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2 2 10 - 8*x - 4*y + 2*y + 4*x = 0
$$4 x^{2} - 8 x + 2 y^{2} - 4 y + 10 = 0$$
4*x^2 - 8*x + 2*y^2 - 4*y + 10 = 0
Detail solution
Given line equation of 2-order:
$$4 x^{2} - 8 x + 2 y^{2} - 4 y + 10 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = 2$$
$$a_{23} = -2$$
$$a_{33} = 10$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & 2\end{matrix}\right|$$
$$\Delta = 8$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
substitute coefficients
$$4 x_{0} - 4 = 0$$
$$2 y_{0} - 2 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = 1$$
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
$$a'_{33} = - 4 x_{0} - 2 y_{0} + 10$$
$$a'_{33} = 4$$
then equation turns into
$$4 x'^{2} + 2 y'^{2} + 4 = 0$$
Given equation is imaginary ellipse
$$\frac{\tilde x^{2}}{1^{2}} + \frac{\tilde y^{2}}{\left(\sqrt{2}\right)^{2}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$4 x^{2} - 8 x + 2 y^{2} - 4 y + 10 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = 2$$
$$a_{23} = -2$$
$$a_{33} = 10$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & 2\end{matrix}\right|$$
$$\Delta = 8$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
substitute coefficients
$$4 x_{0} - 4 = 0$$
$$2 y_{0} - 2 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = 1$$
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
$$a'_{33} = - 4 x_{0} - 2 y_{0} + 10$$
$$a'_{33} = 4$$
then equation turns into
$$4 x'^{2} + 2 y'^{2} + 4 = 0$$
Given equation is imaginary ellipse
$$\frac{\tilde x^{2}}{1^{2}} + \frac{\tilde y^{2}}{\left(\sqrt{2}\right)^{2}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(1, 1)
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:
$$4 x^{2} - 8 x + 2 y^{2} - 4 y + 10 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = 2$$
$$a_{23} = -2$$
$$a_{33} = 10$$
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
$$I_{1} = 6$$
$$I_{3} = \left|\begin{matrix}4 & 0 & -4\\0 & 2 & -2\\-4 & -2 & 10\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0\\0 & 2 - \lambda\end{matrix}\right|$$
$$I_{1} = 6$$
$$I_{2} = 8$$
$$I_{3} = 32$$
$$I{\left(\lambda \right)} = \lambda^{2} - 6 \lambda + 8$$
$$K_{2} = 40$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} > 0$$
then by line type:
this equation is of type : imaginary ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 6 \lambda + 8 = 0$$
Solve this equation
$$\lambda_{1} = 4$$
$$\lambda_{2} = 2$$
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
$$4 \tilde x^{2} + 2 \tilde y^{2} + 4 = 0$$
$$\frac{\tilde x^{2}}{1^{2}} + \frac{\tilde y^{2}}{\left(\sqrt{2}\right)^{2}} = -1$$
- reduced to canonical form
$$4 x^{2} - 8 x + 2 y^{2} - 4 y + 10 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = 2$$
$$a_{23} = -2$$
$$a_{33} = 10$$
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
$$I_{1} = 6$$
|4 0|
I2 = | |
|0 2|$$I_{3} = \left|\begin{matrix}4 & 0 & -4\\0 & 2 & -2\\-4 & -2 & 10\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0\\0 & 2 - \lambda\end{matrix}\right|$$
|4 -4| |2 -2|
K2 = | | + | |
|-4 10| |-2 10|$$I_{1} = 6$$
$$I_{2} = 8$$
$$I_{3} = 32$$
$$I{\left(\lambda \right)} = \lambda^{2} - 6 \lambda + 8$$
$$K_{2} = 40$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} > 0$$
then by line type:
this equation is of type : imaginary ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 6 \lambda + 8 = 0$$
Solve this equation
$$\lambda_{1} = 4$$
$$\lambda_{2} = 2$$
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
$$4 \tilde x^{2} + 2 \tilde y^{2} + 4 = 0$$
$$\frac{\tilde x^{2}}{1^{2}} + \frac{\tilde y^{2}}{\left(\sqrt{2}\right)^{2}} = -1$$
- reduced to canonical form
The graph