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:
- 2*z^2-7*x*y
- ((x^2)/4)+((y^2)/9)=1
- 2y^2+x-y-1=0
-
x^2–6x–8y=0
- Plot:
-
((3*x^2-y^2+36*x+107)*(x^2-y^2+5*x+6)-(2*x+5)*(x^3-3*x*y^2+18*x^2-18*y^2+107*x+210))/((x^2-y^2+5*x+6)^2+4*x^2*y^2+25*y^2+20*x*y^2)=0
- (7/10)*x^2+(y-sqrt((|x|)))^2=1
-
x^3+y^3-3xy=0
- x^2+(y-3*sqrt(x)^2)^2=1
-
Identical expressions
- ((x^ two)/ four)+((y^ two)/ nine)= one
- ((x squared ) divide by 4) plus ((y squared ) divide by 9) equally 1
- ((x to the power of two) divide by four) plus ((y to the power of two) divide by nine) equally one
- ((x2)/4)+((y2)/9)=1
- x2/4+y2/9=1
- ((x²)/4)+((y²)/9)=1
- ((x to the power of 2)/4)+((y to the power of 2)/9)=1
- x^2/4+y^2/9=1
- ((x^2) divide by 4)+((y^2) divide by 9)=1
-
Similar expressions
- ((x^2)/4)-((y^2)/9)=1
((x^2)/4)+((y^2)/9)=1 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
2 2
x y
-1 + -- + -- = 0
4 9
$$\frac{x^{2}}{4} + \frac{y^{2}}{9} - 1 = 0$$
x^2/4 + y^2/9 - 1 = 0
Detail solution
Given line equation of 2-order:
$$\frac{x^{2}}{4} + \frac{y^{2}}{9} - 1 = 0$$
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
$$a_{11} = \frac{1}{4}$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = \frac{1}{9}$$
$$a_{23} = 0$$
$$a_{33} = -1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}\frac{1}{4} & 0\\0 & \frac{1}{9}\end{matrix}\right|$$
$$\Delta = \frac{1}{36}$$
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
$$\frac{x_{0}}{4} = 0$$
$$\frac{y_{0}}{9} = 0$$
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
$$\frac{x'^{2}}{4} + \frac{y'^{2}}{9} - 1 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{2}{1}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{3}{1}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$\frac{x^{2}}{4} + \frac{y^{2}}{9} - 1 = 0$$
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
$$a_{11} = \frac{1}{4}$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = \frac{1}{9}$$
$$a_{23} = 0$$
$$a_{33} = -1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}\frac{1}{4} & 0\\0 & \frac{1}{9}\end{matrix}\right|$$
$$\Delta = \frac{1}{36}$$
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
$$\frac{x_{0}}{4} = 0$$
$$\frac{y_{0}}{9} = 0$$
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
$$\frac{x'^{2}}{4} + \frac{y'^{2}}{9} - 1 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{2}{1}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{3}{1}\right)^{2}} = 1$$
- 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:
$$\frac{x^{2}}{4} + \frac{y^{2}}{9} - 1 = 0$$
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
$$a_{11} = \frac{1}{4}$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = \frac{1}{9}$$
$$a_{23} = 0$$
$$a_{33} = -1$$
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
$$I_{1} = \frac{13}{36}$$
$$I_{3} = \left|\begin{matrix}\frac{1}{4} & 0 & 0\\0 & \frac{1}{9} & 0\\0 & 0 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}\frac{1}{4} - \lambda & 0\\0 & \frac{1}{9} - \lambda\end{matrix}\right|$$
$$I_{1} = \frac{13}{36}$$
$$I_{2} = \frac{1}{36}$$
$$I_{3} = - \frac{1}{36}$$
$$I{\left(\lambda \right)} = \lambda^{2} - \frac{13 \lambda}{36} + \frac{1}{36}$$
$$K_{2} = - \frac{13}{36}$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} < 0$$
then by line type:
this equation is of type : ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - \frac{13 \lambda}{36} + \frac{1}{36} = 0$$
$$\lambda_{1} = \frac{1}{4}$$
$$\lambda_{2} = \frac{1}{9}$$
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
$$\frac{\tilde x^{2}}{4} + \frac{\tilde y^{2}}{9} - 1 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{2}{1}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{3}{1}\right)^{2}} = 1$$
- reduced to canonical form
$$\frac{x^{2}}{4} + \frac{y^{2}}{9} - 1 = 0$$
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
$$a_{11} = \frac{1}{4}$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = \frac{1}{9}$$
$$a_{23} = 0$$
$$a_{33} = -1$$
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
$$I_{1} = \frac{13}{36}$$
|1/4 0 |
I2 = | |
| 0 1/9|$$I_{3} = \left|\begin{matrix}\frac{1}{4} & 0 & 0\\0 & \frac{1}{9} & 0\\0 & 0 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}\frac{1}{4} - \lambda & 0\\0 & \frac{1}{9} - \lambda\end{matrix}\right|$$
|1/4 0 | |1/9 0 |
K2 = | | + | |
| 0 -1| | 0 -1|$$I_{1} = \frac{13}{36}$$
$$I_{2} = \frac{1}{36}$$
$$I_{3} = - \frac{1}{36}$$
$$I{\left(\lambda \right)} = \lambda^{2} - \frac{13 \lambda}{36} + \frac{1}{36}$$
$$K_{2} = - \frac{13}{36}$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} < 0$$
then by line type:
this equation is of type : ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - \frac{13 \lambda}{36} + \frac{1}{36} = 0$$
$$\lambda_{1} = \frac{1}{4}$$
$$\lambda_{2} = \frac{1}{9}$$
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
$$\frac{\tilde x^{2}}{4} + \frac{\tilde y^{2}}{9} - 1 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{2}{1}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{3}{1}\right)^{2}} = 1$$
- reduced to canonical form