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How to use it?
- Canonical form:
- 8*x^2-4*x*y+5*y^2-52*x+22*y+53=0
- -5x^2+11y^2-121y+22=0
- 2y^2(6y-1)+3y(y-4y^2)
- sin(x)/cos(x)
- Plot:
- (x-2)*(y-3)=0
- x^2+y^2=1
- x^2+y^2=4
-
3*cos(x)*sin(y)-sin(x)*cos(y)-4*cos(y)=3*sqrt(3)
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Identical expressions
- -5x^ two +11y^ two -121y+ twenty-two = zero
- minus 5x squared plus 11y squared minus 121y plus 22 equally 0
- minus 5x to the power of two plus 11y to the power of two minus 121y plus twenty minus two equally zero
- -5x2+11y2-121y+22=0
- -5x²+11y²-121y+22=0
- -5x to the power of 2+11y to the power of 2-121y+22=0
- -5x^2+11y^2-121y+22=O
-
Similar expressions
- -5x^2+11y^2+121y+22=0
- -5x^2-11y^2-121y+22=0
- 5x^2+11y^2-121y+22=0
- -5x^2+11y^2-121y-22=0
-5x^2+11y^2-121y+22=0 canonical form
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The solution
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2 2 22 - 121*y - 5*x + 11*y = 0
$$- 5 x^{2} + 11 y^{2} - 121 y + 22 = 0$$
-5*x^2 + 11*y^2 - 121*y + 22 = 0
Detail solution
Given line equation of 2-order:
$$- 5 x^{2} + 11 y^{2} - 121 y + 22 = 0$$
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
$$a_{11} = -5$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 11$$
$$a_{23} = - \frac{121}{2}$$
$$a_{33} = 22$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-5 & 0\\0 & 11\end{matrix}\right|$$
$$\Delta = -55$$
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
$$- 5 x_{0} = 0$$
$$11 y_{0} - \frac{121}{2} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = \frac{11}{2}$$
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
$$a'_{33} = 22 - \frac{121 y_{0}}{2}$$
$$a'_{33} = - \frac{1243}{4}$$
then equation turns into
$$- 5 x'^{2} + 11 y'^{2} - \frac{1243}{4} = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{\frac{1243}{20}} - \frac{\tilde y^{2}}{\frac{113}{4}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$- 5 x^{2} + 11 y^{2} - 121 y + 22 = 0$$
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
$$a_{11} = -5$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 11$$
$$a_{23} = - \frac{121}{2}$$
$$a_{33} = 22$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-5 & 0\\0 & 11\end{matrix}\right|$$
$$\Delta = -55$$
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
$$- 5 x_{0} = 0$$
$$11 y_{0} - \frac{121}{2} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = \frac{11}{2}$$
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
$$a'_{33} = 22 - \frac{121 y_{0}}{2}$$
$$a'_{33} = - \frac{1243}{4}$$
then equation turns into
$$- 5 x'^{2} + 11 y'^{2} - \frac{1243}{4} = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{\frac{1243}{20}} - \frac{\tilde y^{2}}{\frac{113}{4}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 11/2)
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:
$$- 5 x^{2} + 11 y^{2} - 121 y + 22 = 0$$
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
$$a_{11} = -5$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 11$$
$$a_{23} = - \frac{121}{2}$$
$$a_{33} = 22$$
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
$$I_{1} = 6$$
$$I_{3} = \left|\begin{matrix}-5 & 0 & 0\\0 & 11 & - \frac{121}{2}\\0 & - \frac{121}{2} & 22\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 5 & 0\\0 & 11 - \lambda\end{matrix}\right|$$
$$I_{1} = 6$$
$$I_{2} = -55$$
$$I_{3} = \frac{68365}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 6 \lambda - 55$$
$$K_{2} = - \frac{14113}{4}$$
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
$$\lambda^{2} - 6 \lambda - 55 = 0$$
$$\lambda_{1} = 11$$
$$\lambda_{2} = -5$$
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
$$11 \tilde x^{2} - 5 \tilde y^{2} - \frac{1243}{4} = 0$$
$$\frac{\tilde x^{2}}{\frac{113}{4}} - \frac{\tilde y^{2}}{\frac{1243}{20}} = 1$$
- reduced to canonical form
$$- 5 x^{2} + 11 y^{2} - 121 y + 22 = 0$$
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
$$a_{11} = -5$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 11$$
$$a_{23} = - \frac{121}{2}$$
$$a_{33} = 22$$
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
$$I_{1} = 6$$
|-5 0 |
I2 = | |
|0 11|$$I_{3} = \left|\begin{matrix}-5 & 0 & 0\\0 & 11 & - \frac{121}{2}\\0 & - \frac{121}{2} & 22\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 5 & 0\\0 & 11 - \lambda\end{matrix}\right|$$
|-5 0 | | 11 -121/2|
K2 = | | + | |
|0 22| |-121/2 22 |$$I_{1} = 6$$
$$I_{2} = -55$$
$$I_{3} = \frac{68365}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 6 \lambda - 55$$
$$K_{2} = - \frac{14113}{4}$$
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
$$\lambda^{2} - 6 \lambda - 55 = 0$$
$$\lambda_{1} = 11$$
$$\lambda_{2} = -5$$
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
$$11 \tilde x^{2} - 5 \tilde y^{2} - \frac{1243}{4} = 0$$
$$\frac{\tilde x^{2}}{\frac{113}{4}} - \frac{\tilde y^{2}}{\frac{1243}{20}} = 1$$
- reduced to canonical form