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:
- 8*x^2-4*x*y+5*y^2-52*x+22*y+53=0
- x^2+y^2-ax+by+c=0
- x^2/6-y^2/5+z^2-1=0
-
x^2-y^2
- Plot:
- x^2+y^2=1
-
x^2*y^2=(x-1)*(x-2)
- sin(x)+cos(x)=0
- ((1-cos(x^(2)+y^(2))))/((x^(2)+y^(2))x^(2)y^(2))=z
- Inequation:
- x^2-y^2
- Factor polynomial:
- x^2-y^2
- Limit of the function:
- x^2-y^2
-
Identical expressions
- x^ two -y^ two
- x squared minus y squared
- x to the power of two minus y to the power of two
- x2-y2
- x²-y²
- x to the power of 2-y to the power of 2
-
Similar expressions
- x^2+y^2
- x^2-y^2+z^2=0
- 4x^2-y^2+8x-2y+7=0
- x^2-y^2-6x-2y+4=0
x^2-y^2 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$x^{2} - y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -1$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & -1\end{matrix}\right|$$
$$\Delta = -1$$
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
$$x_{0} = 0$$
$$- y_{0} = 0$$
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
$$a'_{33} = 0$$
$$a'_{33} = 0$$
then equation turns into
$$x'^{2} - y'^{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{1^{2}} - \frac{\tilde y^{2}}{1^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$x^{2} - y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -1$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & -1\end{matrix}\right|$$
$$\Delta = -1$$
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
$$x_{0} = 0$$
$$- y_{0} = 0$$
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
$$a'_{33} = 0$$
$$a'_{33} = 0$$
then equation turns into
$$x'^{2} - y'^{2} = 0$$
Given equation is degenerate hyperbole
$$\frac{\tilde x^{2}}{1^{2}} - \frac{\tilde y^{2}}{1^{2}} = 0$$
- 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:
$$x^{2} - y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -1$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 0$$
$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & -1 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda - 1\end{matrix}\right|$$
$$I_{1} = 0$$
$$I_{2} = -1$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 1$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} < 0$$
then by line type:
this equation is of type : degenerate hyperbole
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 1 = 0$$
Solve this equation
$$\lambda_{1} = -1$$
$$\lambda_{2} = 1$$
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
$$- \tilde x^{2} + \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{1^{2}} - \frac{\tilde y^{2}}{1^{2}} = 0$$
- reduced to canonical form
$$x^{2} - y^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -1$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 0$$
|1 0 |
I2 = | |
|0 -1|$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & -1 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda - 1\end{matrix}\right|$$
|1 0| |-1 0|
K2 = | | + | |
|0 0| |0 0|$$I_{1} = 0$$
$$I_{2} = -1$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 1$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} < 0$$
then by line type:
this equation is of type : degenerate hyperbole
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 1 = 0$$
Solve this equation
$$\lambda_{1} = -1$$
$$\lambda_{2} = 1$$
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
$$- \tilde x^{2} + \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{1^{2}} - \frac{\tilde y^{2}}{1^{2}} = 0$$
- reduced to canonical form
The graph