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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:
- 3z^(2)+4y^(2)-x=5
- x^2-2xy+8y^2-12x+64y+148=0
- 9x^2+52y^2-36x+50y-164=0
- x^2-2x-z^2+z=y^2
- Plot:
- y^2-4*x+4*y-20=0
- (x^2+y^2)^2=a^2*x*y
- d*(x^2*y^2+3*x^2*y^2-3*x*y^2)=0
-
(x-2)y^2=1
-
Identical expressions
- 9x^ two +5 two y^2-36x+5 zero y- one hundred and sixty-four =0
- 9x squared plus 52y squared minus 36x plus 50y minus 164 equally 0
- 9x to the power of two plus 5 two y squared minus 36x plus 5 zero y minus one hundred and sixty minus four equally 0
- 9x2+52y2-36x+50y-164=0
- 9x²+52y²-36x+50y-164=0
- 9x to the power of 2+52y to the power of 2-36x+50y-164=0
- 9x^2+52y^2-36x+50y-164=O
-
Similar expressions
- 9x^2+52y^2-36x+50y+164=0
- 9x^2-52y^2-36x+50y-164=0
- 9x^2+52y^2-36x-50y-164=0
- 9x^2+52y^2+36x+50y-164=0
9x^2+52y^2-36x+50y-164=0 canonical form
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2 2 -164 - 36*x + 9*x + 50*y + 52*y = 0
$$9 x^{2} - 36 x + 52 y^{2} + 50 y - 164 = 0$$
9*x^2 - 36*x + 52*y^2 + 50*y - 164 = 0
Detail solution
Given line equation of 2-order:
$$9 x^{2} - 36 x + 52 y^{2} + 50 y - 164 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -18$$
$$a_{22} = 52$$
$$a_{23} = 25$$
$$a_{33} = -164$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 0\\0 & 52\end{matrix}\right|$$
$$\Delta = 468$$
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
$$9 x_{0} - 18 = 0$$
$$52 y_{0} + 25 = 0$$
then
$$x_{0} = 2$$
$$y_{0} = - \frac{25}{52}$$
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
$$a'_{33} = - 18 x_{0} + 25 y_{0} - 164$$
$$a'_{33} = - \frac{11025}{52}$$
then equation turns into
$$9 x'^{2} + 52 y'^{2} - \frac{11025}{52} = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{3 \frac{2 \sqrt{13}}{105}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{26} \sqrt{13}}{\frac{2}{105} \sqrt{13}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$9 x^{2} - 36 x + 52 y^{2} + 50 y - 164 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -18$$
$$a_{22} = 52$$
$$a_{23} = 25$$
$$a_{33} = -164$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 0\\0 & 52\end{matrix}\right|$$
$$\Delta = 468$$
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
$$9 x_{0} - 18 = 0$$
$$52 y_{0} + 25 = 0$$
then
$$x_{0} = 2$$
$$y_{0} = - \frac{25}{52}$$
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
$$a'_{33} = - 18 x_{0} + 25 y_{0} - 164$$
$$a'_{33} = - \frac{11025}{52}$$
then equation turns into
$$9 x'^{2} + 52 y'^{2} - \frac{11025}{52} = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{3 \frac{2 \sqrt{13}}{105}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{26} \sqrt{13}}{\frac{2}{105} \sqrt{13}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
-25
(2, ----)
52 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:
$$9 x^{2} - 36 x + 52 y^{2} + 50 y - 164 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -18$$
$$a_{22} = 52$$
$$a_{23} = 25$$
$$a_{33} = -164$$
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
$$I_{1} = 61$$
$$I_{3} = \left|\begin{matrix}9 & 0 & -18\\0 & 52 & 25\\-18 & 25 & -164\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 0\\0 & 52 - \lambda\end{matrix}\right|$$
$$I_{1} = 61$$
$$I_{2} = 468$$
$$I_{3} = -99225$$
$$I{\left(\lambda \right)} = \lambda^{2} - 61 \lambda + 468$$
$$K_{2} = -10953$$
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
$$\lambda^{2} - 61 \lambda + 468 = 0$$
$$\lambda_{1} = 52$$
$$\lambda_{2} = 9$$
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
$$52 \tilde x^{2} + 9 \tilde y^{2} - \frac{11025}{52} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{26} \sqrt{13}}{\frac{2}{105} \sqrt{13}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{3 \frac{2 \sqrt{13}}{105}}\right)^{2}} = 1$$
- reduced to canonical form
$$9 x^{2} - 36 x + 52 y^{2} + 50 y - 164 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -18$$
$$a_{22} = 52$$
$$a_{23} = 25$$
$$a_{33} = -164$$
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
$$I_{1} = 61$$
|9 0 |
I2 = | |
|0 52|$$I_{3} = \left|\begin{matrix}9 & 0 & -18\\0 & 52 & 25\\-18 & 25 & -164\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 0\\0 & 52 - \lambda\end{matrix}\right|$$
| 9 -18 | |52 25 |
K2 = | | + | |
|-18 -164| |25 -164|$$I_{1} = 61$$
$$I_{2} = 468$$
$$I_{3} = -99225$$
$$I{\left(\lambda \right)} = \lambda^{2} - 61 \lambda + 468$$
$$K_{2} = -10953$$
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
$$\lambda^{2} - 61 \lambda + 468 = 0$$
$$\lambda_{1} = 52$$
$$\lambda_{2} = 9$$
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
$$52 \tilde x^{2} + 9 \tilde y^{2} - \frac{11025}{52} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{26} \sqrt{13}}{\frac{2}{105} \sqrt{13}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{3 \frac{2 \sqrt{13}}{105}}\right)^{2}} = 1$$
- reduced to canonical form