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
- x^2-2x+y^2+6y-2=0
- y^2-5*x-8*y-14=0
- 2x^2=y^2+1
- Plot:
-
23*x-4*y+37=-x-sqrt(x^2-4*y)
-
2*x-4*y+3=-x-sqrt(x^2-4*y)
-
10+15x-18y+2x^2-5xy+3y^2
- (|x|)*(x-1)-5*x*y=(|x|)*(x-1)-5*x
- Integral of d{x}:
-
2x^2
-
y^2+1
- Derivative of:
-
2x^2
- y^2+1
- Graphing y =:
- 2x^2
- y^2+1
-
Identical expressions
- two x^ two =y^2+ one
- 2x squared equally y squared plus 1
- two x to the power of two equally y squared plus one
- 2x2=y2+1
- 2x²=y²+1
- 2x to the power of 2=y to the power of 2+1
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Similar expressions
- 2x^2+5y^2+4z^2-6xy+4xz-8yz-2x-2y+z=-7
- 3x^2+4xy+3y^2+14x+16y+18=0
- 25x^2+16y^2+100x-96y-156=0
- 2x^2+3y^2-6x+4y-10=0
- y^2+16x-6y+41=0
- 2x^2+6xy+5y^2+2xz+2yz+z^2=0
- 2x^2+2xy+2y^2+2^(1/2)x-2^(1/2)y-5=0
- 2x^2=y^2-1
- 2x^2+4y^2+4x-8y-10=0
- 9x^2+4y^2+108x+16y+304=0
2x^2=y^2+1 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$2 x^{2} - y^{2} - 1 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -1$$
$$a_{23} = 0$$
$$a_{33} = -1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}2 & 0\\0 & -1\end{matrix}\right|$$
$$\Delta = -2$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
substitute coefficients
$$2 x_{0} = 0$$
$$- y_{0} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = 0$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = -1$$
$$a'_{33} = -1$$
then equation turns into
$$2 x'^{2} - y'^{2} - 1 = 0$$
Given equation is hyperbole
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$2 x^{2} - y^{2} - 1 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -1$$
$$a_{23} = 0$$
$$a_{33} = -1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}2 & 0\\0 & -1\end{matrix}\right|$$
$$\Delta = -2$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
substitute coefficients
$$2 x_{0} = 0$$
$$- y_{0} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = 0$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = -1$$
$$a'_{33} = -1$$
then equation turns into
$$2 x'^{2} - y'^{2} - 1 = 0$$
Given equation is hyperbole
2 2 \tilde x \tilde y --------- - --------- = 1 1/2 1
- reduced to canonical form
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:
$$2 x^{2} - y^{2} - 1 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -1$$
$$a_{23} = 0$$
$$a_{33} = -1$$
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
$$I_{1} = 1$$
$$I_{3} = \left|\begin{matrix}2 & 0 & 0\\0 & -1 & 0\\0 & 0 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 0\\0 & - \lambda - 1\end{matrix}\right|$$
$$I_{1} = 1$$
$$I_{2} = -2$$
$$I_{3} = 2$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda - 2$$
$$K_{2} = -1$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - \lambda - 2 = 0$$
$$\lambda_{1} = 2$$
$$\lambda_{2} = -1$$
then the canonical form of the equation will be
$$\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0$$
or
$$2 \tilde x^{2} - \tilde y^{2} - 1 = 0$$
- reduced to canonical form
$$2 x^{2} - y^{2} - 1 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -1$$
$$a_{23} = 0$$
$$a_{33} = -1$$
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
$$I_{1} = 1$$
|2 0 |
I2 = | |
|0 -1|$$I_{3} = \left|\begin{matrix}2 & 0 & 0\\0 & -1 & 0\\0 & 0 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 0\\0 & - \lambda - 1\end{matrix}\right|$$
|2 0 | |-1 0 |
K2 = | | + | |
|0 -1| |0 -1|$$I_{1} = 1$$
$$I_{2} = -2$$
$$I_{3} = 2$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda - 2$$
$$K_{2} = -1$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - \lambda - 2 = 0$$
$$\lambda_{1} = 2$$
$$\lambda_{2} = -1$$
then the canonical form of the equation will be
$$\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0$$
or
$$2 \tilde x^{2} - \tilde y^{2} - 1 = 0$$
2 2 \tilde x \tilde y --------- - --------- = 1 1/2 1
- reduced to canonical form