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- 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:
- 3x^2+0xy+4y^2-6x+16y+7=0
- y=-3-sqrt(-3x-21)
- 3*x-4*y+4=0
- x^2+4*y^2+2*x-32*y+64=0
- Plot:
- (x^2+y^2-1)^3-x^2*y^3=0
-
y^2=x(1-x^2)^3
-
y^2=(x^4+83*x^2-408+60*x+10*x^3)/(21+2*x^2+10*x-y^2)
- y=-0,8x;y-2,5x;y=0,4x
-
Identical expressions
- 3x^ two + zero xy+4y^ two -6x+16y+ seven =0
- 3x squared plus 0xy plus 4y squared minus 6x plus 16y plus 7 equally 0
- 3x to the power of two plus zero xy plus 4y to the power of two minus 6x plus 16y plus seven equally 0
- 3x2+0xy+4y2-6x+16y+7=0
- 3x²+0xy+4y²-6x+16y+7=0
- 3x to the power of 2+0xy+4y to the power of 2-6x+16y+7=0
- 3x^2+0xy+4y^2-6x+16y+7=O
-
Similar expressions
- 3x^2-0xy+4y^2-6x+16y+7=0
- 3x^2+0xy-4y^2-6x+16y+7=0
- 3x^2+0xy+4y^2-6x-16y+7=0
- 3x^2+0xy+4y^2+6x+16y+7=0
- 3x^2+0xy+4y^2-6x+16y-7=0
3x^2+0xy+4y^2-6x+16y+7=0 canonical form
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The solution
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2 2 7 - 6*x + 3*x + 4*y + 16*y = 0
$$3 x^{2} - 6 x + 4 y^{2} + 16 y + 7 = 0$$
3*x^2 - 6*x + 4*y^2 + 16*y + 7 = 0
Detail solution
Given line equation of 2-order:
$$3 x^{2} - 6 x + 4 y^{2} + 16 y + 7 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -3$$
$$a_{22} = 4$$
$$a_{23} = 8$$
$$a_{33} = 7$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 0\\0 & 4\end{matrix}\right|$$
$$\Delta = 12$$
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
$$3 x_{0} - 3 = 0$$
$$4 y_{0} + 8 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = -2$$
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
$$a'_{33} = - 3 x_{0} + 8 y_{0} + 7$$
$$a'_{33} = -12$$
then equation turns into
$$3 x'^{2} + 4 y'^{2} - 12 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{6} \sqrt{3}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{2 \frac{\sqrt{3}}{6}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$3 x^{2} - 6 x + 4 y^{2} + 16 y + 7 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -3$$
$$a_{22} = 4$$
$$a_{23} = 8$$
$$a_{33} = 7$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 0\\0 & 4\end{matrix}\right|$$
$$\Delta = 12$$
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
$$3 x_{0} - 3 = 0$$
$$4 y_{0} + 8 = 0$$
then
$$x_{0} = 1$$
$$y_{0} = -2$$
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
$$a'_{33} = - 3 x_{0} + 8 y_{0} + 7$$
$$a'_{33} = -12$$
then equation turns into
$$3 x'^{2} + 4 y'^{2} - 12 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{6} \sqrt{3}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{2 \frac{\sqrt{3}}{6}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(1, -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:
$$3 x^{2} - 6 x + 4 y^{2} + 16 y + 7 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -3$$
$$a_{22} = 4$$
$$a_{23} = 8$$
$$a_{33} = 7$$
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
$$I_{1} = 7$$
$$I_{3} = \left|\begin{matrix}3 & 0 & -3\\0 & 4 & 8\\-3 & 8 & 7\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0\\0 & 4 - \lambda\end{matrix}\right|$$
$$I_{1} = 7$$
$$I_{2} = 12$$
$$I_{3} = -144$$
$$I{\left(\lambda \right)} = \lambda^{2} - 7 \lambda + 12$$
$$K_{2} = -24$$
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
$$\lambda^{2} - 7 \lambda + 12 = 0$$
$$\lambda_{1} = 4$$
$$\lambda_{2} = 3$$
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
$$4 \tilde x^{2} + 3 \tilde y^{2} - 12 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{2 \frac{\sqrt{3}}{6}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{6} \sqrt{3}}\right)^{2}} = 1$$
- reduced to canonical form
$$3 x^{2} - 6 x + 4 y^{2} + 16 y + 7 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = -3$$
$$a_{22} = 4$$
$$a_{23} = 8$$
$$a_{33} = 7$$
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
$$I_{1} = 7$$
|3 0|
I2 = | |
|0 4|$$I_{3} = \left|\begin{matrix}3 & 0 & -3\\0 & 4 & 8\\-3 & 8 & 7\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0\\0 & 4 - \lambda\end{matrix}\right|$$
|3 -3| |4 8|
K2 = | | + | |
|-3 7 | |8 7|$$I_{1} = 7$$
$$I_{2} = 12$$
$$I_{3} = -144$$
$$I{\left(\lambda \right)} = \lambda^{2} - 7 \lambda + 12$$
$$K_{2} = -24$$
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
$$\lambda^{2} - 7 \lambda + 12 = 0$$
$$\lambda_{1} = 4$$
$$\lambda_{2} = 3$$
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
$$4 \tilde x^{2} + 3 \tilde y^{2} - 12 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{2 \frac{\sqrt{3}}{6}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{6} \sqrt{3}}\right)^{2}} = 1$$
- reduced to canonical form