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
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How to use it?
- Canonical form:
- x^2 – 6y^2 + 54y = 0
- x^2/25-y^2/16+x^2/4=1
- 5x^2+9x^2+50x-18y+44=0
- -6x^2-2y^2-24=0
- Plot:
- (z*cos(pi/4)-x*sin(pi/4))^2=(z*sin(pi/4)+(x-2)*cos(pi/4))^2+(y-5)^2
- x^2+y^2=z^2
- (y-x)*x^4+8=0
-
(x^3)*(y-x)-x^3-1=0
-
Identical expressions
- x^ two – 6y^ two + 54y = zero
- x squared – 6y squared plus 54y equally 0
- x to the power of two – 6y to the power of two plus 54y equally zero
- x2 – 6y2 + 54y = 0
- x² – 6y² + 54y = 0
- x to the power of 2 – 6y to the power of 2 + 54y = 0
-
Similar expressions
- x^2 – 6y^2 - 54y = 0
x^2 – 6y^2 + 54y = 0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$x^{2} - 6 y^{2} + 54 y = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -6$$
$$a_{23} = 27$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & -6\end{matrix}\right|$$
$$\Delta = -6$$
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
$$x_{0} = 0$$
$$27 - 6 y_{0} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = \frac{9}{2}$$
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
$$a'_{33} = 27 y_{0}$$
$$a'_{33} = \frac{243}{2}$$
then equation turns into
$$x'^{2} - 6 y'^{2} + \frac{243}{2} = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{\frac{243}{2}} - \frac{\tilde y^{2}}{\frac{81}{4}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$x^{2} - 6 y^{2} + 54 y = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -6$$
$$a_{23} = 27$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & -6\end{matrix}\right|$$
$$\Delta = -6$$
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
$$x_{0} = 0$$
$$27 - 6 y_{0} = 0$$
then
$$x_{0} = 0$$
$$y_{0} = \frac{9}{2}$$
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
$$a'_{33} = 27 y_{0}$$
$$a'_{33} = \frac{243}{2}$$
then equation turns into
$$x'^{2} - 6 y'^{2} + \frac{243}{2} = 0$$
Given equation is hyperbole
$$\frac{\tilde x^{2}}{\frac{243}{2}} - \frac{\tilde y^{2}}{\frac{81}{4}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 9/2)
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:
$$x^{2} - 6 y^{2} + 54 y = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -6$$
$$a_{23} = 27$$
$$a_{33} = 0$$
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
$$I_{1} = -5$$
$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & -6 & 27\\0 & 27 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda - 6\end{matrix}\right|$$
$$I_{1} = -5$$
$$I_{2} = -6$$
$$I_{3} = -729$$
$$I{\left(\lambda \right)} = \lambda^{2} + 5 \lambda - 6$$
$$K_{2} = -729$$
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
$$\lambda^{2} + 5 \lambda - 6 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = -6$$
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
$$\tilde x^{2} - 6 \tilde y^{2} + \frac{243}{2} = 0$$
$$\frac{\tilde x^{2}}{\frac{243}{2}} - \frac{\tilde y^{2}}{\frac{81}{4}} = -1$$
- reduced to canonical form
$$x^{2} - 6 y^{2} + 54 y = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = -6$$
$$a_{23} = 27$$
$$a_{33} = 0$$
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
$$I_{1} = -5$$
|1 0 |
I2 = | |
|0 -6|$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & -6 & 27\\0 & 27 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda - 6\end{matrix}\right|$$
|1 0| |-6 27|
K2 = | | + | |
|0 0| |27 0 |$$I_{1} = -5$$
$$I_{2} = -6$$
$$I_{3} = -729$$
$$I{\left(\lambda \right)} = \lambda^{2} + 5 \lambda - 6$$
$$K_{2} = -729$$
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
$$\lambda^{2} + 5 \lambda - 6 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = -6$$
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
$$\tilde x^{2} - 6 \tilde y^{2} + \frac{243}{2} = 0$$
$$\frac{\tilde x^{2}}{\frac{243}{2}} - \frac{\tilde y^{2}}{\frac{81}{4}} = -1$$
- reduced to canonical form