Mister Exam
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How to use it?
- Canonical form:
- 32y-2(1+8y)^2
- sin5x
- x1x2+x1x3−x2x3
- factorial(x+1)
- Plot:
-
x^2+x*y^2=1
- z=3/sqrt(x^2+y-9)
-
sin(x^e)+cos(y^e)=1
-
x^2+4y^2
-
Identical expressions
- 3 two y- two (one +8y)^2
- 32y minus 2(1 plus 8y) squared
- 3 two y minus two (one plus 8y) squared
- 32y-2(1+8y)2
- 32y-21+8y2
- 32y-2(1+8y)²
- 32y-2(1+8y) to the power of 2
- 32y-21+8y^2
-
Similar expressions
- 32y-2(1-8y)^2
- 32y+2(1+8y)^2
32y-2(1+8y)^2 canonical form
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The solution
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2 - 2*(1 + 8*y) + 32*y = 0
$$32 y - 2 \left(8 y + 1\right)^{2} = 0$$
32*y - 2*(8*y + 1)^2 = 0
Detail solution
Given line equation of 2-order:
$$32 y - 2 \left(8 y + 1\right)^{2} = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -128$$
$$a_{23} = 0$$
$$a_{33} = -2$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & -128\end{matrix}\right|$$
$$\Delta = 0$$
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
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$32 y - 2 \left(8 y + 1\right)^{2} = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -128$$
$$a_{23} = 0$$
$$a_{33} = -2$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & -128\end{matrix}\right|$$
$$\Delta = 0$$
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:
$$32 y - 2 \left(8 y + 1\right)^{2} = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -128$$
$$a_{23} = 0$$
$$a_{33} = -2$$
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
$$I_{1} = -128$$
$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & -128 & 0\\0 & 0 & -2\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & - \lambda - 128\end{matrix}\right|$$
$$I_{1} = -128$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + 128 \lambda$$
$$K_{2} = 256$$
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 imaginary parallel lines
$$I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
$$- 128 \tilde y^{2} - 2 = 0$$
- reduced to canonical form
$$32 y - 2 \left(8 y + 1\right)^{2} = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -128$$
$$a_{23} = 0$$
$$a_{33} = -2$$
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
$$I_{1} = -128$$
|0 0 |
I2 = | |
|0 -128|$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & -128 & 0\\0 & 0 & -2\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & - \lambda - 128\end{matrix}\right|$$
|0 0 | |-128 0 |
K2 = | | + | |
|0 -2| | 0 -2|$$I_{1} = -128$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + 128 \lambda$$
$$K_{2} = 256$$
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 imaginary parallel lines
$$I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
$$- 128 \tilde y^{2} - 2 = 0$$
None
- reduced to canonical form