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:
- (x-4)*(x+2)-(x-5)*(x-6)
- ((x-1)^2/49)+((y+2)^2/16)=1
- 9x^2-4y^2-18x-16y-7=0
- 7x^2-2xy+7y^2-48x-48y+144=0
- Plot:
-
x^2+(x/y)-y^2=1
- x^(sin(y^(2)))+y^(cos(x))=1
- x^2+y^2+z^-x*z-y*z+2*x+2*y+2*z-2=0
-
x=(x^2-y)^(1/2)
- Derivative of:
-
x^2-5x+3
-
Identical expressions
- x^ two -5x+ three
- x squared minus 5x plus 3
- x to the power of two minus 5x plus three
- x2-5x+3
- x²-5x+3
- x to the power of 2-5x+3
-
Similar expressions
- x^2-5x-3
- x^2+5x+3
x^2-5x+3 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$x^{2} - 5 x + 3 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = - \frac{5}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 3$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\left(\tilde x - \frac{5}{2}\right)^{2} = \frac{13}{4}$$
$$\tilde x'^{2} = \frac{13}{4}$$
Given equation is two parallel straight lines
- reduced to canonical form
where replacement made
$$\tilde x' = \tilde x - \frac{5}{2}$$
$$\tilde y' = \tilde y$$
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0 + \frac{5}{2}$$
$$y_{0} = \frac{0 \cdot 5}{2}$$
$$x_{0} = \frac{5}{2}$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$x^{2} - 5 x + 3 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = - \frac{5}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 3$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\left(\tilde x - \frac{5}{2}\right)^{2} = \frac{13}{4}$$
$$\tilde x'^{2} = \frac{13}{4}$$
Given equation is two parallel straight lines
- reduced to canonical form
where replacement made
$$\tilde x' = \tilde x - \frac{5}{2}$$
$$\tilde y' = \tilde y$$
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0 + \frac{5}{2}$$
$$y_{0} = \frac{0 \cdot 5}{2}$$
$$x_{0} = \frac{5}{2}$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
(5/2, 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:
$$x^{2} - 5 x + 3 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = - \frac{5}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 3$$
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
$$I_{1} = 1$$
$$I_{3} = \left|\begin{matrix}1 & 0 & - \frac{5}{2}\\0 & 0 & 0\\- \frac{5}{2} & 0 & 3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda$$
$$K_{2} = - \frac{13}{4}$$
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
$$\tilde y^{2} - \frac{13}{4} = 0$$
- reduced to canonical form
$$x^{2} - 5 x + 3 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = - \frac{5}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 3$$
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
$$I_{1} = 1$$
|1 0|
I2 = | |
|0 0|$$I_{3} = \left|\begin{matrix}1 & 0 & - \frac{5}{2}\\0 & 0 & 0\\- \frac{5}{2} & 0 & 3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
| 1 -5/2| |0 0|
K2 = | | + | |
|-5/2 3 | |0 3|$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda$$
$$K_{2} = - \frac{13}{4}$$
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
$$\tilde y^{2} - \frac{13}{4} = 0$$
None
- reduced to canonical form