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- 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:
- x^2-y^2+z^2=0
- (x^2)/6-(y^2)/4-(z^2)/1=0
- 3x^2+6x-15y+15=0
- x^2-8xy+7y^2
- Plot:
- x^2+(y-sqrt((|x|)))^2=1
-
x^2-4xy+3y^2=6
- x^2+y^2=9
- x^2/a^2+y^2/b^2=1
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Identical expressions
- two xy+ two sqrt(2)x-4sqrt(2)y- nine = zero
- 2xy plus 2 square root of (2)x minus 4 square root of (2)y minus 9 equally 0
- two xy plus two square root of (2)x minus 4 square root of (2)y minus nine equally zero
- 2xy+2√(2)x-4√(2)y-9=0
- 2xy+2sqrt2x-4sqrt2y-9=0
- 2xy+2sqrt(2)x-4sqrt(2)y-9=O
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Similar expressions
- 2xy+2sqrt(2)x+4sqrt(2)y-9=0
- 2xy+2sqrt(2)x-4sqrt(2)y+9=0
- 2xy-2sqrt(2)x-4sqrt(2)y-9=0
2xy+2sqrt(2)x-4sqrt(2)y-9=0 canonical form
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The solution
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___ ___ -9 - 4*y*\/ 2 + 2*x*y + 2*x*\/ 2 = 0
$$2 x y + 2 \sqrt{2} x - 4 \sqrt{2} y - 9 = 0$$
2*x*y + 2*sqrt(2)*x - 4*sqrt(2)*y - 9 = 0
Detail solution
Given line equation of 2-order:
$$2 x y + 2 \sqrt{2} x - 4 \sqrt{2} y - 9 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 1$$
$$a_{13} = \sqrt{2}$$
$$a_{22} = 0$$
$$a_{23} = - 2 \sqrt{2}$$
$$a_{33} = -9$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 1\\1 & 0\end{matrix}\right|$$
$$\Delta = -1$$
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
$$y_{0} + \sqrt{2} = 0$$
$$x_{0} - 2 \sqrt{2} = 0$$
then
$$x_{0} = 2 \sqrt{2}$$
$$y_{0} = - \sqrt{2}$$
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
$$a'_{33} = \sqrt{2} x_{0} - 2 \sqrt{2} y_{0} - 9$$
$$a'_{33} = -1$$
then equation turns into
$$2 x' y' - 1 = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
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
$$x' = \frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}$$
$$y' = \frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}$$
then the equation turns from
$$2 x' y' - 1 = 0$$
to
$$2 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) - 1 = 0$$
simplify
$$\tilde x^{2} - \tilde y^{2} - 1 = 0$$
$$- \tilde x^{2} + \tilde y^{2} + 1 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{1} - \frac{\tilde y^{2}}{1} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$2 x y + 2 \sqrt{2} x - 4 \sqrt{2} y - 9 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 1$$
$$a_{13} = \sqrt{2}$$
$$a_{22} = 0$$
$$a_{23} = - 2 \sqrt{2}$$
$$a_{33} = -9$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 1\\1 & 0\end{matrix}\right|$$
$$\Delta = -1$$
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
$$y_{0} + \sqrt{2} = 0$$
$$x_{0} - 2 \sqrt{2} = 0$$
then
$$x_{0} = 2 \sqrt{2}$$
$$y_{0} = - \sqrt{2}$$
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
$$a'_{33} = \sqrt{2} x_{0} - 2 \sqrt{2} y_{0} - 9$$
$$a'_{33} = -1$$
then equation turns into
$$2 x' y' - 1 = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
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
$$x' = \frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}$$
$$y' = \frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}$$
then the equation turns from
$$2 x' y' - 1 = 0$$
to
$$2 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) - 1 = 0$$
simplify
$$\tilde x^{2} - \tilde y^{2} - 1 = 0$$
$$- \tilde x^{2} + \tilde y^{2} + 1 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{1} - \frac{\tilde y^{2}}{1} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
___ ___ (2*\/ 2, -\/ 2 )
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:
$$2 x y + 2 \sqrt{2} x - 4 \sqrt{2} y - 9 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 1$$
$$a_{13} = \sqrt{2}$$
$$a_{22} = 0$$
$$a_{23} = - 2 \sqrt{2}$$
$$a_{33} = -9$$
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
$$I_{1} = 0$$
$$I_{3} = \left|\begin{matrix}0 & 1 & \sqrt{2}\\1 & 0 & - 2 \sqrt{2}\\\sqrt{2} & - 2 \sqrt{2} & -9\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 1\\1 & - \lambda\end{matrix}\right|$$
$$I_{1} = 0$$
$$I_{2} = -1$$
$$I_{3} = 1$$
$$I{\left(\lambda \right)} = \lambda^{2} - 1$$
$$K_{2} = -10$$
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
$$\lambda^{2} - 1 = 0$$
$$\lambda_{1} = -1$$
$$\lambda_{2} = 1$$
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
$$- \tilde x^{2} + \tilde y^{2} - 1 = 0$$
$$\frac{\tilde x^{2}}{1} - \frac{\tilde y^{2}}{1} = -1$$
- reduced to canonical form
$$2 x y + 2 \sqrt{2} x - 4 \sqrt{2} y - 9 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 1$$
$$a_{13} = \sqrt{2}$$
$$a_{22} = 0$$
$$a_{23} = - 2 \sqrt{2}$$
$$a_{33} = -9$$
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
$$I_{1} = 0$$
|0 1|
I2 = | |
|1 0|$$I_{3} = \left|\begin{matrix}0 & 1 & \sqrt{2}\\1 & 0 & - 2 \sqrt{2}\\\sqrt{2} & - 2 \sqrt{2} & -9\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 1\\1 & - \lambda\end{matrix}\right|$$
| ___| | ___|
| 0 \/ 2 | | 0 -2*\/ 2 |
K2 = | | + | |
| ___ | | ___ |
|\/ 2 -9 | |-2*\/ 2 -9 |$$I_{1} = 0$$
$$I_{2} = -1$$
$$I_{3} = 1$$
$$I{\left(\lambda \right)} = \lambda^{2} - 1$$
$$K_{2} = -10$$
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
$$\lambda^{2} - 1 = 0$$
$$\lambda_{1} = -1$$
$$\lambda_{2} = 1$$
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
$$- \tilde x^{2} + \tilde y^{2} - 1 = 0$$
$$\frac{\tilde x^{2}}{1} - \frac{\tilde y^{2}}{1} = -1$$
- reduced to canonical form