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How to use it?
- Canonical form:
-
9x^2-25y^2-18x-100y-316=0
- ax+by+cz+d=0
- -x1^2-2x2^2-2x1x2-2x1x3
- x+y
- Plot:
- f*(x)=100-x*y-x*y*z
-
x^2+4y^2
- tan(sin(e−1x))−x=0
- (9,27-(x+y+z))*x*y*z=9,27
-
Identical expressions
- 9x^ two - two 5y^2-18x-1 zero 0y- three hundred and sixteen =0
- 9x squared minus 25y squared minus 18x minus 100y minus 316 equally 0
- 9x to the power of two minus two 5y squared minus 18x minus 1 zero 0y minus three hundred and sixteen equally 0
- 9x2-25y2-18x-100y-316=0
- 9x²-25y²-18x-100y-316=0
- 9x to the power of 2-25y to the power of 2-18x-100y-316=0
- 9x^2-25y^2-18x-100y-316=O
-
Similar expressions
- 9x^2-25y^2+18x-100y-316=0
- 9x^2-25y^2-18x+100y-316=0
- 9x^2-25y^2-18x-100y+316=0
- 9x^2+25y^2-18x-100y-316=0
9x^2-25y^2-18x-100y-316=0 canonical form
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The solution
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2 2 -316 - 100*y - 25*y - 18*x + 9*x = 0
$$9 x^{2} - 25 y^{2} - 18 x - 100 y - 316 = 0$$
9*x^2 - 18*x - 25*y^2 - 100*y - 316 = 0
Detail solution
Given line equation of 2-order:
$$9 x^{2} - 25 y^{2} - 18 x - 100 y - 316 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -9$$
$$a_{22} = -25$$
$$a_{23} = -50$$
$$a_{33} = -316$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 0\\0 & -25\end{matrix}\right|$$
$$\Delta = -225$$
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
$$9 x_{0} - 9 = 0$$
$$- 25 y_{0} - 50 = 0$$
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
$$a'_{33} = - 9 x_{0} - 50 y_{0} - 316$$
$$a'_{33} = -225$$
then The equation is transformed to
$$9 x'^{2} - 25 y'^{2} - 225 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{25} - \frac{\tilde y^{2}}{9} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$9 x^{2} - 25 y^{2} - 18 x - 100 y - 316 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -9$$
$$a_{22} = -25$$
$$a_{23} = -50$$
$$a_{33} = -316$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 0\\0 & -25\end{matrix}\right|$$
$$\Delta = -225$$
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
$$9 x_{0} - 9 = 0$$
$$- 25 y_{0} - 50 = 0$$
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
$$a'_{33} = - 9 x_{0} - 50 y_{0} - 316$$
$$a'_{33} = -225$$
then The equation is transformed to
$$9 x'^{2} - 25 y'^{2} - 225 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{25} - \frac{\tilde y^{2}}{9} = 1$$
- 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:
$$9 x^{2} - 25 y^{2} - 18 x - 100 y - 316 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -9$$
$$a_{22} = -25$$
$$a_{23} = -50$$
$$a_{33} = -316$$
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
$$I_{1} = -16$$
$$I_{3} = \left|\begin{matrix}9 & 0 & -9\\0 & -25 & -50\\-9 & -50 & -316\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 9 & 0\\0 & - \lambda - 25\end{matrix}\right|$$
$$I_{1} = -16$$
$$I_{2} = -225$$
$$I_{3} = 50625$$
$$I{\left(\lambda \right)} = \lambda^{2} + 16 \lambda - 225$$
$$K_{2} = 2475$$
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
$$\lambda^{2} + 16 \lambda - 225 = 0$$
Solve this equation
$$\lambda_{1} = 9$$
$$\lambda_{2} = -25$$
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
$$9 \tilde x^{2} - 25 \tilde y^{2} - 225 = 0$$
$$\frac{\tilde x^{2}}{25} - \frac{\tilde y^{2}}{9} = 1$$
- reduced to canonical form
$$9 x^{2} - 25 y^{2} - 18 x - 100 y - 316 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = -9$$
$$a_{22} = -25$$
$$a_{23} = -50$$
$$a_{33} = -316$$
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
$$I_{1} = -16$$
|9 0 |
I2 = | |
|0 -25|$$I_{3} = \left|\begin{matrix}9 & 0 & -9\\0 & -25 & -50\\-9 & -50 & -316\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 9 & 0\\0 & - \lambda - 25\end{matrix}\right|$$
|9 -9 | |-25 -50 |
K2 = | | + | |
|-9 -316| |-50 -316|$$I_{1} = -16$$
$$I_{2} = -225$$
$$I_{3} = 50625$$
$$I{\left(\lambda \right)} = \lambda^{2} + 16 \lambda - 225$$
$$K_{2} = 2475$$
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
$$\lambda^{2} + 16 \lambda - 225 = 0$$
Solve this equation
$$\lambda_{1} = 9$$
$$\lambda_{2} = -25$$
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
$$9 \tilde x^{2} - 25 \tilde y^{2} - 225 = 0$$
$$\frac{\tilde x^{2}}{25} - \frac{\tilde y^{2}}{9} = 1$$
- reduced to canonical form
The graph