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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:
- 5x^2+4xy+8y^2+8x+z+14y+5=0
- 3x^2-3.46xy+5y^2=6
- 𝒇(𝒙)=𝟏𝟐(𝒙−𝟏)^𝟐−𝟐
-
k=4x^2+2sqrt(15)xx+2x^2
- Plot:
- x^2=-y^2+1
-
log(sqrt(x)+sqrt(y))=1/2
- x^2+y^2=1
- x^2+3*y=5
-
Identical expressions
- 𝒇(𝒙)= twelve (𝒙− one)^ two − two
- 𝒇(𝒙) equally 𝟏𝟐(𝒙−𝟏) to the power of 𝟐−𝟐
- 𝒇(𝒙) equally twelve (𝒙− one) to the power of two − two
- 𝒇(𝒙)=𝟏𝟐(𝒙−𝟏)𝟐−𝟐
- 𝒇𝒙=𝟏𝟐𝒙−𝟏𝟐−𝟐
- 𝒇𝒙=𝟏𝟐𝒙−𝟏^𝟐−𝟐
𝒇(𝒙)=𝟏𝟐(𝒙−𝟏)^𝟐−𝟐 canonical form
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The solution
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2 2 - 12*(-1 + x) + f*x = 0
$$f x - 12 \left(x - 1\right)^{2} + 2 = 0$$
f*x - 12*(x - 1)^2 + 2 = 0
Detail solution
Given line equation of 2-order:
$$f x - 12 \left(x - 1\right)^{2} + 2 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} f x + 2 a_{13} x + a_{22} f^{2} + 2 a_{23} f + a_{33} = 0$$
where
$$a_{11} = -12$$
$$a_{12} = \frac{1}{2}$$
$$a_{13} = 12$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = -10$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-12 & \frac{1}{2}\\\frac{1}{2} & 0\end{matrix}\right|$$
$$\Delta = - \frac{1}{4}$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} f_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} f_{0} + a_{23} = 0$$
substitute coefficients
$$\frac{f_{0}}{2} - 12 x_{0} + 12 = 0$$
$$\frac{x_{0}}{2} = 0$$
then
$$x_{0} = 0$$
$$f_{0} = -24$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} f' x' + a_{22} f'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} f_{0} + a_{33}$$
or
$$a'_{33} = 12 x_{0} - 10$$
$$a'_{33} = -10$$
then equation turns into
$$f' x' - 12 x'^{2} - 10 = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = - \tilde f \sin{\left(\phi \right)} + \tilde x \cos{\left(\phi \right)}$$
$$f' = \tilde f \cos{\left(\phi \right)} + \tilde x \sin{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = -12$$
then
$$\phi = - \frac{\operatorname{acot}{\left(12 \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{\sqrt{145}}{145}$$
$$\cos{\left(2 \phi \right)} = \frac{12 \sqrt{145}}{145}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}}$$
substitute coefficients
$$x' = \tilde f \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}} + \tilde x \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}}$$
$$f' = \tilde f \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}} - \tilde x \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}}$$
then the equation turns from
$$f' x' - 12 x'^{2} - 10 = 0$$
to
$$- 12 \left(\tilde f \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}} + \tilde x \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}}\right)^{2} + \left(\tilde f \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}} + \tilde x \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}}\right) \left(\tilde f \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}} - \tilde x \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}}\right) - 10 = 0$$
simplify
$$- 6 \tilde f^{2} + \tilde f^{2} \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}} \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}} + \frac{72 \sqrt{145} \tilde f^{2}}{145} - 24 \tilde f \tilde x \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}} \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}} + \frac{12 \sqrt{145} \tilde f \tilde x}{145} - 6 \tilde x^{2} - \frac{72 \sqrt{145} \tilde x^{2}}{145} - \tilde x^{2} \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}} \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}} - 10 = 0$$
$$- \frac{\sqrt{145} \tilde f^{2}}{2} + 6 \tilde f^{2} + 6 \tilde x^{2} + \frac{\sqrt{145} \tilde x^{2}}{2} + 10 = 0$$
Given equation is hyperbole
$$- \frac{\tilde f^{2}}{10 \frac{1}{-6 + \frac{\sqrt{145}}{2}}} + \frac{\tilde x^{2}}{10 \frac{1}{6 + \frac{\sqrt{145}}{2}}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}}\right)$$
$$\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}}, \ \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}}\right)$$
$$f x - 12 \left(x - 1\right)^{2} + 2 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} f x + 2 a_{13} x + a_{22} f^{2} + 2 a_{23} f + a_{33} = 0$$
where
$$a_{11} = -12$$
$$a_{12} = \frac{1}{2}$$
$$a_{13} = 12$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = -10$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-12 & \frac{1}{2}\\\frac{1}{2} & 0\end{matrix}\right|$$
$$\Delta = - \frac{1}{4}$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} f_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} f_{0} + a_{23} = 0$$
substitute coefficients
$$\frac{f_{0}}{2} - 12 x_{0} + 12 = 0$$
$$\frac{x_{0}}{2} = 0$$
then
$$x_{0} = 0$$
$$f_{0} = -24$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} f' x' + a_{22} f'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} f_{0} + a_{33}$$
or
$$a'_{33} = 12 x_{0} - 10$$
$$a'_{33} = -10$$
then equation turns into
$$f' x' - 12 x'^{2} - 10 = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = - \tilde f \sin{\left(\phi \right)} + \tilde x \cos{\left(\phi \right)}$$
