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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-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
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tan(y)^2=tan(x)
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x^2+x*y^2=1
- atan(y)=-atan(x)+3
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Identical expressions
- 9x^ two -36y^ two +4z^ two -18x+72y+16z- forty-seven = zero
- 9x squared minus 36y squared plus 4z squared minus 18x plus 72y plus 16z minus 47 equally 0
- 9x to the power of two minus 36y to the power of two plus 4z to the power of two minus 18x plus 72y plus 16z minus forty minus seven equally zero
- 9x2-36y2+4z2-18x+72y+16z-47=0
- 9x²-36y²+4z²-18x+72y+16z-47=0
- 9x to the power of 2-36y to the power of 2+4z to the power of 2-18x+72y+16z-47=0
- 9x^2-36y^2+4z^2-18x+72y+16z-47=O
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Similar expressions
- 9x^2-36y^2+4z^2-18x+72y-16z-47=0
- 9x^2-36y^2-4z^2-18x+72y+16z-47=0
- 9x^2-36y^2+4z^2+18x+72y+16z-47=0
- 9x^2-36y^2+4z^2-18x-72y+16z-47=0
- 9x^2-36y^2+4z^2-18x+72y+16z+47=0
- 9x^2+36y^2+4z^2-18x+72y+16z-47=0
9x^2-36y^2+4z^2-18x+72y+16z-47=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
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2 2 2 -47 - 36*y - 18*x + 4*z + 9*x + 16*z + 72*y = 0
$$9 x^{2} - 18 x - 36 y^{2} + 72 y + 4 z^{2} + 16 z - 47 = 0$$
9*x^2 - 18*x - 36*y^2 + 72*y + 4*z^2 + 16*z - 47 = 0
Invariants method
Given equation of the surface of 2-order:
$$9 x^{2} - 18 x - 36 y^{2} + 72 y + 4 z^{2} + 16 z - 47 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0$$
where
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = -9$$
$$a_{22} = -36$$
$$a_{23} = 0$$
$$a_{24} = 36$$
$$a_{33} = 4$$
$$a_{34} = 8$$
$$a_{44} = -47$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$
substitute coefficients
$$I_{1} = -23$$
$$I_{3} = \left|\begin{matrix}9 & 0 & 0\\0 & -36 & 0\\0 & 0 & 4\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}9 & 0 & 0 & -9\\0 & -36 & 0 & 36\\0 & 0 & 4 & 8\\-9 & 36 & 8 & -47\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 0 & 0\\0 & - \lambda - 36 & 0\\0 & 0 & 4 - \lambda\end{matrix}\right|$$
$$I_{1} = -23$$
$$I_{2} = -432$$
$$I_{3} = -1296$$
$$I_{4} = 46656$$
$$I{\left(\lambda \right)} = - \lambda^{3} - 23 \lambda^{2} + 432 \lambda - 1296$$
$$K_{2} = -360$$
$$K_{3} = 7776$$
Because
then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0$$
or
$$\lambda^{3} + 23 \lambda^{2} - 432 \lambda + 1296 = 0$$
Solve this equation
$$\lambda_{1} = 9$$
$$\lambda_{2} = 4$$
$$\lambda_{3} = -36$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$9 \tilde x^{2} + 4 \tilde y^{2} - 36 \tilde z^{2} - 36 = 0$$
$$- \frac{\tilde z^{2}}{1^{2}} + \left(\frac{\tilde x^{2}}{2^{2}} + \frac{\tilde y^{2}}{3^{2}}\right) = 1$$
this equation is fora type one-sided hyperboloid
- reduced to canonical form
$$9 x^{2} - 18 x - 36 y^{2} + 72 y + 4 z^{2} + 16 z - 47 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0$$
where
$$a_{11} = 9$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = -9$$
$$a_{22} = -36$$
$$a_{23} = 0$$
$$a_{24} = 36$$
$$a_{33} = 4$$
$$a_{34} = 8$$
$$a_{44} = -47$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
|a11 a12| |a22 a23| |a11 a13|
I2 = | | + | | + | |
|a12 a22| |a23 a33| |a13 a33|$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$
|a11 a14| |a22 a24| |a33 a34|
K2 = | | + | | + | |
|a14 a44| |a24 a44| |a34 a44| |a11 a12 a14| |a22 a23 a24| |a11 a13 a14|
| | | | | |
K3 = |a12 a22 a24| + |a23 a33 a34| + |a13 a33 a34|
| | | | | |
|a14 a24 a44| |a24 a34 a44| |a14 a34 a44|substitute coefficients
$$I_{1} = -23$$
|9 0 | |-36 0| |9 0|
I2 = | | + | | + | |
|0 -36| | 0 4| |0 4|$$I_{3} = \left|\begin{matrix}9 & 0 & 0\\0 & -36 & 0\\0 & 0 & 4\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}9 & 0 & 0 & -9\\0 & -36 & 0 & 36\\0 & 0 & 4 & 8\\-9 & 36 & 8 & -47\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 0 & 0\\0 & - \lambda - 36 & 0\\0 & 0 & 4 - \lambda\end{matrix}\right|$$
|9 -9 | |-36 36 | |4 8 |
K2 = | | + | | + | |
|-9 -47| |36 -47| |8 -47| |9 0 -9 | |-36 0 36 | |9 0 -9 |
| | | | | |
K3 = |0 -36 36 | + | 0 4 8 | + |0 4 8 |
| | | | | |
|-9 36 -47| |36 8 -47| |-9 8 -47|$$I_{1} = -23$$
$$I_{2} = -432$$
$$I_{3} = -1296$$
$$I_{4} = 46656$$
$$I{\left(\lambda \right)} = - \lambda^{3} - 23 \lambda^{2} + 432 \lambda - 1296$$
$$K_{2} = -360$$
$$K_{3} = 7776$$
Because
I3 != 0
then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0$$
or
$$\lambda^{3} + 23 \lambda^{2} - 432 \lambda + 1296 = 0$$
Solve this equation
$$\lambda_{1} = 9$$
$$\lambda_{2} = 4$$
$$\lambda_{3} = -36$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$9 \tilde x^{2} + 4 \tilde y^{2} - 36 \tilde z^{2} - 36 = 0$$
$$- \frac{\tilde z^{2}}{1^{2}} + \left(\frac{\tilde x^{2}}{2^{2}} + \frac{\tilde y^{2}}{3^{2}}\right) = 1$$
this equation is fora type one-sided hyperboloid
- reduced to canonical form