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How to use it?
- Canonical form:
-
4x^2-5y^2+32x+30y-1=0
-
3x^2+10xy+3y^2-2x-14y=13
- 5x^2-2xy+5y^2+10x-2y+1=0
- (5y-2x)(5y+2x)-2(75y+16x-30)=-1
- Plot:
- z=3/sqrt(x^2+y-9)
- tan(sin(e−1x))−x=0
- (9,27-(x+y+z))*x*y*z=9,27
-
4x^2+4y^2+8x*y+4x+4y+1==0
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Identical expressions
- 4x^ two -5y^ two +32x+3 zero y- one =0
- 4x squared minus 5y squared plus 32x plus 30y minus 1 equally 0
- 4x to the power of two minus 5y to the power of two plus 32x plus 3 zero y minus one equally 0
- 4x2-5y2+32x+30y-1=0
- 4x²-5y²+32x+30y-1=0
- 4x to the power of 2-5y to the power of 2+32x+30y-1=0
- 4x^2-5y^2+32x+30y-1=O
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Similar expressions
- 4x^2-5y^2+32x-30y-1=0
- 4x^2+5y^2+32x+30y-1=0
- 4x^2-5y^2+32x+30y+1=0
- 4x^2-5y^2-32x+30y-1=0
4x^2-5y^2+32x+30y-1=0 canonical form
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The solution
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2 2 -1 - 5*y + 4*x + 30*y + 32*x = 0
$$4 x^{2} - 5 y^{2} + 32 x + 30 y - 1 = 0$$
4*x^2 + 32*x - 5*y^2 + 30*y - 1 = 0
Detail solution
Given line equation of 2-order:
$$4 x^{2} - 5 y^{2} + 32 x + 30 y - 1 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 16$$
$$a_{22} = -5$$
$$a_{23} = 15$$
$$a_{33} = -1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & -5\end{matrix}\right|$$
$$\Delta = -20$$
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
$$4 x_{0} + 16 = 0$$
$$- 5 y_{0} + 15 = 0$$
then
$$x_{0} = -4$$
$$y_{0} = 3$$
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
$$a'_{33} = 16 x_{0} + 15 y_{0} - 1$$
$$a'_{33} = -20$$
then The equation is transformed to
$$4 x'^{2} - 5 y'^{2} - 20 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{5} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$4 x^{2} - 5 y^{2} + 32 x + 30 y - 1 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 16$$
$$a_{22} = -5$$
$$a_{23} = 15$$
$$a_{33} = -1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}4 & 0\\0 & -5\end{matrix}\right|$$
$$\Delta = -20$$
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
$$4 x_{0} + 16 = 0$$
$$- 5 y_{0} + 15 = 0$$
then
$$x_{0} = -4$$
$$y_{0} = 3$$
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
$$a'_{33} = 16 x_{0} + 15 y_{0} - 1$$
$$a'_{33} = -20$$
then The equation is transformed to
$$4 x'^{2} - 5 y'^{2} - 20 = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{5} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(-4, 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:
$$4 x^{2} - 5 y^{2} + 32 x + 30 y - 1 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 16$$
$$a_{22} = -5$$
$$a_{23} = 15$$
$$a_{33} = -1$$
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
$$I_{1} = -1$$
$$I_{3} = \left|\begin{matrix}4 & 0 & 16\\0 & -5 & 15\\16 & 15 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 4 & 0\\0 & - \lambda - 5\end{matrix}\right|$$
$$I_{1} = -1$$
$$I_{2} = -20$$
$$I_{3} = 400$$
$$I{\left(\lambda \right)} = \lambda^{2} + \lambda - 20$$
$$K_{2} = -480$$
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
$$\lambda^{2} + \lambda - 20 = 0$$
Solve this equation
$$\lambda_{1} = 4$$
$$\lambda_{2} = -5$$
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
$$4 \tilde x^{2} - 5 \tilde y^{2} - 20 = 0$$
$$\frac{\tilde x^{2}}{5} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
$$4 x^{2} - 5 y^{2} + 32 x + 30 y - 1 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 16$$
$$a_{22} = -5$$
$$a_{23} = 15$$
$$a_{33} = -1$$
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
$$I_{1} = -1$$
|4 0 |
I2 = | |
|0 -5|$$I_{3} = \left|\begin{matrix}4 & 0 & 16\\0 & -5 & 15\\16 & 15 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 4 & 0\\0 & - \lambda - 5\end{matrix}\right|$$
|4 16| |-5 15|
K2 = | | + | |
|16 -1| |15 -1|$$I_{1} = -1$$
$$I_{2} = -20$$
$$I_{3} = 400$$
$$I{\left(\lambda \right)} = \lambda^{2} + \lambda - 20$$
$$K_{2} = -480$$
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
$$\lambda^{2} + \lambda - 20 = 0$$
Solve this equation
$$\lambda_{1} = 4$$
$$\lambda_{2} = -5$$
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
$$4 \tilde x^{2} - 5 \tilde y^{2} - 20 = 0$$
$$\frac{\tilde x^{2}}{5} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
The graph