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- Canonical form of a elliptical paraboloid
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- Canonical form of a imaginary ellipsoid
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- 4x2+9y2−40x+36y+100=0
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x^2+(y-(x^(2/3)))^2=1
- x^2+(y-sqrt(|x|))^2=10
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Identical expressions
- two x^ two +y^2-2 zero x+16y+ one hundred and ten =0
- 2x squared plus y squared minus 20x plus 16y plus 110 equally 0
- two x to the power of two plus y squared minus 2 zero x plus 16y plus one hundred and ten equally 0
- 2x2+y2-20x+16y+110=0
- 2x²+y²-20x+16y+110=0
- 2x to the power of 2+y to the power of 2-20x+16y+110=0
- 2x^2+y^2-20x+16y+110=O
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Similar expressions
- 2x^2+y^2-20x+16y-110=0
- 2x^2+y^2+20x+16y+110=0
- 2x^2+y^2-20x-16y+110=0
- 2x^2-y^2-20x+16y+110=0
2x^2+y^2-20x+16y+110=0 canonical form
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2 2 110 + y - 20*x + 2*x + 16*y = 0
$$2 x^{2} - 20 x + y^{2} + 16 y + 110 = 0$$
2*x^2 - 20*x + y^2 + 16*y + 110 = 0
Detail solution
Given line equation of 2-order:
$$2 x^{2} - 20 x + y^{2} + 16 y + 110 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = -10$$
$$a_{22} = 1$$
$$a_{23} = 8$$
$$a_{33} = 110$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}2 & 0\\0 & 1\end{matrix}\right|$$
$$\Delta = 2$$
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
$$2 x_{0} - 10 = 0$$
$$y_{0} + 8 = 0$$
then
$$x_{0} = 5$$
$$y_{0} = -8$$
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
$$a'_{33} = - 10 x_{0} + 8 y_{0} + 110$$
$$a'_{33} = -4$$
then equation turns into
$$2 x'^{2} + y'^{2} - 4 = 0$$
Given equation is ellipse
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$2 x^{2} - 20 x + y^{2} + 16 y + 110 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = -10$$
$$a_{22} = 1$$
$$a_{23} = 8$$
$$a_{33} = 110$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}2 & 0\\0 & 1\end{matrix}\right|$$
$$\Delta = 2$$
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
$$2 x_{0} - 10 = 0$$
$$y_{0} + 8 = 0$$
then
$$x_{0} = 5$$
$$y_{0} = -8$$
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
$$a'_{33} = - 10 x_{0} + 8 y_{0} + 110$$
$$a'_{33} = -4$$
then equation turns into
$$2 x'^{2} + y'^{2} - 4 = 0$$
Given equation is ellipse
2 2
\tilde x \tilde y
---------- + --------- = 1
2 2
// ___\\ / 1\
||\/ 2 || \2 /
||-----||
|\ 2 /|
|-------|
\ 1/2 / - reduced to canonical form
The center of canonical coordinate system at point O
(5, -8)
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:
$$2 x^{2} - 20 x + y^{2} + 16 y + 110 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = -10$$
$$a_{22} = 1$$
$$a_{23} = 8$$
$$a_{33} = 110$$
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
$$I_{1} = 3$$
$$I_{3} = \left|\begin{matrix}2 & 0 & -10\\0 & 1 & 8\\-10 & 8 & 110\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 3$$
$$I_{2} = 2$$
$$I_{3} = -8$$
$$I{\left(\lambda \right)} = \lambda^{2} - 3 \lambda + 2$$
$$K_{2} = 166$$
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
$$\lambda^{2} - 3 \lambda + 2 = 0$$
$$\lambda_{1} = 2$$
$$\lambda_{2} = 1$$
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
$$2 \tilde x^{2} + \tilde y^{2} - 4 = 0$$
- reduced to canonical form
$$2 x^{2} - 20 x + y^{2} + 16 y + 110 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = -10$$
$$a_{22} = 1$$
$$a_{23} = 8$$
$$a_{33} = 110$$
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
$$I_{1} = 3$$
|2 0|
I2 = | |
|0 1|$$I_{3} = \left|\begin{matrix}2 & 0 & -10\\0 & 1 & 8\\-10 & 8 & 110\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|$$
| 2 -10| |1 8 |
K2 = | | + | |
|-10 110| |8 110|$$I_{1} = 3$$
$$I_{2} = 2$$
$$I_{3} = -8$$
$$I{\left(\lambda \right)} = \lambda^{2} - 3 \lambda + 2$$
$$K_{2} = 166$$
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
$$\lambda^{2} - 3 \lambda + 2 = 0$$
$$\lambda_{1} = 2$$
$$\lambda_{2} = 1$$
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
$$2 \tilde x^{2} + \tilde y^{2} - 4 = 0$$
2 2
\tilde x \tilde y
---------- + --------- = 1
2 2
// ___\\ / 1\
||\/ 2 || \2 /
||-----||
|\ 2 /|
|-------|
\ 1/2 / - reduced to canonical form