Mister Exam
Other calculators
- 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:
- y^2+z^2+2*x*y+2*y*z+2*z*x+4*x-4*z=0
- f=2x^2+x^2-4xx
- ((x+y+z+t)/2)^2-((x-y-z-t)/2)^2+(z+t)^2-(z-t)^2=0
- x^2-3z^2-2xy+2xz-6yz=0
- Plot:
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(x+1)^2/100+(y-2)^2/0.1=1
- (x^2+y^2+2*a*x)^2-4*a^2*(x^2+y^2)
-
|y-5|=(x-3)^2
- 𝑦+4𝑦+3𝑦+𝑦=𝑢+4𝑢+3�
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Identical expressions
- f= two x^ two +x^2-4xx
- f equally 2x squared plus x squared minus 4xx
- f equally two x to the power of two plus x squared minus 4xx
- f=2x2+x2-4xx
- f=2x²+x²-4xx
- f=2x to the power of 2+x to the power of 2-4xx
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Similar expressions
- f=2x^2+x^2+4xx
- f=2x^2-x^2-4xx
f=2x^2+x^2-4xx canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$f + x^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\tilde x^{2} = - \tilde f$$
$$\tilde x'^{2} = - \tilde f'$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = - \tilde f \sin{\left(\phi \right)} + \tilde x \cos{\left(\phi \right)}$$
$$f_{0} = \tilde f \cos{\left(\phi \right)} + \tilde x \sin{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$f_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$f_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$f + x^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\tilde x^{2} = - \tilde f$$
$$\tilde x'^{2} = - \tilde f'$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = - \tilde f \sin{\left(\phi \right)} + \tilde x \cos{\left(\phi \right)}$$
$$f_{0} = \tilde f \cos{\left(\phi \right)} + \tilde x \sin{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$f_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$f_{0} = 0$$
The center of canonical coordinate system at point O
(0, 0)
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:
$$f + x^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 0$$
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
$$I_{1} = 1$$
$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 0 & \frac{1}{2}\\0 & \frac{1}{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = - \frac{1}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda$$
$$K_{2} = - \frac{1}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : parabola
$$I_{1} \tilde f^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0$$
or
$$\tilde f^{2} + \tilde x = 0$$
$$\tilde f^{2} = \tilde x$$
- reduced to canonical form
$$f + x^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 0$$
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
$$I_{1} = 1$$
|1 0|
I2 = | |
|0 0|$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 0 & \frac{1}{2}\\0 & \frac{1}{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
|1 0| | 0 1/2|
K2 = | | + | |
|0 0| |1/2 0 |$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = - \frac{1}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda$$
$$K_{2} = - \frac{1}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : parabola
$$I_{1} \tilde f^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0$$
or
$$\tilde f^{2} + \tilde x = 0$$
$$\tilde f^{2} = \tilde x$$
- reduced to canonical form