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:
- 4*x^2+4*y^2-4x+8y+59=0
- 4x^2-12xy+13y^2-36x+62y+69=0
- 3(x+2)^2+7(y-1)=36
- 0x^2+5y^2-3z^2+14yz−2y+10z-3=0.
- Plot:
-
x^2+(y-|x|^(1/2))^2=1
- x^2/36+y^2/36=1
- x^4-y^4=x*y
-
((x+5/2)^2-(y-11^(1/2)/2)^2+2.5)*((x+5/2)^2-(y+11^(1/2)/2)^2+2.5)=444-36*x^2
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Identical expressions
- four *x^ two + four *y^ two -4x+8y+ fifty-nine = zero
- 4 multiply by x squared plus 4 multiply by y squared minus 4x plus 8y plus 59 equally 0
- four multiply by x to the power of two plus four multiply by y to the power of two minus 4x plus 8y plus fifty minus nine equally zero
- 4*x2+4*y2-4x+8y+59=0
- 4*x²+4*y²-4x+8y+59=0
- 4*x to the power of 2+4*y to the power of 2-4x+8y+59=0
- 4x^2+4y^2-4x+8y+59=0
- 4x2+4y2-4x+8y+59=0
- 4*x^2+4*y^2-4x+8y+59=O
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Similar expressions
- 4*x^2-4*y^2-4x+8y+59=0
- 4*x^2+4*y^2-4x-8y+59=0
- 4*x^2+4*y^2+4x+8y+59=0
- 4*x^2+4*y^2-4x+8y-59=0
4*x^2+4*y^2-4x+8y+59=0 canonical form
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The solution
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2 2 59 - 4*x + 4*x + 4*y + 8*y = 0
$$4 x^{2} - 4 x + 4 y^{2} + 8 y + 59 = 0$$
4*x^2 - 4*x + 4*y^2 + 8*y + 59 = 0
Detail solution
Given line equation of 2-order:
$$4 x^{2} - 4 x + 4 y^{2} + 8 y + 59 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -2$$
$$a_{22} = 4$$
$$a_{23} = 4$$
$$a_{33} = 59$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & 4\end{matrix}\right|$$
$$\Delta = 16$$
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
$$4 x_{0} - 2 = 0$$
$$4 y_{0} + 4 = 0$$
then
$$x_{0} = \frac{1}{2}$$
$$y_{0} = -1$$
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
$$a'_{33} = - 2 x_{0} + 4 y_{0} + 59$$
$$a'_{33} = 54$$
then equation turns into
$$4 x'^{2} + 4 y'^{2} + 54 = 0$$
Given equation is imaginary ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{2 \frac{\sqrt{6}}{18}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{2 \frac{\sqrt{6}}{18}}\right)^{2}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$4 x^{2} - 4 x + 4 y^{2} + 8 y + 59 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -2$$
$$a_{22} = 4$$
$$a_{23} = 4$$
$$a_{33} = 59$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & 4\end{matrix}\right|$$
$$\Delta = 16$$
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
$$4 x_{0} - 2 = 0$$
$$4 y_{0} + 4 = 0$$
then
$$x_{0} = \frac{1}{2}$$
$$y_{0} = -1$$
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
$$a'_{33} = - 2 x_{0} + 4 y_{0} + 59$$
$$a'_{33} = 54$$
then equation turns into
$$4 x'^{2} + 4 y'^{2} + 54 = 0$$
Given equation is imaginary ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{2 \frac{\sqrt{6}}{18}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{2 \frac{\sqrt{6}}{18}}\right)^{2}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(1/2, -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:
$$4 x^{2} - 4 x + 4 y^{2} + 8 y + 59 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -2$$
$$a_{22} = 4$$
$$a_{23} = 4$$
$$a_{33} = 59$$
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
$$I_{1} = 8$$
$$I_{3} = \left|\begin{matrix}4 & 0 & -2\\0 & 4 & 4\\-2 & 4 & 59\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0\\0 & 4 - \lambda\end{matrix}\right|$$
$$I_{1} = 8$$
$$I_{2} = 16$$
$$I_{3} = 864$$
$$I{\left(\lambda \right)} = \lambda^{2} - 8 \lambda + 16$$
$$K_{2} = 452$$
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
$$\lambda^{2} - 8 \lambda + 16 = 0$$
$$\lambda_{1} = 4$$
$$\lambda_{2} = 4$$
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
$$4 \tilde x^{2} + 4 \tilde y^{2} + 54 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{2 \frac{\sqrt{6}}{18}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{2 \frac{\sqrt{6}}{18}}\right)^{2}} = -1$$
- reduced to canonical form
$$4 x^{2} - 4 x + 4 y^{2} + 8 y + 59 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -2$$
$$a_{22} = 4$$
$$a_{23} = 4$$
$$a_{33} = 59$$
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
$$I_{1} = 8$$
|4 0|
I2 = | |
|0 4|$$I_{3} = \left|\begin{matrix}4 & 0 & -2\\0 & 4 & 4\\-2 & 4 & 59\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0\\0 & 4 - \lambda\end{matrix}\right|$$
|4 -2| |4 4 |
K2 = | | + | |
|-2 59| |4 59|$$I_{1} = 8$$
$$I_{2} = 16$$
$$I_{3} = 864$$
$$I{\left(\lambda \right)} = \lambda^{2} - 8 \lambda + 16$$
$$K_{2} = 452$$
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
$$\lambda^{2} - 8 \lambda + 16 = 0$$
$$\lambda_{1} = 4$$
$$\lambda_{2} = 4$$
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
$$4 \tilde x^{2} + 4 \tilde y^{2} + 54 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{2 \frac{\sqrt{6}}{18}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{2 \frac{\sqrt{6}}{18}}\right)^{2}} = -1$$
- reduced to canonical form