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- Canonical form of a parabola
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How to use it?
- Canonical form:
- xy+2x+y+5/2=0
- 4x^2-9y^2-24x+54y-189=0
- 4x^2+25y^2-8x+200y+304=0
- x²+y²-6x+2y-15=0
- Plot:
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x^2+(y+(x)^(2/3))^2=1
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tg(x)^arctg(x)=e^((y(x))/x)
- x^2+(y-x^2/3)^2=1
-
-x^2*y^3+(x^2+y^2-1)^3=0
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Identical expressions
- 9y^ two -8x-54y+ seventeen = zero
- 9y squared minus 8x minus 54y plus 17 equally 0
- 9y to the power of two minus 8x minus 54y plus seventeen equally zero
- 9y2-8x-54y+17=0
- 9y²-8x-54y+17=0
- 9y to the power of 2-8x-54y+17=0
- 9y^2-8x-54y+17=O
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Similar expressions
- 9y^2-8x+54y+17=0
- 9y^2+8x-54y+17=0
- 9y^2-8x-54y-17=0
9y^2-8x-54y+17=0 canonical form
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The solution
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2 17 - 54*y - 8*x + 9*y = 0
$$- 8 x + 9 y^{2} - 54 y + 17 = 0$$
-8*x + 9*y^2 - 54*y + 17 = 0
Detail solution
Given line equation of 2-order:
$$- 8 x + 9 y^{2} - 54 y + 17 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = 9$$
$$a_{23} = -27$$
$$a_{33} = 17$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 9\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\left(3 \tilde y - 9\right)^{2} = 8 \tilde x + 64$$
$$\left(\tilde y - 3\right)^{2} = \frac{8 \tilde x}{9} + \frac{64}{9}$$
$$\tilde y'^{2} = \frac{8 \tilde x}{9} + \frac{64}{9}$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$y_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$- 8 x + 9 y^{2} - 54 y + 17 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = 9$$
$$a_{23} = -27$$
$$a_{33} = 17$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 9\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\left(3 \tilde y - 9\right)^{2} = 8 \tilde x + 64$$
$$\left(\tilde y - 3\right)^{2} = \frac{8 \tilde x}{9} + \frac{64}{9}$$
$$\tilde y'^{2} = \frac{8 \tilde x}{9} + \frac{64}{9}$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$y_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$y_{0} = 0$$
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:
$$- 8 x + 9 y^{2} - 54 y + 17 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = 9$$
$$a_{23} = -27$$
$$a_{33} = 17$$
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
$$I_{1} = 9$$
$$I_{3} = \left|\begin{matrix}0 & 0 & -4\\0 & 9 & -27\\-4 & -27 & 17\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 9 - \lambda\end{matrix}\right|$$
$$I_{1} = 9$$
$$I_{2} = 0$$
$$I_{3} = -144$$
$$I{\left(\lambda \right)} = \lambda^{2} - 9 \lambda$$
$$K_{2} = -592$$
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
$$8 \tilde x + 9 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{8 \tilde x}{9}$$
- reduced to canonical form
$$- 8 x + 9 y^{2} - 54 y + 17 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = -4$$
$$a_{22} = 9$$
$$a_{23} = -27$$
$$a_{33} = 17$$
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
$$I_{1} = 9$$
|0 0|
I2 = | |
|0 9|$$I_{3} = \left|\begin{matrix}0 & 0 & -4\\0 & 9 & -27\\-4 & -27 & 17\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 9 - \lambda\end{matrix}\right|$$
|0 -4| | 9 -27|
K2 = | | + | |
|-4 17| |-27 17 |$$I_{1} = 9$$
$$I_{2} = 0$$
$$I_{3} = -144$$
$$I{\left(\lambda \right)} = \lambda^{2} - 9 \lambda$$
$$K_{2} = -592$$
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
$$8 \tilde x + 9 \tilde y^{2} = 0$$
$$\tilde y^{2} = \frac{8 \tilde x}{9}$$
- reduced to canonical form