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:
- 4x^2+8x-16y+16=0
- x^2+y^2-2x+1=0
- 5x^2–6y^2–20x–36y–64=0
- 6*x-7*y-4*z-2
- Plot:
- z=x^2+y^2+3*a*x*y
- z=x^2+2*y^2+x*y-x
- z=x+y
- z=sqrt(x^2+25*y^2-25)
-
Identical expressions
- ((y^ two +2y)/ four)-x= zero
- ((y squared plus 2y) divide by 4) minus x equally 0
- ((y to the power of two plus 2y) divide by four) minus x equally zero
- ((y2+2y)/4)-x=0
- y2+2y/4-x=0
- ((y²+2y)/4)-x=0
- ((y to the power of 2+2y)/4)-x=0
- y^2+2y/4-x=0
- ((y^2+2y)/4)-x=O
- ((y^2+2y) divide by 4)-x=0
-
Similar expressions
- ((y^2+2y)/4)+x=0
- ((y^2-2y)/4)-x=0
((y^2+2y)/4)-x=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
2 y y - - x + -- = 0 2 4
$$- x + \frac{y^{2}}{4} + \frac{y}{2} = 0$$
-x + y^2/4 + y/2 = 0
Detail solution
Given line equation of 2-order:
$$- x + \frac{y^{2}}{4} + \frac{y}{2} = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = - \frac{1}{2}$$
$$a_{22} = \frac{1}{4}$$
$$a_{23} = \frac{1}{4}$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & \frac{1}{4}\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\left(\frac{\tilde y}{2} + \frac{1}{2}\right)^{2} = \tilde x + \frac{1}{4}$$
$$\left(\tilde y + 1\right)^{2} = 4 \tilde x + 1$$
$$\tilde y'^{2} = 4 \tilde x + 1$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 1 + 0 \cdot 0$$
$$y_{0} = 0 \cdot 0 + 0 \cdot 1$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$- x + \frac{y^{2}}{4} + \frac{y}{2} = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = - \frac{1}{2}$$
$$a_{22} = \frac{1}{4}$$
$$a_{23} = \frac{1}{4}$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & \frac{1}{4}\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\left(\frac{\tilde y}{2} + \frac{1}{2}\right)^{2} = \tilde x + \frac{1}{4}$$
$$\left(\tilde y + 1\right)^{2} = 4 \tilde x + 1$$
$$\tilde y'^{2} = 4 \tilde x + 1$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 1 + 0 \cdot 0$$
$$y_{0} = 0 \cdot 0 + 0 \cdot 1$$
$$x_{0} = 0$$
$$y_{0} = 0$$
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:
$$- x + \frac{y^{2}}{4} + \frac{y}{2} = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = - \frac{1}{2}$$
$$a_{22} = \frac{1}{4}$$
$$a_{23} = \frac{1}{4}$$
$$a_{33} = 0$$
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
$$I_{1} = \frac{1}{4}$$
$$I_{3} = \left|\begin{matrix}0 & 0 & - \frac{1}{2}\\0 & \frac{1}{4} & \frac{1}{4}\\- \frac{1}{2} & \frac{1}{4} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & \frac{1}{4} - \lambda\end{matrix}\right|$$
$$I_{1} = \frac{1}{4}$$
$$I_{2} = 0$$
$$I_{3} = - \frac{1}{16}$$
$$I{\left(\lambda \right)} = \lambda^{2} - \frac{\lambda}{4}$$
$$K_{2} = - \frac{5}{16}$$
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
$$\tilde x + \frac{\tilde y^{2}}{4} = 0$$
$$\tilde y^{2} = 4 \tilde x$$
- reduced to canonical form
$$- x + \frac{y^{2}}{4} + \frac{y}{2} = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = - \frac{1}{2}$$
$$a_{22} = \frac{1}{4}$$
$$a_{23} = \frac{1}{4}$$
$$a_{33} = 0$$
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
$$I_{1} = \frac{1}{4}$$
|0 0 |
I2 = | |
|0 1/4|$$I_{3} = \left|\begin{matrix}0 & 0 & - \frac{1}{2}\\0 & \frac{1}{4} & \frac{1}{4}\\- \frac{1}{2} & \frac{1}{4} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & \frac{1}{4} - \lambda\end{matrix}\right|$$
| 0 -1/2| |1/4 1/4|
K2 = | | + | |
|-1/2 0 | |1/4 0 |$$I_{1} = \frac{1}{4}$$
$$I_{2} = 0$$
$$I_{3} = - \frac{1}{16}$$
$$I{\left(\lambda \right)} = \lambda^{2} - \frac{\lambda}{4}$$
$$K_{2} = - \frac{5}{16}$$
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
$$\tilde x + \frac{\tilde y^{2}}{4} = 0$$
$$\tilde y^{2} = 4 \tilde x$$
- reduced to canonical form
The graph