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^2-2xy+y^2+10x-6y+12=0
- 5*x^2+10*x-2*y+5=0
- 13x^2+18xy+37y^2
- x^3-x^2+x-1
- Plot:
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x=449.4*(y-1)*(5/(1.4*(6*y+1)))^(1/2)
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x^2*y^2+(x-2)^2+(y-2)^2=0
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4x^2-y^2-8x-6y-9=0
-
x^2+(y-|x|^(2/3))^2=1
- 5x^2+3y^2
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Identical expressions
- 5x^ two +3y^ two
- 5x squared plus 3y squared
- 5x to the power of two plus 3y to the power of two
- 5x2+3y2
- 5x²+3y²
- 5x to the power of 2+3y to the power of 2
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Similar expressions
- 5x^2-3y^2
5x^2+3y^2 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$5 x^{2} + 3 y^{2} = 0$$
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
$$a_{11} = 5$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}5 & 0\\0 & 3\end{matrix}\right|$$
$$\Delta = 15$$
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
$$5 x_{0} = 0$$
$$3 y_{0} = 0$$
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
$$5 x'^{2} + 3 y'^{2} = 0$$
Given equation is degenerate ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{5}}{5}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$5 x^{2} + 3 y^{2} = 0$$
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
$$a_{11} = 5$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}5 & 0\\0 & 3\end{matrix}\right|$$
$$\Delta = 15$$
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
$$5 x_{0} = 0$$
$$3 y_{0} = 0$$
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
$$5 x'^{2} + 3 y'^{2} = 0$$
Given equation is degenerate ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{5}}{5}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} = 0$$
- 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:
$$5 x^{2} + 3 y^{2} = 0$$
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
$$a_{11} = 5$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 8$$
$$I_{3} = \left|\begin{matrix}5 & 0 & 0\\0 & 3 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}5 - \lambda & 0\\0 & 3 - \lambda\end{matrix}\right|$$
$$I_{1} = 8$$
$$I_{2} = 15$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 8 \lambda + 15$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} > 0$$
then by line type:
this equation is of type : degenerate ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 8 \lambda + 15 = 0$$
$$\lambda_{1} = 5$$
$$\lambda_{2} = 3$$
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
$$5 \tilde x^{2} + 3 \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{5}}{5}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} = 0$$
- reduced to canonical form
$$5 x^{2} + 3 y^{2} = 0$$
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
$$a_{11} = 5$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = 8$$
|5 0|
I2 = | |
|0 3|$$I_{3} = \left|\begin{matrix}5 & 0 & 0\\0 & 3 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}5 - \lambda & 0\\0 & 3 - \lambda\end{matrix}\right|$$
|5 0| |3 0|
K2 = | | + | |
|0 0| |0 0|$$I_{1} = 8$$
$$I_{2} = 15$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} - 8 \lambda + 15$$
$$K_{2} = 0$$
Because
$$I_{3} = 0 \wedge I_{2} > 0$$
then by line type:
this equation is of type : degenerate ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 8 \lambda + 15 = 0$$
$$\lambda_{1} = 5$$
$$\lambda_{2} = 3$$
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
$$5 \tilde x^{2} + 3 \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{5}}{5}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} = 0$$
- reduced to canonical form