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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:
- 3x^2+0xy+4y^2-6x+16y+7=0
- y=-3-sqrt(-3x-21)
- 3*x-4*y+4=0
- x^2+4*y^2+2*x-32*y+64=0
- Plot:
- (x^2+y^2-1)^3-x^2*y^3=0
-
y^2=x(1-x^2)^3
-
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
-
Identical expressions
- 9x^ two +y^ two + one 8x+2y+1= zero
- 9x squared plus y squared plus 18x plus 2y plus 1 equally 0
- 9x to the power of two plus y to the power of two plus one 8x plus 2y plus 1 equally zero
- 9x2+y2+18x+2y+1=0
- 9x²+y²+18x+2y+1=0
- 9x to the power of 2+y to the power of 2+18x+2y+1=0
- 9x^2+y^2+18x+2y+1=O
-
Similar expressions
- 9x^2+y^2-18x+2y+1=0
- 9x^2+y^2+18x+2y-1=0
- 9x^2-y^2+18x+2y+1=0
- 9x^2+y^2+18x-2y+1=0
9x^2+y^2+18x+2y+1=0 canonical form
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2 2 1 + y + 2*y + 9*x + 18*x = 0
$$9 x^{2} + 18 x + y^{2} + 2 y + 1 = 0$$
9*x^2 + 18*x + y^2 + 2*y + 1 = 0
Detail solution
Given line equation of 2-order:
$$9 x^{2} + 18 x + y^{2} + 2 y + 1 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = 9$$
$$a_{22} = 1$$
$$a_{23} = 1$$
$$a_{33} = 1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 0\\0 & 1\end{matrix}\right|$$
$$\Delta = 9$$
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
$$9 x_{0} + 9 = 0$$
$$y_{0} + 1 = 0$$
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
$$a'_{33} = 9 x_{0} + y_{0} + 1$$
$$a'_{33} = -9$$
then equation turns into
$$9 x'^{2} + y'^{2} - 9 = 0$$
Given equation is ellipse
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$9 x^{2} + 18 x + y^{2} + 2 y + 1 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = 9$$
$$a_{22} = 1$$
$$a_{23} = 1$$
$$a_{33} = 1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 0\\0 & 1\end{matrix}\right|$$
$$\Delta = 9$$
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
$$9 x_{0} + 9 = 0$$
$$y_{0} + 1 = 0$$
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
$$a'_{33} = 9 x_{0} + y_{0} + 1$$
$$a'_{33} = -9$$
then equation turns into
$$9 x'^{2} + y'^{2} - 9 = 0$$
Given equation is ellipse
2 2
\tilde x \tilde y
--------- + --------- = 1
2 2
/ 1 \ / 1\
|-----| \3 /
\3*1/3/ - 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:
$$9 x^{2} + 18 x + y^{2} + 2 y + 1 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = 9$$
$$a_{22} = 1$$
$$a_{23} = 1$$
$$a_{33} = 1$$
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
$$I_{1} = 10$$
$$I_{3} = \left|\begin{matrix}9 & 0 & 9\\0 & 1 & 1\\9 & 1 & 1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 10$$
$$I_{2} = 9$$
$$I_{3} = -81$$
$$I{\left(\lambda \right)} = \lambda^{2} - 10 \lambda + 9$$
$$K_{2} = -72$$
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
$$\lambda^{2} - 10 \lambda + 9 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 1$$
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
$$9 \tilde x^{2} + \tilde y^{2} - 9 = 0$$
- reduced to canonical form
$$9 x^{2} + 18 x + y^{2} + 2 y + 1 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = 9$$
$$a_{22} = 1$$
$$a_{23} = 1$$
$$a_{33} = 1$$
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
$$I_{1} = 10$$
|9 0|
I2 = | |
|0 1|$$I_{3} = \left|\begin{matrix}9 & 0 & 9\\0 & 1 & 1\\9 & 1 & 1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|$$
|9 9| |1 1|
K2 = | | + | |
|9 1| |1 1|$$I_{1} = 10$$
$$I_{2} = 9$$
$$I_{3} = -81$$
$$I{\left(\lambda \right)} = \lambda^{2} - 10 \lambda + 9$$
$$K_{2} = -72$$
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
$$\lambda^{2} - 10 \lambda + 9 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 1$$
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
$$9 \tilde x^{2} + \tilde y^{2} - 9 = 0$$
2 2
\tilde x \tilde y
--------- + --------- = 1
2 2
/ 1 \ / 1\
|-----| \3 /
\3*1/3/ - reduced to canonical form