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- Canonical form of a double hyperboloid
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- Canonical form of a degenerate ellipse
- Canonical form of a parabola
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How to use it?
- Canonical form:
- x^2+y^2-14x-8y+40=0
- 4x^2-9y^2-18y+16z^2-72z=-27
- 2x^2+2y^2+14x-10y+35=0
- 5*x^2+y^2+4*z^2+6*x*y+8*x*z+4*y*z-4*x-4*y-4*z=0
- Plot:
-
sqrt((x-4)^2+y^2)+sqrt(x^2+(y+4)^2)=8*sqrt(2)
- sin(999*x)+y=1-y
- 3*x-3*y-3=6
- 3*x+2*y=12
-
Identical expressions
- 4x^ two +9y^ two -8x-36y- sixty-eight = zero
- 4x squared plus 9y squared minus 8x minus 36y minus 68 equally 0
- 4x to the power of two plus 9y to the power of two minus 8x minus 36y minus sixty minus eight equally zero
- 4x2+9y2-8x-36y-68=0
- 4x²+9y²-8x-36y-68=0
- 4x to the power of 2+9y to the power of 2-8x-36y-68=0
- 4x^2+9y^2-8x-36y-68=O
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Similar expressions
- 4x^2-9y^2-8x-36y-68=0
- 4x^2+9y^2-8x+36y-68=0
- 4x^2+9y^2-8x-36y+68=0
- 4x^2+9y^2+8x-36y-68=0
4x^2+9y^2-8x-36y-68=0 canonical form
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2 2 -68 - 36*y - 8*x + 4*x + 9*y = 0
$$4 x^{2} - 8 x + 9 y^{2} - 36 y - 68 = 0$$
4*x^2 - 8*x + 9*y^2 - 36*y - 68 = 0
Detail solution
Given line equation of 2-order:
$$4 x^{2} - 8 x + 9 y^{2} - 36 y - 68 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = 9$$
$$a_{23} = -18$$
$$a_{33} = -68$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & 9\end{matrix}\right|$$
$$\Delta = 36$$
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
$$4 x_{0} - 4 = 0$$
$$9 y_{0} - 18 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = 2$$
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
$$a'_{33} = - 4 x_{0} - 18 y_{0} - 68$$
$$a'_{33} = -108$$
then equation turns into
$$4 x'^{2} + 9 y'^{2} - 108 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{2 \frac{\sqrt{3}}{18}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{3 \frac{\sqrt{3}}{18}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$4 x^{2} - 8 x + 9 y^{2} - 36 y - 68 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = 9$$
$$a_{23} = -18$$
$$a_{33} = -68$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & 9\end{matrix}\right|$$
$$\Delta = 36$$
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
$$4 x_{0} - 4 = 0$$
$$9 y_{0} - 18 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = 2$$
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
$$a'_{33} = - 4 x_{0} - 18 y_{0} - 68$$
$$a'_{33} = -108$$
then equation turns into
$$4 x'^{2} + 9 y'^{2} - 108 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{2 \frac{\sqrt{3}}{18}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{3 \frac{\sqrt{3}}{18}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(1, 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:
$$4 x^{2} - 8 x + 9 y^{2} - 36 y - 68 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = 9$$
$$a_{23} = -18$$
$$a_{33} = -68$$
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
$$I_{1} = 13$$
$$I_{3} = \left|\begin{matrix}4 & 0 & -4\\0 & 9 & -18\\-4 & -18 & -68\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0\\0 & 9 - \lambda\end{matrix}\right|$$
$$I_{1} = 13$$
$$I_{2} = 36$$
$$I_{3} = -3888$$
$$I{\left(\lambda \right)} = \lambda^{2} - 13 \lambda + 36$$
$$K_{2} = -1224$$
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
$$\lambda^{2} - 13 \lambda + 36 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 4$$
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
$$9 \tilde x^{2} + 4 \tilde y^{2} - 108 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{3 \frac{\sqrt{3}}{18}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{2 \frac{\sqrt{3}}{18}}\right)^{2}} = 1$$
- reduced to canonical form
$$4 x^{2} - 8 x + 9 y^{2} - 36 y - 68 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = 9$$
$$a_{23} = -18$$
$$a_{33} = -68$$
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
$$I_{1} = 13$$
|4 0|
I2 = | |
|0 9|$$I_{3} = \left|\begin{matrix}4 & 0 & -4\\0 & 9 & -18\\-4 & -18 & -68\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0\\0 & 9 - \lambda\end{matrix}\right|$$
|4 -4 | | 9 -18|
K2 = | | + | |
|-4 -68| |-18 -68|$$I_{1} = 13$$
$$I_{2} = 36$$
$$I_{3} = -3888$$
$$I{\left(\lambda \right)} = \lambda^{2} - 13 \lambda + 36$$
$$K_{2} = -1224$$
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
$$\lambda^{2} - 13 \lambda + 36 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 4$$
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
$$9 \tilde x^{2} + 4 \tilde y^{2} - 108 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{3 \frac{\sqrt{3}}{18}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{2 \frac{\sqrt{3}}{18}}\right)^{2}} = 1$$
- reduced to canonical form