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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:
-
9x^2+4y^2-18x+24y+9=0
- x1x2+x1x3+x1x4+x2x3+x2x4+x3x4
- -2c^3(3+16c^2)-(c^2+4)×3c^3+6(3c^3+1)
- (x2+y2–1)3–x2y3=0
- Plot:
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y=x^4-26x^2+25
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xy=0
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xy=6
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4x^2+4y^2=25
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Identical expressions
- nine x^ two +4y^ two -18x+24y+9= zero
- 9x squared plus 4y squared minus 18x plus 24y plus 9 equally 0
- nine x to the power of two plus 4y to the power of two minus 18x plus 24y plus 9 equally zero
- 9x2+4y2-18x+24y+9=0
- 9x²+4y²-18x+24y+9=0
- 9x to the power of 2+4y to the power of 2-18x+24y+9=0
- 9x^2+4y^2-18x+24y+9=O
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Similar expressions
- 9x^2+4y^2-18x+24y-9=0
- 9x^2-4y^2-18x+24y+9=0
- 9x^2+4y^2+18x+24y+9=0
- 9x^2+4y^2-18x-24y+9=0
9x^2+4y^2-18x+24y+9=0 canonical form
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The solution
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2 2 9 - 18*x + 4*y + 9*x + 24*y = 0
$$9 x^{2} - 18 x + 4 y^{2} + 24 y + 9 = 0$$
9*x^2 - 18*x + 4*y^2 + 24*y + 9 = 0
Detail solution
Given line equation of 2-order:
$$9 x^{2} - 18 x + 4 y^{2} + 24 y + 9 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -9$$
$$a_{22} = 4$$
$$a_{23} = 12$$
$$a_{33} = 9$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 0\\0 & 4\end{matrix}\right|$$
$$\Delta = 36$$
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
$$9 x_{0} - 9 = 0$$
$$4 y_{0} + 12 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = -3$$
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
$$a'_{33} = - 9 x_{0} + 12 y_{0} + 9$$
$$a'_{33} = -36$$
then equation turns into
$$9 x'^{2} + 4 y'^{2} - 36 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{2^{2}} + \frac{\tilde y^{2}}{3^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$9 x^{2} - 18 x + 4 y^{2} + 24 y + 9 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -9$$
$$a_{22} = 4$$
$$a_{23} = 12$$
$$a_{33} = 9$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 0\\0 & 4\end{matrix}\right|$$
$$\Delta = 36$$
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
$$9 x_{0} - 9 = 0$$
$$4 y_{0} + 12 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = -3$$
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
$$a'_{33} = - 9 x_{0} + 12 y_{0} + 9$$
$$a'_{33} = -36$$
then equation turns into
$$9 x'^{2} + 4 y'^{2} - 36 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{2^{2}} + \frac{\tilde y^{2}}{3^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(1, -3)
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:
$$9 x^{2} - 18 x + 4 y^{2} + 24 y + 9 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -9$$
$$a_{22} = 4$$
$$a_{23} = 12$$
$$a_{33} = 9$$
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
$$I_{1} = 13$$
$$I_{3} = \left|\begin{matrix}9 & 0 & -9\\0 & 4 & 12\\-9 & 12 & 9\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 0\\0 & 4 - \lambda\end{matrix}\right|$$
$$I_{1} = 13$$
$$I_{2} = 36$$
$$I_{3} = -1296$$
$$I{\left(\lambda \right)} = \lambda^{2} - 13 \lambda + 36$$
$$K_{2} = -108$$
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
$$\lambda^{2} - 13 \lambda + 36 = 0$$
Solve this equation
$$\lambda_{1} = 9$$
$$\lambda_{2} = 4$$
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
$$9 \tilde x^{2} + 4 \tilde y^{2} - 36 = 0$$
$$\frac{\tilde x^{2}}{2^{2}} + \frac{\tilde y^{2}}{3^{2}} = 1$$
- reduced to canonical form
$$9 x^{2} - 18 x + 4 y^{2} + 24 y + 9 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -9$$
$$a_{22} = 4$$
$$a_{23} = 12$$
$$a_{33} = 9$$
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
$$I_{1} = 13$$
|9 0|
I2 = | |
|0 4|$$I_{3} = \left|\begin{matrix}9 & 0 & -9\\0 & 4 & 12\\-9 & 12 & 9\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 0\\0 & 4 - \lambda\end{matrix}\right|$$
|9 -9| |4 12|
K2 = | | + | |
|-9 9 | |12 9 |$$I_{1} = 13$$
$$I_{2} = 36$$
$$I_{3} = -1296$$
$$I{\left(\lambda \right)} = \lambda^{2} - 13 \lambda + 36$$
$$K_{2} = -108$$
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
$$\lambda^{2} - 13 \lambda + 36 = 0$$
Solve this equation
$$\lambda_{1} = 9$$
$$\lambda_{2} = 4$$
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
$$9 \tilde x^{2} + 4 \tilde y^{2} - 36 = 0$$
$$\frac{\tilde x^{2}}{2^{2}} + \frac{\tilde y^{2}}{3^{2}} = 1$$
- reduced to canonical form
The graph