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How to use it?
- Canonical form:
- 16x^2-9y^2+64x+54y+127=0
-
14x^2+24xy+21y^2-4x-18y-139=0
-
9x^2+4y^2-18x+24y+9=0
- x1x2+x1x3+x1x4+x2x3+x2x4+x3x4
- Plot:
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y=x^4-26x^2+25
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xy=0
-
xy=6
-
4x^2+4y^2=25
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Identical expressions
- 16x^ two -9y^ two +64x+54y+ one hundred and twenty-seven = zero
- 16x squared minus 9y squared plus 64x plus 54y plus 127 equally 0
- 16x to the power of two minus 9y to the power of two plus 64x plus 54y plus one hundred and twenty minus seven equally zero
- 16x2-9y2+64x+54y+127=0
- 16x²-9y²+64x+54y+127=0
- 16x to the power of 2-9y to the power of 2+64x+54y+127=0
- 16x^2-9y^2+64x+54y+127=O
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Similar expressions
- 16x^2-9y^2+64x+54y-127=0
- 16x^2+9y^2+64x+54y+127=0
- 16x^2-9y^2-64x+54y+127=0
- 16x^2-9y^2+64x-54y+127=0
16x^2-9y^2+64x+54y+127=0 canonical form
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2 2 127 - 9*y + 16*x + 54*y + 64*x = 0
$$16 x^{2} + 64 x - 9 y^{2} + 54 y + 127 = 0$$
16*x^2 + 64*x - 9*y^2 + 54*y + 127 = 0
Detail solution
Given line equation of 2-order:
$$16 x^{2} + 64 x - 9 y^{2} + 54 y + 127 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = 0$$
$$a_{13} = 32$$
$$a_{22} = -9$$
$$a_{23} = 27$$
$$a_{33} = 127$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}16 & 0\\0 & -9\end{matrix}\right|$$
$$\Delta = -144$$
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
$$16 x_{0} + 32 = 0$$
$$27 - 9 y_{0} = 0$$
then
$$x_{0} = -2$$
$$y_{0} = 3$$
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
$$a'_{33} = 32 x_{0} + 27 y_{0} + 127$$
$$a'_{33} = 144$$
then equation turns into
$$16 x'^{2} - 9 y'^{2} + 144 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{9} - \frac{\tilde y^{2}}{16} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$16 x^{2} + 64 x - 9 y^{2} + 54 y + 127 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = 0$$
$$a_{13} = 32$$
$$a_{22} = -9$$
$$a_{23} = 27$$
$$a_{33} = 127$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}16 & 0\\0 & -9\end{matrix}\right|$$
$$\Delta = -144$$
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
$$16 x_{0} + 32 = 0$$
$$27 - 9 y_{0} = 0$$
then
$$x_{0} = -2$$
$$y_{0} = 3$$
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
$$a'_{33} = 32 x_{0} + 27 y_{0} + 127$$
$$a'_{33} = 144$$
then equation turns into
$$16 x'^{2} - 9 y'^{2} + 144 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{9} - \frac{\tilde y^{2}}{16} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(-2, 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:
$$16 x^{2} + 64 x - 9 y^{2} + 54 y + 127 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = 0$$
$$a_{13} = 32$$
$$a_{22} = -9$$
$$a_{23} = 27$$
$$a_{33} = 127$$
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
$$I_{1} = 7$$
$$I_{3} = \left|\begin{matrix}16 & 0 & 32\\0 & -9 & 27\\32 & 27 & 127\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}16 - \lambda & 0\\0 & - \lambda - 9\end{matrix}\right|$$
$$I_{1} = 7$$
$$I_{2} = -144$$
$$I_{3} = -20736$$
$$I{\left(\lambda \right)} = \lambda^{2} - 7 \lambda - 144$$
$$K_{2} = -864$$
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
$$\lambda^{2} - 7 \lambda - 144 = 0$$
$$\lambda_{1} = 16$$
$$\lambda_{2} = -9$$
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
$$16 \tilde x^{2} - 9 \tilde y^{2} + 144 = 0$$
$$\frac{\tilde x^{2}}{9} - \frac{\tilde y^{2}}{16} = -1$$
- reduced to canonical form
$$16 x^{2} + 64 x - 9 y^{2} + 54 y + 127 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = 0$$
$$a_{13} = 32$$
$$a_{22} = -9$$
$$a_{23} = 27$$
$$a_{33} = 127$$
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
$$I_{1} = 7$$
|16 0 |
I2 = | |
|0 -9|$$I_{3} = \left|\begin{matrix}16 & 0 & 32\\0 & -9 & 27\\32 & 27 & 127\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}16 - \lambda & 0\\0 & - \lambda - 9\end{matrix}\right|$$
|16 32 | |-9 27 |
K2 = | | + | |
|32 127| |27 127|$$I_{1} = 7$$
$$I_{2} = -144$$
$$I_{3} = -20736$$
$$I{\left(\lambda \right)} = \lambda^{2} - 7 \lambda - 144$$
$$K_{2} = -864$$
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
$$\lambda^{2} - 7 \lambda - 144 = 0$$
$$\lambda_{1} = 16$$
$$\lambda_{2} = -9$$
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
$$16 \tilde x^{2} - 9 \tilde y^{2} + 144 = 0$$
$$\frac{\tilde x^{2}}{9} - \frac{\tilde y^{2}}{16} = -1$$
- reduced to canonical form