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:
- x^2-8x-9
- 10x^2+12xy+y^2-104x-52y+263=0
- 9x^(2)+6y^(2)+4xy+40x+20y-50=0
- (x-1)^2/25-y^2/4=1
- Plot:
- (x^(2)+y^(2)-1)^(3)-x^(2)*y^(3)=0
- y=x^3+8*y^3-6*x*y+5
- y-x=2
-
y/(3*10^8)*1.6*10^(-19)/(6.63*10^(-34))*2*3.14*sqrt((246/25-8281/(100*y^2))*1/((246/25-8281/(100*y^2))+1))=x
-
Identical expressions
- (2x- three)(five -x)+ five =y
- (2x minus 3)(5 minus x) plus 5 equally y
- (2x minus three)(five minus x) plus five equally y
- 2x-35-x+5=y
-
Similar expressions
- (2x+3)(5-x)+5=y
- (2x-3)(5+x)+5=y
- (2x-3)(5-x)-5=y
(2x-3)(5-x)+5=y canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
5 - y + (-3 + 2*x)*(5 - x) = 0
$$- y + \left(5 - x\right) \left(2 x - 3\right) + 5 = 0$$
-y + (5 - x)*(2*x - 3) + 5 = 0
Detail solution
Given line equation of 2-order:
$$- y + \left(5 - x\right) \left(2 x - 3\right) + 5 = 0$$
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
$$a_{11} = -2$$
$$a_{12} = 0$$
$$a_{13} = \frac{13}{2}$$
$$a_{22} = 0$$
$$a_{23} = - \frac{1}{2}$$
$$a_{33} = -10$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-2 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
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
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$- y + \left(5 - x\right) \left(2 x - 3\right) + 5 = 0$$
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
$$a_{11} = -2$$
$$a_{12} = 0$$
$$a_{13} = \frac{13}{2}$$
$$a_{22} = 0$$
$$a_{23} = - \frac{1}{2}$$
$$a_{33} = -10$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-2 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
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:
$$- y + \left(5 - x\right) \left(2 x - 3\right) + 5 = 0$$
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
$$a_{11} = -2$$
$$a_{12} = 0$$
$$a_{13} = \frac{13}{2}$$
$$a_{22} = 0$$
$$a_{23} = - \frac{1}{2}$$
$$a_{33} = -10$$
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
$$I_{1} = -2$$
$$I_{3} = \left|\begin{matrix}-2 & 0 & \frac{13}{2}\\0 & 0 & - \frac{1}{2}\\\frac{13}{2} & - \frac{1}{2} & -10\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 2 & 0\\0 & - \lambda\end{matrix}\right|$$
$$I_{1} = -2$$
$$I_{2} = 0$$
$$I_{3} = \frac{1}{2}$$
$$I{\left(\lambda \right)} = \lambda^{2} + 2 \lambda$$
$$K_{2} = - \frac{45}{2}$$
Because
$$I_{2} = 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : parabola
$$I_{1} \tilde y^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0$$
or
$$\tilde x - 2 \tilde y^{2} = 0$$
$$\tilde y^{2} = - \frac{\tilde x}{2}$$
- reduced to canonical form
$$- y + \left(5 - x\right) \left(2 x - 3\right) + 5 = 0$$
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
$$a_{11} = -2$$
$$a_{12} = 0$$
$$a_{13} = \frac{13}{2}$$
$$a_{22} = 0$$
$$a_{23} = - \frac{1}{2}$$
$$a_{33} = -10$$
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
$$I_{1} = -2$$
|-2 0|
I2 = | |
|0 0|$$I_{3} = \left|\begin{matrix}-2 & 0 & \frac{13}{2}\\0 & 0 & - \frac{1}{2}\\\frac{13}{2} & - \frac{1}{2} & -10\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 2 & 0\\0 & - \lambda\end{matrix}\right|$$
| -2 13/2| | 0 -1/2|
K2 = | | + | |
|13/2 -10 | |-1/2 -10 |$$I_{1} = -2$$
$$I_{2} = 0$$
$$I_{3} = \frac{1}{2}$$
$$I{\left(\lambda \right)} = \lambda^{2} + 2 \lambda$$
$$K_{2} = - \frac{45}{2}$$
Because
$$I_{2} = 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : parabola
$$I_{1} \tilde y^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0$$
or
$$\tilde x - 2 \tilde y^{2} = 0$$
$$\tilde y^{2} = - \frac{\tilde x}{2}$$
- reduced to canonical form