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:
- sqrt(x+1)-11/(10*x)=0
- 24xy+24y+8xy2 +8y2
- 6x1^2+5x2^2+7x3^2-4x1x2+4x1x3=0
- y(x^2+1)
- Plot:
- 27*x=(3*y+x)^3
- z=/sqrt(3*x+2*y+2)
- y^2=4*x
-
x^4*y-x*y^3=-2
-
Identical expressions
- (x^ two)/ sixteen +(y^ two)/ four -(z^ two)/ thirty-six = one
- (x squared ) divide by 16 plus (y squared ) divide by 4 minus (z squared ) divide by 36 equally 1
- (x to the power of two) divide by sixteen plus (y to the power of two) divide by four minus (z to the power of two) divide by thirty minus six equally one
- (x2)/16+(y2)/4-(z2)/36=1
- x2/16+y2/4-z2/36=1
- (x²)/16+(y²)/4-(z²)/36=1
- (x to the power of 2)/16+(y to the power of 2)/4-(z to the power of 2)/36=1
- x^2/16+y^2/4-z^2/36=1
- (x^2) divide by 16+(y^2) divide by 4-(z^2) divide by 36=1
-
Similar expressions
- (x^2)/16-(y^2)/4-(z^2)/36=1
- (x^2)/16+(y^2)/4+(z^2)/36=1
(x^2)/16+(y^2)/4-(z^2)/36=1 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
2 2 2
z y x
-1 - -- + -- + -- = 0
36 4 16
$$\frac{x^{2}}{16} + \frac{y^{2}}{4} - \frac{z^{2}}{36} - 1 = 0$$
x^2/16 + y^2/4 - z^2/36 - 1 = 0
Invariants method
Given equation of the surface of 2-order:
$$\frac{x^{2}}{16} + \frac{y^{2}}{4} - \frac{z^{2}}{36} - 1 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + a_{22} y^{2} + 2 a_{23} y z + a_{33} z^{2} + 2 a_{14} x + 2 a_{24} y + 2 a_{34} z + a_{44} = 0$$
where
$$a_{11} = \frac{1}{16}$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = \frac{1}{4}$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = - \frac{1}{36}$$
$$a_{34} = 0$$
$$a_{44} = -1$$
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
$$I_{1} = \frac{41}{144}$$
$$I_{3} = \left|\begin{matrix}\frac{1}{16} & 0 & 0\\0 & \frac{1}{4} & 0\\0 & 0 & - \frac{1}{36}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}\frac{1}{16} & 0 & 0 & 0\\0 & \frac{1}{4} & 0 & 0\\0 & 0 & - \frac{1}{36} & 0\\0 & 0 & 0 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + \frac{1}{16} & 0 & 0\\0 & - \lambda + \frac{1}{4} & 0\\0 & 0 & - \lambda - \frac{1}{36}\end{matrix}\right|$$
$$I_{1} = \frac{41}{144}$$
$$I_{2} = \frac{1}{144}$$
$$I_{3} = - \frac{1}{2304}$$
$$I_{4} = \frac{1}{2304}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{41 \lambda^{2}}{144} - \frac{\lambda}{144} - \frac{1}{2304}$$
$$K_{2} = - \frac{41}{144}$$
$$K_{3} = - \frac{1}{144}$$
Because
then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + \lambda^{3} + I_{2} \lambda - I_{3} = 0$$
or
$$\lambda^{3} - \frac{41 \lambda^{2}}{144} + \frac{\lambda}{144} + \frac{1}{2304} = 0$$
Solve this equation
$$\lambda_{1} = \frac{1}{4}$$
$$\lambda_{2} = \frac{1}{16}$$
$$\lambda_{3} = - \frac{1}{36}$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\frac{\tilde x^{2}}{4} + \frac{\tilde y^{2}}{16} - \frac{\tilde z^{2}}{36} - 1 = 0$$
$$- \frac{\tilde z^{2}}{6^{2}} + \left(\frac{\tilde x^{2}}{2^{2}} + \frac{\tilde y^{2}}{4^{2}}\right) = 1$$
this equation is fora type one-sided hyperboloid
- reduced to canonical form
$$\frac{x^{2}}{16} + \frac{y^{2}}{4} - \frac{z^{2}}{36} - 1 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + a_{22} y^{2} + 2 a_{23} y z + a_{33} z^{2} + 2 a_{14} x + 2 a_{24} y + 2 a_{34} z + a_{44} = 0$$
where
$$a_{11} = \frac{1}{16}$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = \frac{1}{4}$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = - \frac{1}{36}$$
$$a_{34} = 0$$
$$a_{44} = -1$$
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
$$I_{1} = \frac{41}{144}$$
|1/16 0 | |1/4 0 | |1/16 0 |
I2 = | | + | | + | |
| 0 1/4| | 0 -1/36| | 0 -1/36|$$I_{3} = \left|\begin{matrix}\frac{1}{16} & 0 & 0\\0 & \frac{1}{4} & 0\\0 & 0 & - \frac{1}{36}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}\frac{1}{16} & 0 & 0 & 0\\0 & \frac{1}{4} & 0 & 0\\0 & 0 & - \frac{1}{36} & 0\\0 & 0 & 0 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + \frac{1}{16} & 0 & 0\\0 & - \lambda + \frac{1}{4} & 0\\0 & 0 & - \lambda - \frac{1}{36}\end{matrix}\right|$$
|1/16 0 | |1/4 0 | |-1/36 0 |
K2 = | | + | | + | |
| 0 -1| | 0 -1| | 0 -1| |1/16 0 0 | |1/4 0 0 | |1/16 0 0 |
| | | | | |
K3 = | 0 1/4 0 | + | 0 -1/36 0 | + | 0 -1/36 0 |
| | | | | |
| 0 0 -1| | 0 0 -1| | 0 0 -1|$$I_{1} = \frac{41}{144}$$
$$I_{2} = \frac{1}{144}$$
$$I_{3} = - \frac{1}{2304}$$
$$I_{4} = \frac{1}{2304}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{41 \lambda^{2}}{144} - \frac{\lambda}{144} - \frac{1}{2304}$$
$$K_{2} = - \frac{41}{144}$$
$$K_{3} = - \frac{1}{144}$$
Because
I3 != 0
then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + \lambda^{3} + I_{2} \lambda - I_{3} = 0$$
or
$$\lambda^{3} - \frac{41 \lambda^{2}}{144} + \frac{\lambda}{144} + \frac{1}{2304} = 0$$
Solve this equation
$$\lambda_{1} = \frac{1}{4}$$
$$\lambda_{2} = \frac{1}{16}$$
$$\lambda_{3} = - \frac{1}{36}$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\frac{\tilde x^{2}}{4} + \frac{\tilde y^{2}}{16} - \frac{\tilde z^{2}}{36} - 1 = 0$$
$$- \frac{\tilde z^{2}}{6^{2}} + \left(\frac{\tilde x^{2}}{2^{2}} + \frac{\tilde y^{2}}{4^{2}}\right) = 1$$
this equation is fora type one-sided hyperboloid
- reduced to canonical form