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How to use it?
- Canonical form:
- -9x^2+4y^2-72x-8y-176=0
- x^2+26x+y^2+24y+z=12
- x^2+6x+16y^2+4z^2-8z+2=0
- 9x^2+4y^2=36
- Plot:
- u=1,2v
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2*x^3-3*x^2*y+4*y^3=1
- 10*x+5*y+15=0
- -x^3+3*x-2=y
-
Identical expressions
- 9x^ two +1 two xy+4y^2-24x-16y+ three = zero
- 9x squared plus 12xy plus 4y squared minus 24x minus 16y plus 3 equally 0
- 9x to the power of two plus 1 two xy plus 4y squared minus 24x minus 16y plus three equally zero
- 9x2+12xy+4y2-24x-16y+3=0
- 9x²+12xy+4y²-24x-16y+3=0
- 9x to the power of 2+12xy+4y to the power of 2-24x-16y+3=0
- 9x^2+12xy+4y^2-24x-16y+3=O
-
Similar expressions
- 9x^2+12xy+4y^2-24x-16y-3=0
- 9x^2+12xy+4y^2-24x+16y+3=0
- 9x^2+12xy+4y^2+24x-16y+3=0
- 9x^2-12xy+4y^2-24x-16y+3=0
- 9x^2+12xy-4y^2-24x-16y+3=0
9x^2+12xy+4y^2-24x-16y+3=0 canonical form
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The solution
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2 2 3 - 24*x - 16*y + 4*y + 9*x + 12*x*y = 0
$$9 x^{2} + 12 x y - 24 x + 4 y^{2} - 16 y + 3 = 0$$
9*x^2 + 12*x*y - 24*x + 4*y^2 - 16*y + 3 = 0
Detail solution
Given line equation of 2-order:
$$9 x^{2} + 12 x y - 24 x + 4 y^{2} - 16 y + 3 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 6$$
$$a_{13} = -12$$
$$a_{22} = 4$$
$$a_{23} = -8$$
$$a_{33} = 3$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 6\\6 & 4\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
Rotate the resulting coordinate system by an angle φ
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = \frac{5}{12}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{5}{12} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{12}{13}$$
$$\cos{\left(2 \phi \right)} = \frac{5}{13}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{3 \sqrt{13}}{13}$$
$$\sin{\left(\phi \right)} = \frac{2 \sqrt{13}}{13}$$
substitute coefficients
$$x' = \frac{3 \sqrt{13} \tilde x}{13} - \frac{2 \sqrt{13} \tilde y}{13}$$
$$y' = \frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}$$
then the equation turns from
$$9 x'^{2} + 12 x' y' - 24 x' + 4 y'^{2} - 16 y' + 3 = 0$$
to
$$4 \left(\frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}\right)^{2} + 12 \left(\frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}\right) \left(\frac{3 \sqrt{13} \tilde x}{13} - \frac{2 \sqrt{13} \tilde y}{13}\right) - 16 \left(\frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}\right) + 9 \left(\frac{3 \sqrt{13} \tilde x}{13} - \frac{2 \sqrt{13} \tilde y}{13}\right)^{2} - 24 \left(\frac{3 \sqrt{13} \tilde x}{13} - \frac{2 \sqrt{13} \tilde y}{13}\right) + 3 = 0$$
simplify
$$13 \tilde x^{2} - 8 \sqrt{13} \tilde x + 3 = 0$$
$$13 \tilde x^{2} - 8 \sqrt{13} \tilde x = -3$$
$$\left(\sqrt{13} \tilde x - 4\right)^{2} = 16$$
$$\left(\tilde x - \frac{4 \sqrt{13}}{13}\right)^{2} = \frac{16}{13}$$
$$\tilde x'^{2} = \frac{16}{13}$$
Given equation is two parallel straight lines
- reduced to canonical form
where replacement made
$$\tilde x' = \tilde x - \frac{4 \sqrt{13}}{13}$$
$$\tilde y' = \tilde y$$
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \frac{2 \sqrt{13}}{13} + \frac{3 \sqrt{13}}{13} \frac{4 \sqrt{13}}{13}$$
$$y_{0} = 0 \frac{3 \sqrt{13}}{13} + \frac{2 \sqrt{13}}{13} \frac{4 \sqrt{13}}{13}$$
$$x_{0} = \frac{12}{13}$$
$$y_{0} = \frac{8}{13}$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{3 \sqrt{13}}{13}, \ \frac{2 \sqrt{13}}{13}\right)$$
$$\vec e_2 = \left( - \frac{2 \sqrt{13}}{13}, \ \frac{3 \sqrt{13}}{13}\right)$$
$$9 x^{2} + 12 x y - 24 x + 4 y^{2} - 16 y + 3 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 6$$
$$a_{13} = -12$$
$$a_{22} = 4$$
$$a_{23} = -8$$
$$a_{33} = 3$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 6\\6 & 4\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
