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- Canonical form of a elliptical paraboloid
- Canonical form of a double hyperboloid
- Canonical form of a imaginary ellipsoid
- Canonical form of a degenerate ellipse
- Canonical form of a parabola
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How to use it?
- Canonical form:
-
4*x-y^2+2*y+7=0
- sqrt(x+1)-11/(10*x)=0
- 111.91x1-42.24x2-42.24x3=593.596
- 16x^2-24xy+9y^2-160x+120y+425=0
- Plot:
- x^2+y^2=1
- x^2+3*y=5
- ((1-cos(x^(2)+y^(2))))/((x^(2)+y^(2))x^(2)y^(2))=z
-
xy=x^2=10
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Identical expressions
- 16x^ two - two 4xy+9y^2-16 zero x+120y+ four hundred and twenty-five =0
- 16x squared minus 24xy plus 9y squared minus 160x plus 120y plus 425 equally 0
- 16x to the power of two minus two 4xy plus 9y squared minus 16 zero x plus 120y plus four hundred and twenty minus five equally 0
- 16x2-24xy+9y2-160x+120y+425=0
- 16x²-24xy+9y²-160x+120y+425=0
- 16x to the power of 2-24xy+9y to the power of 2-160x+120y+425=0
- 16x^2-24xy+9y^2-160x+120y+425=O
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Similar expressions
- 16x^2-24xy-9y^2-160x+120y+425=0
- 16x^2-24xy+9y^2-160x-120y+425=0
- 16x^2-24xy+9y^2+160x+120y+425=0
- 16x^2+24xy+9y^2-160x+120y+425=0
- 16x^2-24xy+9y^2-160x+120y-425=0
16x^2-24xy+9y^2-160x+120y+425=0 canonical form
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The solution
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2 2 425 - 160*x + 9*y + 16*x + 120*y - 24*x*y = 0
$$16 x^{2} - 24 x y - 160 x + 9 y^{2} + 120 y + 425 = 0$$
16*x^2 - 24*x*y - 160*x + 9*y^2 + 120*y + 425 = 0
Detail solution
Given line equation of 2-order:
$$16 x^{2} - 24 x y - 160 x + 9 y^{2} + 120 y + 425 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = -12$$
$$a_{13} = -80$$
$$a_{22} = 9$$
$$a_{23} = 60$$
$$a_{33} = 425$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}16 & -12\\-12 & 9\end{matrix}\right|$$
$$\Delta = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{7}{24}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{7}{24} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{24}{25}$$
$$\cos{\left(2 \phi \right)} = \frac{7}{25}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{4}{5}$$
$$\sin{\left(\phi \right)} = - \frac{3}{5}$$
substitute coefficients
$$x' = \frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}$$
$$y' = - \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}$$
then the equation turns from
$$16 x'^{2} - 24 x' y' - 160 x' + 9 y'^{2} + 120 y' + 425 = 0$$
to
$$9 \left(- \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right)^{2} - 24 \left(- \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) \left(\frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}\right) + 120 \left(- \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) + 16 \left(\frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}\right)^{2} - 160 \left(\frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}\right) + 425 = 0$$
simplify
$$25 \tilde x^{2} - 200 \tilde x + 425 = 0$$
$$\tilde x^{2} - 8 \tilde x = -17$$
$$\left(\tilde x - 4\right)^{2} = 16$$
$$\tilde x'^{2} = 16$$
Given equation is two parallel straight lines
- reduced to canonical form
where replacement made
$$\tilde x' = \tilde x - 4$$
$$\tilde y' = \tilde y$$
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = \frac{\left(-3\right) 0}{5} + \frac{4 \cdot 4}{5}$$
$$y_{0} = \frac{\left(-3\right) 4}{5} + \frac{0 \cdot 4}{5}$$
$$x_{0} = \frac{16}{5}$$
$$y_{0} = - \frac{12}{5}$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{4}{5}, \ - \frac{3}{5}\right)$$
$$\vec e_2 = \left( \frac{3}{5}, \ \frac{4}{5}\right)$$
$$16 x^{2} - 24 x y - 160 x + 9 y^{2} + 120 y + 425 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = -12$$
$$a_{13} = -80$$
$$a_{22} = 9$$
$$a_{23} = 60$$
$$a_{33} = 425$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}16 & -12\\-12 & 9\end{matrix}\right|$$
$$\Delta = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{7}{24}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{7}{24} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{24}{25}$$
$$\cos{\left(2 \phi \right)} = \frac{7}{25}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{4}{5}$$
$$\sin{\left(\phi \right)} = - \frac{3}{5}$$
substitute coefficients
$$x' = \frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}$$
$$y' = - \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}$$
then the equation turns from
$$16 x'^{2} - 24 x' y' - 160 x' + 9 y'^{2} + 120 y' + 425 = 0$$
to
$$9 \left(- \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right)^{2} - 24 \left(- \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) \left(\frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}\right) + 120 \left(- \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) + 16 \left(\frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}\right)^{2} - 160 \left(\frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}\right) + 425 = 0$$
simplify
$$25 \tilde x^{2} - 200 \tilde x + 425 = 0$$
$$\tilde x^{2} - 8 \tilde x = -17$$
$$\left(\tilde x - 4\right)^{2} = 16$$
$$\tilde x'^{2} = 16$$
Given equation is two parallel straight lines
- reduced to canonical form
where replacement made
$$\tilde x' = \tilde x - 4$$
$$\tilde y' = \tilde y$$
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = \frac{\left(-3\right) 0}{5} + \frac{4 \cdot 4}{5}$$
$$y_{0} = \frac{\left(-3\right) 4}{5} + \frac{0 \cdot 4}{5}$$
$$x_{0} = \frac{16}{5}$$
$$y_{0} = - \frac{12}{5}$$
The center of canonical coordinate system at point O
(16/5, -12/5)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{4}{5}, \ - \frac{3}{5}\right)$$
$$\vec e_2 = \left( \frac{3}{5}, \ \frac{4}{5}\right)$$
Invariants method
Given line equation of 2-order:
$$16 x^{2} - 24 x y - 160 x + 9 y^{2} + 120 y + 425 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = -12$$
$$a_{13} = -80$$
$$a_{22} = 9$$
$$a_{23} = 60$$
$$a_{33} = 425$$
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
$$I_{1} = 25$$
$$I_{3} = \left|\begin{matrix}16 & -12 & -80\\-12 & 9 & 60\\-80 & 60 & 425\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}16 - \lambda & -12\\-12 & 9 - \lambda\end{matrix}\right|$$
$$I_{1} = 25$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 25 \lambda$$
$$K_{2} = 625$$
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 imaginary parallel lines
$$I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
$$25 \tilde y^{2} + 25 = 0$$
- reduced to canonical form
$$16 x^{2} - 24 x y - 160 x + 9 y^{2} + 120 y + 425 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = -12$$
$$a_{13} = -80$$
$$a_{22} = 9$$
$$a_{23} = 60$$
$$a_{33} = 425$$
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
$$I_{1} = 25$$
|16 -12|
I2 = | |
|-12 9 |$$I_{3} = \left|\begin{matrix}16 & -12 & -80\\-12 & 9 & 60\\-80 & 60 & 425\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}16 - \lambda & -12\\-12 & 9 - \lambda\end{matrix}\right|$$
|16 -80| |9 60 |
K2 = | | + | |
|-80 425| |60 425|$$I_{1} = 25$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 25 \lambda$$
$$K_{2} = 625$$
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 imaginary parallel lines
$$I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
$$25 \tilde y^{2} + 25 = 0$$
None
- reduced to canonical form