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- Canonical form:
- 25x^2+9y^2-9z^2=225
- x^2-4x+2+y^2+2y
- 1x^2+10x-2y+11=0
- x^2-8xy+16y^2+6x-24y+9=0
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- y^2=2*x+1
- z=x^2+y^2/4
- y=y^x/x
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x^2+(y-pow(x*x,1/3))^2=1
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Identical expressions
- two x^ two - two y^2-8sqrt(2)y
- 2x squared minus 2y squared minus 8 square root of (2)y
- two x to the power of two minus two y squared minus 8 square root of (2)y
- 2x^2-2y^2-8√(2)y
- 2x2-2y2-8sqrt(2)y
- 2x2-2y2-8sqrt2y
- 2x²-2y²-8sqrt(2)y
- 2x to the power of 2-2y to the power of 2-8sqrt(2)y
- 2x^2-2y^2-8sqrt2y
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Similar expressions
- 2x^2+2y^2-8sqrt(2)y
- 2x^2-2y^2+8sqrt(2)y
2x^2-2y^2-8sqrt(2)y canonical form
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2 2 ___ - 2*y + 2*x - 8*y*\/ 2 = 0
$$2 x^{2} - 2 y^{2} - 8 \sqrt{2} y = 0$$
2*x^2 - 2*y^2 - 8*sqrt(2)*y = 0
Detail solution
Given line equation of 2-order:
$$2 x^{2} - 2 y^{2} - 8 \sqrt{2} y = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -2$$
$$a_{23} = - 4 \sqrt{2}$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}2 & 0\\0 & -2\end{matrix}\right|$$
$$\Delta = -4$$
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
$$2 x_{0} = 0$$
$$- 2 y_{0} - 4 \sqrt{2} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = - 2 \sqrt{2}$$
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
$$a'_{33} = - 4 \sqrt{2} y_{0}$$
$$a'_{33} = 16$$
then equation turns into
$$2 x'^{2} - 2 y'^{2} + 16 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{8} - \frac{\tilde y^{2}}{8} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$2 x^{2} - 2 y^{2} - 8 \sqrt{2} y = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -2$$
$$a_{23} = - 4 \sqrt{2}$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}2 & 0\\0 & -2\end{matrix}\right|$$
$$\Delta = -4$$
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
$$2 x_{0} = 0$$
$$- 2 y_{0} - 4 \sqrt{2} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = - 2 \sqrt{2}$$
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
$$a'_{33} = - 4 \sqrt{2} y_{0}$$
$$a'_{33} = 16$$
then equation turns into
$$2 x'^{2} - 2 y'^{2} + 16 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{8} - \frac{\tilde y^{2}}{8} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
___ (0, -2*\/ 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:
$$2 x^{2} - 2 y^{2} - 8 \sqrt{2} y = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -2$$
$$a_{23} = - 4 \sqrt{2}$$
$$a_{33} = 0$$
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
$$I_{1} = 0$$
$$I_{3} = \left|\begin{matrix}2 & 0 & 0\\0 & -2 & - 4 \sqrt{2}\\0 & - 4 \sqrt{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 0\\0 & - \lambda - 2\end{matrix}\right|$$
$$I_{1} = 0$$
$$I_{2} = -4$$
$$I_{3} = -64$$
$$I{\left(\lambda \right)} = \lambda^{2} - 4$$
$$K_{2} = -32$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 4 = 0$$
$$\lambda_{1} = -2$$
$$\lambda_{2} = 2$$
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
$$- 2 \tilde x^{2} + 2 \tilde y^{2} + 16 = 0$$
$$\frac{\tilde x^{2}}{8} - \frac{\tilde y^{2}}{8} = 1$$
- reduced to canonical form
$$2 x^{2} - 2 y^{2} - 8 \sqrt{2} y = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -2$$
$$a_{23} = - 4 \sqrt{2}$$
$$a_{33} = 0$$
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
$$I_{1} = 0$$
|2 0 |
I2 = | |
|0 -2|$$I_{3} = \left|\begin{matrix}2 & 0 & 0\\0 & -2 & - 4 \sqrt{2}\\0 & - 4 \sqrt{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 0\\0 & - \lambda - 2\end{matrix}\right|$$
| ___|
|2 0| | -2 -4*\/ 2 |
K2 = | | + | |
|0 0| | ___ |
|-4*\/ 2 0 |$$I_{1} = 0$$
$$I_{2} = -4$$
$$I_{3} = -64$$
$$I{\left(\lambda \right)} = \lambda^{2} - 4$$
$$K_{2} = -32$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 4 = 0$$
$$\lambda_{1} = -2$$
$$\lambda_{2} = 2$$
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
$$- 2 \tilde x^{2} + 2 \tilde y^{2} + 16 = 0$$
$$\frac{\tilde x^{2}}{8} - \frac{\tilde y^{2}}{8} = 1$$
- reduced to canonical form