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
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How to use it?
- Canonical form:
- sqrt(16-x^2-y^2)=z
- y^2-16x-8y+32=0
- 9x^2-36y^2+4z^2-18x+72y+16z-47=0
- 6x^2+3y^2+3z^2+-4xy-2yz+4xz=0
- Plot:
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tan(y)^2=tan(x)
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-1-(36*(x^2-y^2+5*x+6)-(36*x+132)*(2*x+5))/((x^2-y^2+5*x+6)^2+4*x^2*y+25*y^2+20*x*y^2)=0
- tan(x-y+(1/10))-x*y=0
- tan(x*factorial(y))-1=0
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Identical expressions
- y^ two -16x-8y+ thirty-two = zero
- y squared minus 16x minus 8y plus 32 equally 0
- y to the power of two minus 16x minus 8y plus thirty minus two equally zero
- y2-16x-8y+32=0
- y²-16x-8y+32=0
- y to the power of 2-16x-8y+32=0
- y^2-16x-8y+32=O
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Similar expressions
- y^2-16x-8y-32=0
- y^2-16x+8y+32=0
- y^2+16x-8y+32=0
y^2-16x-8y+32=0 canonical form
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The solution
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2 32 + y - 16*x - 8*y = 0
$$- 16 x + y^{2} - 8 y + 32 = 0$$
-16*x + y^2 - 8*y + 32 = 0
Detail solution
Given line equation of 2-order:
$$- 16 x + y^{2} - 8 y + 32 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = -8$$
$$a_{22} = 1$$
$$a_{23} = -4$$
$$a_{33} = 32$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 1\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\left(\tilde y - 4\right)^{2} = 16 \tilde x - 16$$
$$\tilde y'^{2} = 16 \tilde x - 16$$
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
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$- 16 x + y^{2} - 8 y + 32 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = -8$$
$$a_{22} = 1$$
$$a_{23} = -4$$
$$a_{33} = 32$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 1\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\left(\tilde y - 4\right)^{2} = 16 \tilde x - 16$$
$$\tilde y'^{2} = 16 \tilde x - 16$$
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:
$$- 16 x + y^{2} - 8 y + 32 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = -8$$
$$a_{22} = 1$$
$$a_{23} = -4$$
$$a_{33} = 32$$
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
$$I_{1} = 1$$
$$I_{3} = \left|\begin{matrix}0 & 0 & -8\\0 & 1 & -4\\-8 & -4 & 32\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = -64$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda$$
$$K_{2} = -48$$
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
$$16 \tilde x + \tilde y^{2} = 0$$
$$\tilde y^{2} = 16 \tilde x$$
- reduced to canonical form
$$- 16 x + y^{2} - 8 y + 32 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = -8$$
$$a_{22} = 1$$
$$a_{23} = -4$$
$$a_{33} = 32$$
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
$$I_{1} = 1$$
|0 0|
I2 = | |
|0 1|$$I_{3} = \left|\begin{matrix}0 & 0 & -8\\0 & 1 & -4\\-8 & -4 & 32\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|$$
|0 -8| |1 -4|
K2 = | | + | |
|-8 32| |-4 32|$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = -64$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda$$
$$K_{2} = -48$$
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
$$16 \tilde x + \tilde y^{2} = 0$$
$$\tilde y^{2} = 16 \tilde x$$
- reduced to canonical form