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- Canonical form of a double hyperboloid
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- Canonical form of a degenerate ellipse
- Canonical form of a parabola
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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
-
Identical expressions
- 16x^ two - two 4xy+8y^2+25x- fifty y+5 zero =0
- 16x squared minus 24xy plus 8y squared plus 25x minus 50y plus 50 equally 0
- 16x to the power of two minus two 4xy plus 8y squared plus 25x minus fifty y plus 5 zero equally 0
- 16x2-24xy+8y2+25x-50y+50=0
- 16x²-24xy+8y²+25x-50y+50=0
- 16x to the power of 2-24xy+8y to the power of 2+25x-50y+50=0
- 16x^2-24xy+8y^2+25x-50y+50=O
-
Similar expressions
- 16x^2-24xy+8y^2-25x-50y+50=0
- 16x^2-24xy+8y^2+25x+50y+50=0
- 16x^2-24xy+8y^2+25x-50y-50=0
- 16x^2-24xy-8y^2+25x-50y+50=0
- 16x^2+24xy+8y^2+25x-50y+50=0
16x^2-24xy+8y^2+25x-50y+50=0 canonical form
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The solution
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2 2 50 - 50*y + 8*y + 16*x + 25*x - 24*x*y = 0
$$16 x^{2} - 24 x y + 25 x + 8 y^{2} - 50 y + 50 = 0$$
16*x^2 - 24*x*y + 25*x + 8*y^2 - 50*y + 50 = 0
Detail solution
Given line equation of 2-order:
$$16 x^{2} - 24 x y + 25 x + 8 y^{2} - 50 y + 50 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = -12$$
$$a_{13} = \frac{25}{2}$$
$$a_{22} = 8$$
$$a_{23} = -25$$
$$a_{33} = 50$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}16 & -12\\-12 & 8\end{matrix}\right|$$
$$\Delta = -16$$
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
$$16 x_{0} - 12 y_{0} + \frac{25}{2} = 0$$
$$- 12 x_{0} + 8 y_{0} - 25 = 0$$
then
$$x_{0} = - \frac{25}{2}$$
$$y_{0} = - \frac{125}{8}$$
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
$$a'_{33} = \frac{25 x_{0}}{2} - 25 y_{0} + 50$$
$$a'_{33} = \frac{2275}{8}$$
then equation turns into
$$16 x'^{2} - 24 x' y' + 8 y'^{2} + \frac{2275}{8} = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{1}{3}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{1}{3} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{3 \sqrt{10}}{10}$$
$$\cos{\left(2 \phi \right)} = \frac{\sqrt{10}}{10}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}}$$
$$y' = - \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}} + \tilde y \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}}$$
then the equation turns from
$$16 x'^{2} - 24 x' y' + 8 y'^{2} + \frac{2275}{8} = 0$$
to
$$8 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}} + \tilde y \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}}\right)^{2} - 24 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}} + \tilde y \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}}\right) + 16 \left(\tilde x \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}}\right)^{2} + \frac{2275}{8} = 0$$
simplify
$$\frac{2 \sqrt{10} \tilde x^{2}}{5} + 24 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}} \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}} + 12 \tilde x^{2} - \frac{12 \sqrt{10} \tilde x \tilde y}{5} + 16 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}} \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}} - 24 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}} \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}} - \frac{2 \sqrt{10} \tilde y^{2}}{5} + 12 \tilde y^{2} + \frac{2275}{8} = 0$$
$$12 \tilde x^{2} + 4 \sqrt{10} \tilde x^{2} - 4 \sqrt{10} \tilde y^{2} + 12 \tilde y^{2} + \frac{2275}{8} = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{\frac{2275}{8} \frac{1}{12 + 4 \sqrt{10}}} - \frac{\tilde y^{2}}{\frac{2275}{8} \frac{1}{-12 + 4 \sqrt{10}}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}}\right)$$
$$\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}}, \ \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}}\right)$$
$$16 x^{2} - 24 x y + 25 x + 8 y^{2} - 50 y + 50 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = -12$$
$$a_{13} = \frac{25}{2}$$
$$a_{22} = 8$$
$$a_{23} = -25$$
$$a_{33} = 50$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}16 & -12\\-12 & 8\end{matrix}\right|$$
$$\Delta = -16$$
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
$$16 x_{0} - 12 y_{0} + \frac{25}{2} = 0$$
$$- 12 x_{0} + 8 y_{0} - 25 = 0$$
then
$$x_{0} = - \frac{25}{2}$$
$$y_{0} = - \frac{125}{8}$$
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
