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:
- 3x^2+2y^2=5
- 3x^2+4xy-6xz+3y^2-2yz+4z^2
- x²-6y²+3z²+8x+12y+1=0
- x^2-2y^2-3z^2=1
- Plot:
- y=x^3-3*x^2+4
-
y=4xx=1y=0
- y-x=0
- y-x*tg(z/2)
- 3x^2+2y^2
-
Identical expressions
- 3x^ two + two y^2= five
- 3x squared plus 2y squared equally 5
- 3x to the power of two plus two y squared equally five
- 3x2+2y2=5
- 3x²+2y²=5
- 3x to the power of 2+2y to the power of 2=5
-
Similar expressions
- 3x^2-2y^2=5
3x^2+2y^2=5 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$3 x^{2} + 2 y^{2} - 5 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 2$$
$$a_{23} = 0$$
$$a_{33} = -5$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 0\\0 & 2\end{matrix}\right|$$
$$\Delta = 6$$
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
$$3 x_{0} = 0$$
$$2 y_{0} = 0$$
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
$$a'_{33} = -5$$
$$a'_{33} = -5$$
then equation turns into
$$3 x'^{2} + 2 y'^{2} - 5 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{5} \sqrt{5}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{5} \sqrt{5}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$3 x^{2} + 2 y^{2} - 5 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 2$$
$$a_{23} = 0$$
$$a_{33} = -5$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 0\\0 & 2\end{matrix}\right|$$
$$\Delta = 6$$
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
$$3 x_{0} = 0$$
$$2 y_{0} = 0$$
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
$$a'_{33} = -5$$
$$a'_{33} = -5$$
then equation turns into
$$3 x'^{2} + 2 y'^{2} - 5 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{5} \sqrt{5}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{5} \sqrt{5}}\right)^{2}} = 1$$
- 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:
$$3 x^{2} + 2 y^{2} - 5 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 2$$
$$a_{23} = 0$$
$$a_{33} = -5$$
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
$$I_{1} = 5$$
$$I_{3} = \left|\begin{matrix}3 & 0 & 0\\0 & 2 & 0\\0 & 0 & -5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0\\0 & 2 - \lambda\end{matrix}\right|$$
$$I_{1} = 5$$
$$I_{2} = 6$$
$$I_{3} = -30$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda + 6$$
$$K_{2} = -25$$
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
$$\lambda^{2} - 5 \lambda + 6 = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = 2$$
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
$$3 \tilde x^{2} + 2 \tilde y^{2} - 5 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{5} \sqrt{5}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{5} \sqrt{5}}\right)^{2}} = 1$$
- reduced to canonical form
$$3 x^{2} + 2 y^{2} - 5 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 2$$
$$a_{23} = 0$$
$$a_{33} = -5$$
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
$$I_{1} = 5$$
|3 0|
I2 = | |
|0 2|$$I_{3} = \left|\begin{matrix}3 & 0 & 0\\0 & 2 & 0\\0 & 0 & -5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0\\0 & 2 - \lambda\end{matrix}\right|$$
|3 0 | |2 0 |
K2 = | | + | |
|0 -5| |0 -5|$$I_{1} = 5$$
$$I_{2} = 6$$
$$I_{3} = -30$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda + 6$$
$$K_{2} = -25$$
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
$$\lambda^{2} - 5 \lambda + 6 = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = 2$$
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
$$3 \tilde x^{2} + 2 \tilde y^{2} - 5 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{5} \sqrt{5}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{5} \sqrt{5}}\right)^{2}} = 1$$
- reduced to canonical form