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:
- 5*x^2+4*y^2-4*z^2-2*x+2*y-4*z-3*y*z-3*x*z-x*y=0
- 8x21+16x1x2+14x1x3−28x2+32x2x3+14x23
- x^2-4*x+9*y^2+18*y-4*z^2-24*z-23=0
- 6y^2-12x+24y-5=0
- Plot:
- 9*(x*cos(a)-y*sin(a))^2+24*(x*cos(a)-y*sin(a))*(y*sin(a)+y*cos(a))+16*(y*sin(a)+y*cos(a))^2+50*(x*cos(a)-y*sin(a))-100*(y*sin(a)+y*cos(a))+25=0
-
y=sin(60*x)
- y*(-x)=x2*e^x
-
x^5+y^5=1
- Sum of series:
-
64
-
Identical expressions
- 16y^ two +4z^ two = sixty-four
- 16y squared plus 4z squared equally 64
- 16y to the power of two plus 4z to the power of two equally sixty minus four
- 16y2+4z2=64
- 16y²+4z²=64
- 16y to the power of 2+4z to the power of 2=64
-
Similar expressions
- 16y^2-4z^2=64
16y^2+4z^2=64 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
2 2 -64 + 4*z + 16*y = 0
$$16 y^{2} + 4 z^{2} - 64 = 0$$
16*y^2 + 4*z^2 - 64 = 0
Invariants method
Given equation of the surface of 2-order:
$$16 y^{2} + 4 z^{2} - 64 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 16$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 4$$
$$a_{34} = 0$$
$$a_{44} = -64$$
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
$$I_{1} = 20$$
$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 16 & 0\\0 & 0 & 4\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & 0\\0 & 16 & 0 & 0\\0 & 0 & 4 & 0\\0 & 0 & 0 & -64\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & 16 - \lambda & 0\\0 & 0 & 4 - \lambda\end{matrix}\right|$$
$$I_{1} = 20$$
$$I_{2} = 64$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 20 \lambda^{2} - 64 \lambda$$
$$K_{2} = -1280$$
$$K_{3} = -4096$$
Because
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
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
$$\lambda^{3} - 20 \lambda^{2} + 64 \lambda = 0$$
$$\lambda_{1} = 16$$
$$\lambda_{2} = 4$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$16 \tilde x^{2} + 4 \tilde y^{2} - 64 = 0$$
this equation is fora type elliptical cylinder
- reduced to canonical form
$$16 y^{2} + 4 z^{2} - 64 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 16$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 4$$
$$a_{34} = 0$$
$$a_{44} = -64$$
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
$$I_{1} = 20$$
|0 0 | |16 0| |0 0|
I2 = | | + | | + | |
|0 16| |0 4| |0 4|$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 16 & 0\\0 & 0 & 4\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & 0\\0 & 16 & 0 & 0\\0 & 0 & 4 & 0\\0 & 0 & 0 & -64\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & 16 - \lambda & 0\\0 & 0 & 4 - \lambda\end{matrix}\right|$$
|0 0 | |16 0 | |4 0 |
K2 = | | + | | + | |
|0 -64| |0 -64| |0 -64| |0 0 0 | |16 0 0 | |0 0 0 |
| | | | | |
K3 = |0 16 0 | + |0 4 0 | + |0 4 0 |
| | | | | |
|0 0 -64| |0 0 -64| |0 0 -64|$$I_{1} = 20$$
$$I_{2} = 64$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 20 \lambda^{2} - 64 \lambda$$
$$K_{2} = -1280$$
$$K_{3} = -4096$$
Because
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
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
$$\lambda^{3} - 20 \lambda^{2} + 64 \lambda = 0$$
$$\lambda_{1} = 16$$
$$\lambda_{2} = 4$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$16 \tilde x^{2} + 4 \tilde y^{2} - 64 = 0$$
2 2
\tilde x \tilde y
--------- + --------- = 1
2 2
/ 1 \ / 1 \
|-----| |-----|
\4*1/8/ \2*1/8/ this equation is fora type elliptical cylinder
- reduced to canonical form