$$a'_{33} = \frac{25 x_{0}}{2} - 25 y_{0} + 50$$
$$a'_{33} = \frac{2275}{8}$$
then equation turns into
$$16 x'^{2} - 24 x' y' + 8 y'^{2} + \frac{2275}{8} = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{1}{3}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{1}{3} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{3 \sqrt{10}}{10}$$
$$\cos{\left(2 \phi \right)} = \frac{\sqrt{10}}{10}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}}$$
$$y' = - \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}} + \tilde y \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}}$$
then the equation turns from
$$16 x'^{2} - 24 x' y' + 8 y'^{2} + \frac{2275}{8} = 0$$
to
$$8 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}} + \tilde y \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}}\right)^{2} - 24 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}} + \tilde y \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}}\right) + 16 \left(\tilde x \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}}\right)^{2} + \frac{2275}{8} = 0$$
simplify
$$\frac{2 \sqrt{10} \tilde x^{2}}{5} + 24 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}} \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}} + 12 \tilde x^{2} - \frac{12 \sqrt{10} \tilde x \tilde y}{5} + 16 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}} \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}} - 24 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}} \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}} - \frac{2 \sqrt{10} \tilde y^{2}}{5} + 12 \tilde y^{2} + \frac{2275}{8} = 0$$
$$12 \tilde x^{2} + 4 \sqrt{10} \tilde x^{2} - 4 \sqrt{10} \tilde y^{2} + 12 \tilde y^{2} + \frac{2275}{8} = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{\frac{2275}{8} \frac{1}{12 + 4 \sqrt{10}}} - \frac{\tilde y^{2}}{\frac{2275}{8} \frac{1}{-12 + 4 \sqrt{10}}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(-25/2, -125/8)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}}\right)$$
$$\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{\sqrt{10}}{20}}, \ \sqrt{\frac{\sqrt{10}}{20} + \frac{1}{2}}\right)$$
Invariants method
Given line equation of 2-order:
$$16 x^{2} - 24 x y + 25 x + 8 y^{2} - 50 y + 50 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = -12$$
$$a_{13} = \frac{25}{2}$$
$$a_{22} = 8$$
$$a_{23} = -25$$
$$a_{33} = 50$$
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
$$I_{1} = 24$$
$$I_{3} = \left|\begin{matrix}16 & -12 & \frac{25}{2}\\-12 & 8 & -25\\\frac{25}{2} & -25 & 50\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}16 - \lambda & -12\\-12 & 8 - \lambda\end{matrix}\right|$$
$$I_{1} = 24$$
$$I_{2} = -16$$
$$I_{3} = -4550$$
$$I{\left(\lambda \right)} = \lambda^{2} - 24 \lambda - 16$$
$$K_{2} = \frac{1675}{4}$$
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
$$\lambda^{2} - 24 \lambda - 16 = 0$$
$$\lambda_{1} = 12 - 4 \sqrt{10}$$
$$\lambda_{2} = 12 + 4 \sqrt{10}$$
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
$$\tilde x^{2} \left(12 - 4 \sqrt{10}\right) + \tilde y^{2} \left(12 + 4 \sqrt{10}\right) + \frac{2275}{8} = 0$$
$$\frac{\tilde x^{2}}{\frac{2275}{8} \frac{1}{-12 + 4 \sqrt{10}}} - \frac{\tilde y^{2}}{\frac{2275}{8} \frac{1}{12 + 4 \sqrt{10}}} = 1$$
- reduced to canonical form
$$16 x^{2} - 24 x y + 25 x + 8 y^{2} - 50 y + 50 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = -12$$
$$a_{13} = \frac{25}{2}$$
$$a_{22} = 8$$
$$a_{23} = -25$$
$$a_{33} = 50$$
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
$$I_{1} = 24$$
|16 -12|
I2 = | |
|-12 8 |$$I_{3} = \left|\begin{matrix}16 & -12 & \frac{25}{2}\\-12 & 8 & -25\\\frac{25}{2} & -25 & 50\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}16 - \lambda & -12\\-12 & 8 - \lambda\end{matrix}\right|$$
| 16 25/2| | 8 -25|
K2 = | | + | |
|25/2 50 | |-25 50 |$$I_{1} = 24$$
$$I_{2} = -16$$
$$I_{3} = -4550$$
$$I{\left(\lambda \right)} = \lambda^{2} - 24 \lambda - 16$$
$$K_{2} = \frac{1675}{4}$$
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
$$\lambda^{2} - 24 \lambda - 16 = 0$$
$$\lambda_{1} = 12 - 4 \sqrt{10}$$
$$\lambda_{2} = 12 + 4 \sqrt{10}$$
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
$$\tilde x^{2} \left(12 - 4 \sqrt{10}\right) + \tilde y^{2} \left(12 + 4 \sqrt{10}\right) + \frac{2275}{8} = 0$$
$$\frac{\tilde x^{2}}{\frac{2275}{8} \frac{1}{-12 + 4 \sqrt{10}}} - \frac{\tilde y^{2}}{\frac{2275}{8} \frac{1}{12 + 4 \sqrt{10}}} = 1$$
- reduced to canonical form