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How to use it?
- Canonical form:
- 4x-y-z+12=0
-
x^2-y^2+4x-10y-25=0
- 16x^2-24xy+8y^2+25x-50y+50=0
- 2x^2+3y^2-4x+12y-12z+2=0
- Plot:
-
e^y/x=x*y+1
- z=sqrt(x*y)
- x^2+y^2+10z^2=100
-
|arctg(y/x)|=pi/12
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Identical expressions
- x^ two -y^ two +4x-1 zero y- twenty-five =0
- x squared minus y squared plus 4x minus 10y minus 25 equally 0
- x to the power of two minus y to the power of two plus 4x minus 1 zero y minus twenty minus five equally 0
- x2-y2+4x-10y-25=0
- x²-y²+4x-10y-25=0
- x to the power of 2-y to the power of 2+4x-10y-25=0
- x^2-y^2+4x-10y-25=O
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Similar expressions
- x^2+y^2+4x-10y-25=0
- x^2-y^2+4x-10y+25=0
- x^2-y^2-4x-10y-25=0
- x^2-y^2+4x+10y-25=0
x^2-y^2+4x-10y-25=0 canonical form
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2 2 -25 + x - y - 10*y + 4*x = 0
$$x^{2} + 4 x - y^{2} - 10 y - 25 = 0$$
x^2 + 4*x - y^2 - 10*y - 25 = 0
Detail solution
Given line equation of 2-order:
$$x^{2} + 4 x - y^{2} - 10 y - 25 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 2$$
$$a_{22} = -1$$
$$a_{23} = -5$$
$$a_{33} = -25$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & -1\end{matrix}\right|$$
$$\Delta = -1$$
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
$$x_{0} + 2 = 0$$
$$- y_{0} - 5 = 0$$
then
$$x_{0} = -2$$
$$y_{0} = -5$$
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
$$a'_{33} = 2 x_{0} - 5 y_{0} - 25$$
$$a'_{33} = -4$$
then equation turns into
$$x'^{2} - y'^{2} - 4 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$x^{2} + 4 x - y^{2} - 10 y - 25 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 2$$
$$a_{22} = -1$$
$$a_{23} = -5$$
$$a_{33} = -25$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & -1\end{matrix}\right|$$
$$\Delta = -1$$
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
$$x_{0} + 2 = 0$$
$$- y_{0} - 5 = 0$$
then
$$x_{0} = -2$$
$$y_{0} = -5$$
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
$$a'_{33} = 2 x_{0} - 5 y_{0} - 25$$
$$a'_{33} = -4$$
then equation turns into
$$x'^{2} - y'^{2} - 4 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(-2, -5)
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:
$$x^{2} + 4 x - y^{2} - 10 y - 25 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 2$$
$$a_{22} = -1$$
$$a_{23} = -5$$
$$a_{33} = -25$$
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
$$I_{1} = 0$$
$$I_{3} = \left|\begin{matrix}1 & 0 & 2\\0 & -1 & -5\\2 & -5 & -25\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda - 1\end{matrix}\right|$$
$$I_{1} = 0$$
$$I_{2} = -1$$
$$I_{3} = 4$$
$$I{\left(\lambda \right)} = \lambda^{2} - 1$$
$$K_{2} = -29$$
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
$$\lambda^{2} - 1 = 0$$
Solve this equation
$$\lambda_{1} = -1$$
$$\lambda_{2} = 1$$
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
$$- \tilde x^{2} + \tilde y^{2} - 4 = 0$$
$$\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{4} = -1$$
- reduced to canonical form
$$x^{2} + 4 x - y^{2} - 10 y - 25 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 2$$
$$a_{22} = -1$$
$$a_{23} = -5$$
$$a_{33} = -25$$
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
$$I_{1} = 0$$
|1 0 |
I2 = | |
|0 -1|$$I_{3} = \left|\begin{matrix}1 & 0 & 2\\0 & -1 & -5\\2 & -5 & -25\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda - 1\end{matrix}\right|$$
|1 2 | |-1 -5 |
K2 = | | + | |
|2 -25| |-5 -25|$$I_{1} = 0$$
$$I_{2} = -1$$
$$I_{3} = 4$$
$$I{\left(\lambda \right)} = \lambda^{2} - 1$$
$$K_{2} = -29$$
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
$$\lambda^{2} - 1 = 0$$
Solve this equation
$$\lambda_{1} = -1$$
$$\lambda_{2} = 1$$
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
$$- \tilde x^{2} + \tilde y^{2} - 4 = 0$$
$$\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{4} = -1$$
- reduced to canonical form
The graph