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How to use it?
- Canonical form:
-
25x^2-16y^2+150x+128y-431=0
- x^2-4xy+4y^2-20x+10y-50=0
- x^2=2y^2+3z^2
- y=-7+2/5*sqrt(16+6x-x^2)
- Plot:
-
xy=4
-
(x^2+y^2-1)^3-(x^2)*y^3=0
- xyz
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Identical expressions
- two 5x^ two -16y^2+15 zero x+128y- four hundred and thirty-one =0
- 25x squared minus 16y squared plus 150x plus 128y minus 431 equally 0
- two 5x to the power of two minus 16y squared plus 15 zero x plus 128y minus four hundred and thirty minus one equally 0
- 25x2-16y2+150x+128y-431=0
- 25x²-16y²+150x+128y-431=0
- 25x to the power of 2-16y to the power of 2+150x+128y-431=0
- 25x^2-16y^2+150x+128y-431=O
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Similar expressions
- 25x^2-16y^2-150x+128y-431=0
- 25x^2+16y^2+150x+128y-431=0
- 25x^2-16y^2+150x+128y+431=0
- 25x^2-16y^2+150x-128y-431=0
25x^2-16y^2+150x+128y-431=0 canonical form
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The solution
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2 2 -431 - 16*y + 25*x + 128*y + 150*x = 0
$$25 x^{2} + 150 x - 16 y^{2} + 128 y - 431 = 0$$
25*x^2 + 150*x - 16*y^2 + 128*y - 431 = 0
Detail solution
Given line equation of 2-order:
$$25 x^{2} + 150 x - 16 y^{2} + 128 y - 431 = 0$$
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
$$a_{11} = 25$$
$$a_{12} = 0$$
$$a_{13} = 75$$
$$a_{22} = -16$$
$$a_{23} = 64$$
$$a_{33} = -431$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}25 & 0\\0 & -16\end{matrix}\right|$$
$$\Delta = -400$$
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
$$25 x_{0} + 75 = 0$$
$$64 - 16 y_{0} = 0$$
then
$$x_{0} = -3$$
$$y_{0} = 4$$
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
$$a'_{33} = 75 x_{0} + 64 y_{0} - 431$$
$$a'_{33} = -400$$
then equation turns into
$$25 x'^{2} - 16 y'^{2} - 400 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{16} - \frac{\tilde y^{2}}{25} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$25 x^{2} + 150 x - 16 y^{2} + 128 y - 431 = 0$$
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
$$a_{11} = 25$$
$$a_{12} = 0$$
$$a_{13} = 75$$
$$a_{22} = -16$$
$$a_{23} = 64$$
$$a_{33} = -431$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}25 & 0\\0 & -16\end{matrix}\right|$$
$$\Delta = -400$$
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
$$25 x_{0} + 75 = 0$$
$$64 - 16 y_{0} = 0$$
then
$$x_{0} = -3$$
$$y_{0} = 4$$
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
$$a'_{33} = 75 x_{0} + 64 y_{0} - 431$$
$$a'_{33} = -400$$
then equation turns into
$$25 x'^{2} - 16 y'^{2} - 400 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{16} - \frac{\tilde y^{2}}{25} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(-3, 4)
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:
$$25 x^{2} + 150 x - 16 y^{2} + 128 y - 431 = 0$$
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
$$a_{11} = 25$$
$$a_{12} = 0$$
$$a_{13} = 75$$
$$a_{22} = -16$$
$$a_{23} = 64$$
$$a_{33} = -431$$
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
$$I_{1} = 9$$
$$I_{3} = \left|\begin{matrix}25 & 0 & 75\\0 & -16 & 64\\75 & 64 & -431\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}25 - \lambda & 0\\0 & - \lambda - 16\end{matrix}\right|$$
$$I_{1} = 9$$
$$I_{2} = -400$$
$$I_{3} = 160000$$
$$I{\left(\lambda \right)} = \lambda^{2} - 9 \lambda - 400$$
$$K_{2} = -13600$$
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
$$\lambda^{2} - 9 \lambda - 400 = 0$$
Solve this equation
$$\lambda_{1} = 25$$
$$\lambda_{2} = -16$$
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
$$25 \tilde x^{2} - 16 \tilde y^{2} - 400 = 0$$
$$\frac{\tilde x^{2}}{16} - \frac{\tilde y^{2}}{25} = 1$$
- reduced to canonical form
$$25 x^{2} + 150 x - 16 y^{2} + 128 y - 431 = 0$$
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
$$a_{11} = 25$$
$$a_{12} = 0$$
$$a_{13} = 75$$
$$a_{22} = -16$$
$$a_{23} = 64$$
$$a_{33} = -431$$
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
$$I_{1} = 9$$
|25 0 |
I2 = | |
|0 -16|$$I_{3} = \left|\begin{matrix}25 & 0 & 75\\0 & -16 & 64\\75 & 64 & -431\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}25 - \lambda & 0\\0 & - \lambda - 16\end{matrix}\right|$$
|25 75 | |-16 64 |
K2 = | | + | |
|75 -431| |64 -431|$$I_{1} = 9$$
$$I_{2} = -400$$
$$I_{3} = 160000$$
$$I{\left(\lambda \right)} = \lambda^{2} - 9 \lambda - 400$$
$$K_{2} = -13600$$
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
$$\lambda^{2} - 9 \lambda - 400 = 0$$
Solve this equation
$$\lambda_{1} = 25$$
$$\lambda_{2} = -16$$
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
$$25 \tilde x^{2} - 16 \tilde y^{2} - 400 = 0$$
$$\frac{\tilde x^{2}}{16} - \frac{\tilde y^{2}}{25} = 1$$
- reduced to canonical form
The graph