Rotate the resulting coordinate system by an angle φ
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = \frac{5}{12}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{5}{12} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{12}{13}$$
$$\cos{\left(2 \phi \right)} = \frac{5}{13}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{3 \sqrt{13}}{13}$$
$$\sin{\left(\phi \right)} = \frac{2 \sqrt{13}}{13}$$
substitute coefficients
$$x' = \frac{3 \sqrt{13} \tilde x}{13} - \frac{2 \sqrt{13} \tilde y}{13}$$
$$y' = \frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}$$
then the equation turns from
$$9 x'^{2} + 12 x' y' - 24 x' + 4 y'^{2} - 16 y' + 3 = 0$$
to
$$4 \left(\frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}\right)^{2} + 12 \left(\frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}\right) \left(\frac{3 \sqrt{13} \tilde x}{13} - \frac{2 \sqrt{13} \tilde y}{13}\right) - 16 \left(\frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}\right) + 9 \left(\frac{3 \sqrt{13} \tilde x}{13} - \frac{2 \sqrt{13} \tilde y}{13}\right)^{2} - 24 \left(\frac{3 \sqrt{13} \tilde x}{13} - \frac{2 \sqrt{13} \tilde y}{13}\right) + 3 = 0$$
simplify
$$13 \tilde x^{2} - 8 \sqrt{13} \tilde x + 3 = 0$$
$$13 \tilde x^{2} - 8 \sqrt{13} \tilde x = -3$$
$$\left(\sqrt{13} \tilde x - 4\right)^{2} = 16$$
$$\left(\tilde x - \frac{4 \sqrt{13}}{13}\right)^{2} = \frac{16}{13}$$
$$\tilde x'^{2} = \frac{16}{13}$$
Given equation is two parallel straight lines
- reduced to canonical form
where replacement made
$$\tilde x' = \tilde x - \frac{4 \sqrt{13}}{13}$$
$$\tilde y' = \tilde y$$
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \frac{2 \sqrt{13}}{13} + \frac{3 \sqrt{13}}{13} \frac{4 \sqrt{13}}{13}$$
$$y_{0} = 0 \frac{3 \sqrt{13}}{13} + \frac{2 \sqrt{13}}{13} \frac{4 \sqrt{13}}{13}$$
$$x_{0} = \frac{12}{13}$$
$$y_{0} = \frac{8}{13}$$
The center of canonical coordinate system at point O
12 (--, 8/13) 13
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{3 \sqrt{13}}{13}, \ \frac{2 \sqrt{13}}{13}\right)$$
$$\vec e_2 = \left( - \frac{2 \sqrt{13}}{13}, \ \frac{3 \sqrt{13}}{13}\right)$$
Invariants method
Given line equation of 2-order:
$$9 x^{2} + 12 x y - 24 x + 4 y^{2} - 16 y + 3 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 6$$
$$a_{13} = -12$$
$$a_{22} = 4$$
$$a_{23} = -8$$
$$a_{33} = 3$$
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
$$I_{1} = 13$$
$$I_{3} = \left|\begin{matrix}9 & 6 & -12\\6 & 4 & -8\\-12 & -8 & 3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 6\\6 & 4 - \lambda\end{matrix}\right|$$
$$I_{1} = 13$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 13 \lambda$$
$$K_{2} = -169$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge K_{2} < 0 \wedge I_{1} \neq 0$$
then by line type:
this equation is of type : two parallel lines
$$I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
$$13 \tilde y^{2} - 13 = 0$$
- reduced to canonical form
$$9 x^{2} + 12 x y - 24 x + 4 y^{2} - 16 y + 3 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 6$$
$$a_{13} = -12$$
$$a_{22} = 4$$
$$a_{23} = -8$$
$$a_{33} = 3$$
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
$$I_{1} = 13$$
|9 6|
I2 = | |
|6 4|$$I_{3} = \left|\begin{matrix}9 & 6 & -12\\6 & 4 & -8\\-12 & -8 & 3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 6\\6 & 4 - \lambda\end{matrix}\right|$$
| 9 -12| |4 -8|
K2 = | | + | |
|-12 3 | |-8 3 |$$I_{1} = 13$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 13 \lambda$$
$$K_{2} = -169$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge K_{2} < 0 \wedge I_{1} \neq 0$$
then by line type:
this equation is of type : two parallel lines
$$I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
$$13 \tilde y^{2} - 13 = 0$$
None
- reduced to canonical form