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- Canonical form of a double hyperboloid
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How to use it?
- Canonical form:
- 25x^2-30xy+9y^2+68x+19=0
- 4x^2-16y^2-8x+96y-204=0
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9x^2-4y^2-18x+16y-43=0
- 5x2+4xy+8y2−32x−56y+80=0
- Plot:
-
sin(x+y)=1.2*x
-
x^2+y^2-1
-
x^2+(y-cbrt(x^2))^2=1
- x^2+(y-(x^2)^(1/3))^2=1
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Identical expressions
- 9x^ two -4y^ two -18x+16y- forty-three = zero
- 9x squared minus 4y squared minus 18x plus 16y minus 43 equally 0
- 9x to the power of two minus 4y to the power of two minus 18x plus 16y minus forty minus three equally zero
- 9x2-4y2-18x+16y-43=0
- 9x²-4y²-18x+16y-43=0
- 9x to the power of 2-4y to the power of 2-18x+16y-43=0
- 9x^2-4y^2-18x+16y-43=O
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Similar expressions
- 9x^2+4y^2-18x+16y-43=0
- 9x^2-4y^2-18x-16y-43=0
- 9x^2-4y^2+18x+16y-43=0
- 9x^2-4y^2-18x+16y+43=0
9x^2-4y^2-18x+16y-43=0 canonical form
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2 2 -43 - 18*x - 4*y + 9*x + 16*y = 0
$$9 x^{2} - 4 y^{2} - 18 x + 16 y - 43 = 0$$
9*x^2 - 18*x - 4*y^2 + 16*y - 43 = 0
Detail solution
Given line equation of 2-order:
$$9 x^{2} - 4 y^{2} - 18 x + 16 y - 43 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -9$$
$$a_{22} = -4$$
$$a_{23} = 8$$
$$a_{33} = -43$$
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
$$9 x_{0} - 9 = 0$$
$$- 4 y_{0} + 8 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = 2$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - 9 x_{0} + 8 y_{0} - 43$$
$$a'_{33} = -36$$
then The equation is transformed to
$$9 x'^{2} - 4 y'^{2} - 36 = 0$$
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
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$9 x^{2} - 4 y^{2} - 18 x + 16 y - 43 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -9$$
$$a_{22} = -4$$
$$a_{23} = 8$$
$$a_{33} = -43$$
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
$$9 x_{0} - 9 = 0$$
$$- 4 y_{0} + 8 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = 2$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - 9 x_{0} + 8 y_{0} - 43$$
$$a'_{33} = -36$$
then The equation is transformed to
$$9 x'^{2} - 4 y'^{2} - 36 = 0$$
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
(1, 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:
$$9 x^{2} - 4 y^{2} - 18 x + 16 y - 43 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -9$$
$$a_{22} = -4$$
$$a_{23} = 8$$
$$a_{33} = -43$$
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
$$I_{1} = 5$$
$$I_{3} = \left|\begin{matrix}9 & 0 & -9\\0 & -4 & 8\\-9 & 8 & -43\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 9 & 0\\0 & - \lambda - 4\end{matrix}\right|$$
$$I_{1} = 5$$
$$I_{2} = -36$$
$$I_{3} = 1296$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda - 36$$
$$K_{2} = -360$$
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:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - 5 \lambda - 36 = 0$$
Solve this equation
$$\lambda_{1} = 9$$
$$\lambda_{2} = -4$$
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
$$9 x^{2} - 4 y^{2} - 18 x + 16 y - 43 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -9$$
$$a_{22} = -4$$
$$a_{23} = 8$$
$$a_{33} = -43$$
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|$$I_{3} = \left|\begin{matrix}9 & 0 & -9\\0 & -4 & 8\\-9 & 8 & -43\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 9 & 0\\0 & - \lambda - 4\end{matrix}\right|$$
|9 -9 | |-4 8 |
K2 = | | + | |
|-9 -43| |8 -43|$$I_{1} = 5$$
$$I_{2} = -36$$
$$I_{3} = 1296$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda - 36$$
$$K_{2} = -360$$
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:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - 5 \lambda - 36 = 0$$
Solve this equation
$$\lambda_{1} = 9$$
$$\lambda_{2} = -4$$
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
The graph