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- 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:
- 4𝑥1𝑥3−6𝑥2𝑥4
- x^2+y^2-6x+4y-12=0
- (5y-2x)(5y+2x)-2(75y+16x-30)=-1
- 3x^2+4xy+3z^2+8xz+8yz=0
- Plot:
- z=3/sqrt(x^2+y-9)
- (9,27-(x+y+z))*x*y*z=9,27
-
x²+(y-x⅔)²=1
-
4x^2+4y^2+8x*y+4x+4y+1==0
- Integral of d{x}:
-
-1
- Derivative of:
-
-1
- Limit of the function:
- -1
-
Identical expressions
- (5y- two x)(5y+2x)-2(75y+ one 6x- thirty)=-1
- (5y minus 2x)(5y plus 2x) minus 2(75y plus 16x minus 30) equally minus 1
- (5y minus two x)(5y plus 2x) minus 2(75y plus one 6x minus thirty) equally minus 1
- 5y-2x5y+2x-275y+16x-30=-1
-
Similar expressions
- (5y+2x)(5y+2x)-2(75y+16x-30)=-1
- (5y-2x)(5y+2x)+2(75y+16x-30)=-1
- (5y-2x)(5y-2x)-2(75y+16x-30)=-1
- (5y-2x)(5y+2x)-2(75y+16x-30)=+1
- (5y-2x)(5y+2x)-2(75y+16x+30)=-1
- (5y-2x)(5y+2x)-2(75y-16x-30)=-1
(5y-2x)(5y+2x)-2(75y+16x-30)=-1 canonical form
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61 - 150*y - 32*x + (-2*x + 5*y)*(2*x + 5*y) = 0
$$- 32 x - 150 y + \left(- 2 x + 5 y\right) \left(2 x + 5 y\right) + 61 = 0$$
-32*x - 150*y + (-2*x + 5*y)*(2*x + 5*y) + 61 = 0
Detail solution
Given line equation of 2-order:
$$- 32 x - 150 y + \left(- 2 x + 5 y\right) \left(2 x + 5 y\right) + 61 = 0$$
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
$$a_{11} = -4$$
$$a_{12} = 0$$
$$a_{13} = -16$$
$$a_{22} = 25$$
$$a_{23} = -75$$
$$a_{33} = 61$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-4 & 0\\0 & 25\end{matrix}\right|$$
$$\Delta = -100$$
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
$$- 4 x_{0} - 16 = 0$$
$$25 y_{0} - 75 = 0$$
then
$$x_{0} = -4$$
$$y_{0} = 3$$
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
$$a'_{33} = - 16 x_{0} - 75 y_{0} + 61$$
$$a'_{33} = -100$$
then equation turns into
$$- 4 x'^{2} + 25 y'^{2} - 100 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{25} - \frac{\tilde y^{2}}{4} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$- 32 x - 150 y + \left(- 2 x + 5 y\right) \left(2 x + 5 y\right) + 61 = 0$$
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
$$a_{11} = -4$$
$$a_{12} = 0$$
$$a_{13} = -16$$
$$a_{22} = 25$$
$$a_{23} = -75$$
$$a_{33} = 61$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-4 & 0\\0 & 25\end{matrix}\right|$$
$$\Delta = -100$$
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
$$- 4 x_{0} - 16 = 0$$
$$25 y_{0} - 75 = 0$$
then
$$x_{0} = -4$$
$$y_{0} = 3$$
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
$$a'_{33} = - 16 x_{0} - 75 y_{0} + 61$$
$$a'_{33} = -100$$
then equation turns into
$$- 4 x'^{2} + 25 y'^{2} - 100 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{25} - \frac{\tilde y^{2}}{4} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(-4, 3)
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:
$$- 32 x - 150 y + \left(- 2 x + 5 y\right) \left(2 x + 5 y\right) + 61 = 0$$
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
$$a_{11} = -4$$
$$a_{12} = 0$$
$$a_{13} = -16$$
$$a_{22} = 25$$
$$a_{23} = -75$$
$$a_{33} = 61$$
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
$$I_{1} = 21$$
$$I_{3} = \left|\begin{matrix}-4 & 0 & -16\\0 & 25 & -75\\-16 & -75 & 61\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 4 & 0\\0 & 25 - \lambda\end{matrix}\right|$$
$$I_{1} = 21$$
$$I_{2} = -100$$
$$I_{3} = 10000$$
$$I{\left(\lambda \right)} = \lambda^{2} - 21 \lambda - 100$$
$$K_{2} = -4600$$
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
$$\lambda^{2} - 21 \lambda - 100 = 0$$
$$\lambda_{1} = 25$$
$$\lambda_{2} = -4$$
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
$$25 \tilde x^{2} - 4 \tilde y^{2} - 100 = 0$$
$$\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{25} = 1$$
- reduced to canonical form
$$- 32 x - 150 y + \left(- 2 x + 5 y\right) \left(2 x + 5 y\right) + 61 = 0$$
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
$$a_{11} = -4$$
$$a_{12} = 0$$
$$a_{13} = -16$$
$$a_{22} = 25$$
$$a_{23} = -75$$
$$a_{33} = 61$$
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
$$I_{1} = 21$$
|-4 0 |
I2 = | |
|0 25|$$I_{3} = \left|\begin{matrix}-4 & 0 & -16\\0 & 25 & -75\\-16 & -75 & 61\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 4 & 0\\0 & 25 - \lambda\end{matrix}\right|$$
|-4 -16| |25 -75|
K2 = | | + | |
|-16 61 | |-75 61 |$$I_{1} = 21$$
$$I_{2} = -100$$
$$I_{3} = 10000$$
$$I{\left(\lambda \right)} = \lambda^{2} - 21 \lambda - 100$$
$$K_{2} = -4600$$
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
$$\lambda^{2} - 21 \lambda - 100 = 0$$
$$\lambda_{1} = 25$$
$$\lambda_{2} = -4$$
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
$$25 \tilde x^{2} - 4 \tilde y^{2} - 100 = 0$$
$$\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{25} = 1$$
- reduced to canonical form