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:
- x^2-2x+10y+21=0
- 3x^2+y^2-12x-2y+4=0
- -x^2+4x-3
- x^2+px+q
- Plot:
- x^2*(5*y/4-sqrt(|x|))^2=1
-
9(1-y)x^2-(3-6y)x+2-3y-|3x+1|=0
-
(-x^2)/9-y^2/4=1
- z=asin(1/(x*y))+acos((x^2+y^2)/4)
-
Identical expressions
- x^ two +px+q
- x squared plus px plus q
- x to the power of two plus px plus q
- x2+px+q
- x²+px+q
- x to the power of 2+px+q
-
Similar expressions
- x^2+px-q
- x^2-px+q
x^2+px+q canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Invariants method
Given equation of the surface of 2-order:
$$p x + q + x^{2} = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} q x + 2 a_{13} p x + 2 a_{14} x + a_{22} q^{2} + 2 a_{23} p q + 2 a_{24} q + a_{33} p^{2} + 2 a_{34} p + a_{44} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = \frac{1}{2}$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = \frac{1}{2}$$
$$a_{33} = 0$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 1$$
$$I_{3} = \left|\begin{matrix}1 & 0 & \frac{1}{2}\\0 & 0 & 0\\\frac{1}{2} & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 0 & \frac{1}{2} & 0\\0 & 0 & 0 & \frac{1}{2}\\\frac{1}{2} & 0 & 0 & 0\\0 & \frac{1}{2} & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0 & \frac{1}{2}\\0 & - \lambda & 0\\\frac{1}{2} & 0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 1$$
$$I_{2} = - \frac{1}{4}$$
$$I_{3} = 0$$
$$I_{4} = \frac{1}{16}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \lambda^{2} + \frac{\lambda}{4}$$
$$K_{2} = - \frac{1}{4}$$
$$K_{3} = - \frac{1}{4}$$
Because
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
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
$$\lambda^{3} - \lambda^{2} - \frac{\lambda}{4} = 0$$
$$\lambda_{1} = \frac{1}{2} - \frac{\sqrt{2}}{2}$$
$$\lambda_{2} = \frac{1}{2} + \frac{\sqrt{2}}{2}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\tilde p 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde q^{2} \lambda_{2} + \tilde x^{2} \lambda_{1}\right) = 0$$
and
$$- \tilde p 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde q^{2} \lambda_{2} + \tilde x^{2} \lambda_{1}\right) = 0$$
$$\tilde p + \tilde q^{2} \left(\frac{1}{2} + \frac{\sqrt{2}}{2}\right) + \tilde x^{2} \left(\frac{1}{2} - \frac{\sqrt{2}}{2}\right) = 0$$
and
$$- \tilde p + \tilde q^{2} \left(\frac{1}{2} + \frac{\sqrt{2}}{2}\right) + \tilde x^{2} \left(\frac{1}{2} - \frac{\sqrt{2}}{2}\right) = 0$$
$$- 2 \tilde p + \left(- \frac{\tilde q^{2}}{\frac{1}{2} \frac{1}{\frac{1}{2} + \frac{\sqrt{2}}{2}}} + \frac{\tilde x^{2}}{\frac{1}{2} \frac{1}{- \frac{1}{2} + \frac{\sqrt{2}}{2}}}\right) = 0$$
and
$$2 \tilde p + \left(- \frac{\tilde q^{2}}{\frac{1}{2} \frac{1}{\frac{1}{2} + \frac{\sqrt{2}}{2}}} + \frac{\tilde x^{2}}{\frac{1}{2} \frac{1}{- \frac{1}{2} + \frac{\sqrt{2}}{2}}}\right) = 0$$
this equation is fora type hyperbolic paraboloid
- reduced to canonical form
$$p x + q + x^{2} = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} q x + 2 a_{13} p x + 2 a_{14} x + a_{22} q^{2} + 2 a_{23} p q + 2 a_{24} q + a_{33} p^{2} + 2 a_{34} p + a_{44} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = \frac{1}{2}$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = \frac{1}{2}$$
$$a_{33} = 0$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 1$$
|1 0| |0 0| | 1 1/2|
I2 = | | + | | + | |
|0 0| |0 0| |1/2 0 |$$I_{3} = \left|\begin{matrix}1 & 0 & \frac{1}{2}\\0 & 0 & 0\\\frac{1}{2} & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 0 & \frac{1}{2} & 0\\0 & 0 & 0 & \frac{1}{2}\\\frac{1}{2} & 0 & 0 & 0\\0 & \frac{1}{2} & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0 & \frac{1}{2}\\0 & - \lambda & 0\\\frac{1}{2} & 0 & - \lambda\end{matrix}\right|$$
|1 0| | 0 1/2| |0 0|
K2 = | | + | | + | |
|0 0| |1/2 0 | |0 0| |1 0 0 | | 0 0 1/2| | 1 1/2 0|
| | | | | |
K3 = |0 0 1/2| + | 0 0 0 | + |1/2 0 0|
| | | | | |
|0 1/2 0 | |1/2 0 0 | | 0 0 0|$$I_{1} = 1$$
$$I_{2} = - \frac{1}{4}$$
$$I_{3} = 0$$
$$I_{4} = \frac{1}{16}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \lambda^{2} + \frac{\lambda}{4}$$
$$K_{2} = - \frac{1}{4}$$
$$K_{3} = - \frac{1}{4}$$
Because
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
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
$$\lambda^{3} - \lambda^{2} - \frac{\lambda}{4} = 0$$
$$\lambda_{1} = \frac{1}{2} - \frac{\sqrt{2}}{2}$$
$$\lambda_{2} = \frac{1}{2} + \frac{\sqrt{2}}{2}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\tilde p 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde q^{2} \lambda_{2} + \tilde x^{2} \lambda_{1}\right) = 0$$
and
$$- \tilde p 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde q^{2} \lambda_{2} + \tilde x^{2} \lambda_{1}\right) = 0$$
$$\tilde p + \tilde q^{2} \left(\frac{1}{2} + \frac{\sqrt{2}}{2}\right) + \tilde x^{2} \left(\frac{1}{2} - \frac{\sqrt{2}}{2}\right) = 0$$
and
$$- \tilde p + \tilde q^{2} \left(\frac{1}{2} + \frac{\sqrt{2}}{2}\right) + \tilde x^{2} \left(\frac{1}{2} - \frac{\sqrt{2}}{2}\right) = 0$$
$$- 2 \tilde p + \left(- \frac{\tilde q^{2}}{\frac{1}{2} \frac{1}{\frac{1}{2} + \frac{\sqrt{2}}{2}}} + \frac{\tilde x^{2}}{\frac{1}{2} \frac{1}{- \frac{1}{2} + \frac{\sqrt{2}}{2}}}\right) = 0$$
and
$$2 \tilde p + \left(- \frac{\tilde q^{2}}{\frac{1}{2} \frac{1}{\frac{1}{2} + \frac{\sqrt{2}}{2}}} + \frac{\tilde x^{2}}{\frac{1}{2} \frac{1}{- \frac{1}{2} + \frac{\sqrt{2}}{2}}}\right) = 0$$
this equation is fora type hyperbolic paraboloid
- reduced to canonical form