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How to use it?
- Canonical form:
- 9x²-4y²-90x-8y+185=0
- 2x^2+10y^2+2z^2+6xy+2xz+6yz=36
- x1x2+x1x3−x2x3
- 4x^2+y^2+4z^2+4y+24z+36=0
- Plot:
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xy=6
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4x^2+4y^2=25
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(x^2+y^2-1)^3-(x^2)*y^3=0
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xy=4
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Identical expressions
- 9x²-4y²-9 zero x-8y+ one hundred and eighty-five =0
- 9x² minus 4y² minus 90x minus 8y plus 185 equally 0
- 9x² minus 4y² minus 9 zero x minus 8y plus one hundred and eighty minus five equally 0
- 9x²-4y²-90x-8y+185=O
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Similar expressions
- 9x²-4y²-90x-8y-185=0
- 9x²-4y²+90x-8y+185=0
- 9x²-4y²-90x+8y+185=0
- 9x²+4y²-90x-8y+185=0
9x²-4y²-90x-8y+185=0 canonical form
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The solution
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2 2 185 - 90*x - 8*y - 4*y + 9*x = 0
$$9 x^{2} - 90 x - 4 y^{2} - 8 y + 185 = 0$$
9*x^2 - 90*x - 4*y^2 - 8*y + 185 = 0
Detail solution
Given line equation of 2-order:
$$9 x^{2} - 90 x - 4 y^{2} - 8 y + 185 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -45$$
$$a_{22} = -4$$
$$a_{23} = -4$$
$$a_{33} = 185$$
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
$$9 x_{0} - 45 = 0$$
$$- 4 y_{0} - 4 = 0$$
then
$$x_{0} = 5$$
$$y_{0} = -1$$
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
$$a'_{33} = - 45 x_{0} - 4 y_{0} + 185$$
$$a'_{33} = -36$$
then equation turns into
$$9 x'^{2} - 4 y'^{2} - 36 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{9} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$9 x^{2} - 90 x - 4 y^{2} - 8 y + 185 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -45$$
$$a_{22} = -4$$
$$a_{23} = -4$$
$$a_{33} = 185$$
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
$$9 x_{0} - 45 = 0$$
$$- 4 y_{0} - 4 = 0$$
then
$$x_{0} = 5$$
$$y_{0} = -1$$
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
$$a'_{33} = - 45 x_{0} - 4 y_{0} + 185$$
$$a'_{33} = -36$$
then equation turns into
$$9 x'^{2} - 4 y'^{2} - 36 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{9} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(5, -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:
$$9 x^{2} - 90 x - 4 y^{2} - 8 y + 185 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -45$$
$$a_{22} = -4$$
$$a_{23} = -4$$
$$a_{33} = 185$$
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
$$I_{1} = 5$$
$$I_{3} = \left|\begin{matrix}9 & 0 & -45\\0 & -4 & -4\\-45 & -4 & 185\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 0\\0 & - \lambda - 4\end{matrix}\right|$$
$$I_{1} = 5$$
$$I_{2} = -36$$
$$I_{3} = 1296$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda - 36$$
$$K_{2} = -1116$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 5 \lambda - 36 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = -4$$
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
$$9 \tilde x^{2} - 4 \tilde y^{2} - 36 = 0$$
$$\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{9} = 1$$
- reduced to canonical form
$$9 x^{2} - 90 x - 4 y^{2} - 8 y + 185 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -45$$
$$a_{22} = -4$$
$$a_{23} = -4$$
$$a_{33} = 185$$
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
$$I_{1} = 5$$
|9 0 |
I2 = | |
|0 -4|$$I_{3} = \left|\begin{matrix}9 & 0 & -45\\0 & -4 & -4\\-45 & -4 & 185\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 0\\0 & - \lambda - 4\end{matrix}\right|$$
| 9 -45| |-4 -4 |
K2 = | | + | |
|-45 185| |-4 185|$$I_{1} = 5$$
$$I_{2} = -36$$
$$I_{3} = 1296$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda - 36$$
$$K_{2} = -1116$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 5 \lambda - 36 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = -4$$
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
$$9 \tilde x^{2} - 4 \tilde y^{2} - 36 = 0$$
$$\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{9} = 1$$
- reduced to canonical form