$$f' = \tilde f \cos{\left(\phi \right)} + \tilde x \sin{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = -12$$
then
$$\phi = - \frac{\operatorname{acot}{\left(12 \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{\sqrt{145}}{145}$$
$$\cos{\left(2 \phi \right)} = \frac{12 \sqrt{145}}{145}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}}$$
substitute coefficients
$$x' = \tilde f \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}} + \tilde x \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}}$$
$$f' = \tilde f \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}} - \tilde x \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}}$$
then the equation turns from
$$f' x' - 12 x'^{2} - 10 = 0$$
to
$$- 12 \left(\tilde f \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}} + \tilde x \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}}\right)^{2} + \left(\tilde f \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}} + \tilde x \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}}\right) \left(\tilde f \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}} - \tilde x \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}}\right) - 10 = 0$$
simplify
$$- 6 \tilde f^{2} + \tilde f^{2} \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}} \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}} + \frac{72 \sqrt{145} \tilde f^{2}}{145} - 24 \tilde f \tilde x \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}} \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}} + \frac{12 \sqrt{145} \tilde f \tilde x}{145} - 6 \tilde x^{2} - \frac{72 \sqrt{145} \tilde x^{2}}{145} - \tilde x^{2} \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}} \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}} - 10 = 0$$
$$- \frac{\sqrt{145} \tilde f^{2}}{2} + 6 \tilde f^{2} + 6 \tilde x^{2} + \frac{\sqrt{145} \tilde x^{2}}{2} + 10 = 0$$
Given equation is hyperbole
$$- \frac{\tilde f^{2}}{10 \frac{1}{-6 + \frac{\sqrt{145}}{2}}} + \frac{\tilde x^{2}}{10 \frac{1}{6 + \frac{\sqrt{145}}{2}}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, -24)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}}\right)$$
$$\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}}, \ \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}}\right)$$
Invariants method
Given line equation of 2-order:
$$f x - 12 \left(x - 1\right)^{2} + 2 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} f x + 2 a_{13} x + a_{22} f^{2} + 2 a_{23} f + a_{33} = 0$$
where
$$a_{11} = -12$$
$$a_{12} = \frac{1}{2}$$
$$a_{13} = 12$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = -10$$
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
$$I_{1} = -12$$
$$I_{3} = \left|\begin{matrix}-12 & \frac{1}{2} & 12\\\frac{1}{2} & 0 & 0\\12 & 0 & -10\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 12 & \frac{1}{2}\\\frac{1}{2} & - \lambda\end{matrix}\right|$$
$$I_{1} = -12$$
$$I_{2} = - \frac{1}{4}$$
$$I_{3} = \frac{5}{2}$$
$$I{\left(\lambda \right)} = \lambda^{2} + 12 \lambda - \frac{1}{4}$$
$$K_{2} = -24$$
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
$$\lambda^{2} + 12 \lambda - \frac{1}{4} = 0$$
$$\lambda_{1} = - \frac{\sqrt{145}}{2} - 6$$
$$\lambda_{2} = -6 + \frac{\sqrt{145}}{2}$$
then the canonical form of the equation will be
$$\tilde f^{2} \lambda_{2} + \tilde x^{2} \lambda_{1} + \frac{I_{3}}{I_{2}} = 0$$
or
$$\tilde f^{2} \left(-6 + \frac{\sqrt{145}}{2}\right) + \tilde x^{2} \left(- \frac{\sqrt{145}}{2} - 6\right) - 10 = 0$$
$$- \frac{\tilde f^{2}}{10 \frac{1}{-6 + \frac{\sqrt{145}}{2}}} + \frac{\tilde x^{2}}{10 \frac{1}{6 + \frac{\sqrt{145}}{2}}} = -1$$
- reduced to canonical form
$$f x - 12 \left(x - 1\right)^{2} + 2 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} f x + 2 a_{13} x + a_{22} f^{2} + 2 a_{23} f + a_{33} = 0$$
where
$$a_{11} = -12$$
$$a_{12} = \frac{1}{2}$$
$$a_{13} = 12$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = -10$$
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
$$I_{1} = -12$$
|-12 1/2|
I2 = | |
|1/2 0 |$$I_{3} = \left|\begin{matrix}-12 & \frac{1}{2} & 12\\\frac{1}{2} & 0 & 0\\12 & 0 & -10\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 12 & \frac{1}{2}\\\frac{1}{2} & - \lambda\end{matrix}\right|$$
|-12 12 | |0 0 |
K2 = | | + | |
|12 -10| |0 -10|$$I_{1} = -12$$
$$I_{2} = - \frac{1}{4}$$
$$I_{3} = \frac{5}{2}$$
$$I{\left(\lambda \right)} = \lambda^{2} + 12 \lambda - \frac{1}{4}$$
$$K_{2} = -24$$
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
$$\lambda^{2} + 12 \lambda - \frac{1}{4} = 0$$
$$\lambda_{1} = - \frac{\sqrt{145}}{2} - 6$$
$$\lambda_{2} = -6 + \frac{\sqrt{145}}{2}$$
then the canonical form of the equation will be
$$\tilde f^{2} \lambda_{2} + \tilde x^{2} \lambda_{1} + \frac{I_{3}}{I_{2}} = 0$$
or
$$\tilde f^{2} \left(-6 + \frac{\sqrt{145}}{2}\right) + \tilde x^{2} \left(- \frac{\sqrt{145}}{2} - 6\right) - 10 = 0$$
$$- \frac{\tilde f^{2}}{10 \frac{1}{-6 + \frac{\sqrt{145}}{2}}} + \frac{\tilde x^{2}}{10 \frac{1}{6 + \frac{\sqrt{145}}{2}}} = -1$$
- reduced to canonical form