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How to use it?
- Canonical form:
- x^2+y^2-6x+4y-12=0
- 6x1^2-3x2^2-3x3^2
- 4𝑥1𝑥3−6𝑥2𝑥4
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- Plot:
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4x^2+4y^2+8x*y+4x+4y+1==0
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Identical expressions
- x^ two +y^ two -6x+4y- twelve = zero
- x squared plus y squared minus 6x plus 4y minus 12 equally 0
- x to the power of two plus y to the power of two minus 6x plus 4y minus twelve equally zero
- x2+y2-6x+4y-12=0
- x²+y²-6x+4y-12=0
- x to the power of 2+y to the power of 2-6x+4y-12=0
- x^2+y^2-6x+4y-12=O
-
Similar expressions
- x^2+y^2-6x+4y+12=0
- x^2+y^2-6x-4y-12=0
- x^2+y^2+6x+4y-12=0
- x^2-y^2-6x+4y-12=0
x^2+y^2-6x+4y-12=0 canonical form
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2 2 -12 + x + y - 6*x + 4*y = 0
$$x^{2} - 6 x + y^{2} + 4 y - 12 = 0$$
x^2 - 6*x + y^2 + 4*y - 12 = 0
Detail solution
Given line equation of 2-order:
$$x^{2} - 6 x + y^{2} + 4 y - 12 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = -3$$
$$a_{22} = 1$$
$$a_{23} = 2$$
$$a_{33} = -12$$
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
$$x_{0} - 3 = 0$$
$$y_{0} + 2 = 0$$
then
$$x_{0} = 3$$
$$y_{0} = -2$$
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
$$a'_{33} = - 3 x_{0} + 2 y_{0} - 12$$
$$a'_{33} = -25$$
then equation turns into
$$x'^{2} + y'^{2} - 25 = 0$$
Given equation is circle
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$x^{2} - 6 x + y^{2} + 4 y - 12 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = -3$$
$$a_{22} = 1$$
$$a_{23} = 2$$
$$a_{33} = -12$$
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
$$x_{0} - 3 = 0$$
$$y_{0} + 2 = 0$$
then
$$x_{0} = 3$$
$$y_{0} = -2$$
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
$$a'_{33} = - 3 x_{0} + 2 y_{0} - 12$$
$$a'_{33} = -25$$
then equation turns into
$$x'^{2} + y'^{2} - 25 = 0$$
Given equation is circle
2 2
\tilde x \tilde y
--------- + --------- = 1
2 2
/ 1\ / 1\
\5 / \5 / - reduced to canonical form
The center of canonical coordinate system at point O
(3, -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:
$$x^{2} - 6 x + y^{2} + 4 y - 12 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = -3$$
$$a_{22} = 1$$
$$a_{23} = 2$$
$$a_{33} = -12$$
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
$$I_{1} = 2$$
$$I_{3} = \left|\begin{matrix}1 & 0 & -3\\0 & 1 & 2\\-3 & 2 & -12\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 2$$
$$I_{2} = 1$$
$$I_{3} = -25$$
$$I{\left(\lambda \right)} = \lambda^{2} - 2 \lambda + 1$$
$$K_{2} = -37$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} < 0$$
then by line type:
this equation is of type : circle
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 2 \lambda + 1 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = 1$$
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
$$\tilde x^{2} + \tilde y^{2} - 25 = 0$$
- reduced to canonical form
$$x^{2} - 6 x + y^{2} + 4 y - 12 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = -3$$
$$a_{22} = 1$$
$$a_{23} = 2$$
$$a_{33} = -12$$
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
$$I_{1} = 2$$
|1 0|
I2 = | |
|0 1|$$I_{3} = \left|\begin{matrix}1 & 0 & -3\\0 & 1 & 2\\-3 & 2 & -12\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|$$
|1 -3 | |1 2 |
K2 = | | + | |
|-3 -12| |2 -12|$$I_{1} = 2$$
$$I_{2} = 1$$
$$I_{3} = -25$$
$$I{\left(\lambda \right)} = \lambda^{2} - 2 \lambda + 1$$
$$K_{2} = -37$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} < 0$$
then by line type:
this equation is of type : circle
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 2 \lambda + 1 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = 1$$
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
$$\tilde x^{2} + \tilde y^{2} - 25 = 0$$
2 2
\tilde x \tilde y
--------- + --------- = 1
2 2
/ 1\ / 1\
\5 / \5 / - reduced to canonical form