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
-
How to use it?
- Canonical form:
- 2x^2+2y^2+2yz+2z^2-1=0
- 3x^2+3y^2-18x-18y+51=0
- 25x^2+9y^2-9z^2=225
- 4x^2-4xy+y^2-14x-8y-2=0
- Plot:
- y=1/3*x^3-2*x^2
- x^4/3-y^4/3=1
-
x^2+(y-pow(x*x,1/3))^2=1
- x^2*(y-x^2/3)^2=1
- Graphing y =:
- -x^2+6x-5
- Integral of d{x}:
-
-x^2+6x-5
- Derivative of:
- -x^2+6x-5
-
Identical expressions
- -x^ two +6x- five
- minus x squared plus 6x minus 5
- minus x to the power of two plus 6x minus five
- -x2+6x-5
- -x²+6x-5
- -x to the power of 2+6x-5
-
Similar expressions
- -x^2+6x+5
- -x^2-6x-5
- x^2+6x-5
-x^2+6x-5 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$- x^{2} + 6 x - 5 = 0$$
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
$$a_{11} = -1$$
$$a_{12} = 0$$
$$a_{13} = 3$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = -5$$
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
Given equation is straight line
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$y_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$- x^{2} + 6 x - 5 = 0$$
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
$$a_{11} = -1$$
$$a_{12} = 0$$
$$a_{13} = 3$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = -5$$
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
Given equation is straight line
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$y_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$y_{0} = 0$$
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:
$$- x^{2} + 6 x - 5 = 0$$
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
$$a_{11} = -1$$
$$a_{12} = 0$$
$$a_{13} = 3$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = -5$$
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
$$I_{1} = -1$$
$$I_{3} = \left|\begin{matrix}-1 & 0 & 3\\0 & 0 & 0\\3 & 0 & -5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 1 & 0\\0 & - \lambda\end{matrix}\right|$$
$$I_{1} = -1$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + \lambda$$
$$K_{2} = -4$$
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
$$4 - \tilde y^{2} = 0$$
- reduced to canonical form
$$- x^{2} + 6 x - 5 = 0$$
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
$$a_{11} = -1$$
$$a_{12} = 0$$
$$a_{13} = 3$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = -5$$
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|$$I_{3} = \left|\begin{matrix}-1 & 0 & 3\\0 & 0 & 0\\3 & 0 & -5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 1 & 0\\0 & - \lambda\end{matrix}\right|$$
|-1 3 | |0 0 |
K2 = | | + | |
|3 -5| |0 -5|$$I_{1} = -1$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + \lambda$$
$$K_{2} = -4$$
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
$$4 - \tilde y^{2} = 0$$
None
- reduced to canonical form