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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
-
How to use it?
- Canonical form:
- x2+y2−4y+3=0
- 2x^2+2y^2-8sqrt(2)y
- 9*x^2+18*x+18xy+22y+13y^2=11
- 3x^2+5y^2+3z^2+2*x*z-2*x*y-2y*z=0
- Plot:
- x^2+y^2=1
- x^2+y^2=4
-
3*cos(x)*sin(y)-sin(x)*cos(y)-4*cos(y)=3*sqrt(3)
- x=x-sqrt(abs(y))
- Derivative of:
- 11
- Limit of the function:
- 11
- Integral of d{x}:
-
11
-
Identical expressions
- nine *x^ two + eighteen *x+18xy+ two 2y+13y^2= eleven
- 9 multiply by x squared plus 18 multiply by x plus 18xy plus 22y plus 13y squared equally 11
- nine multiply by x to the power of two plus eighteen multiply by x plus 18xy plus two 2y plus 13y squared equally eleven
- 9*x2+18*x+18xy+22y+13y2=11
- 9*x²+18*x+18xy+22y+13y²=11
- 9*x to the power of 2+18*x+18xy+22y+13y to the power of 2=11
- 9x^2+18x+18xy+22y+13y^2=11
- 9x2+18x+18xy+22y+13y2=11
-
Similar expressions
- 9*x^2+18*x+18xy+22y-13y^2=11
- 9*x^2+18*x+18xy-22y+13y^2=11
- 9*x^2+18*x-18xy+22y+13y^2=11
- 9*x^2-18*x+18xy+22y+13y^2=11
9*x^2+18*x+18xy+22y+13y^2=11 canonical form
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The solution
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2 2 -11 + 9*x + 13*y + 18*x + 22*y + 18*x*y = 0
$$9 x^{2} + 18 x y + 18 x + 13 y^{2} + 22 y - 11 = 0$$
9*x^2 + 18*x*y + 18*x + 13*y^2 + 22*y - 11 = 0
Detail solution
Given line equation of 2-order:
$$9 x^{2} + 18 x y + 18 x + 13 y^{2} + 22 y - 11 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 9$$
$$a_{13} = 9$$
$$a_{22} = 13$$
$$a_{23} = 11$$
$$a_{33} = -11$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 9\\9 & 13\end{matrix}\right|$$
$$\Delta = 36$$
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
$$9 x_{0} + 9 y_{0} + 9 = 0$$
$$9 x_{0} + 13 y_{0} + 11 = 0$$
then
$$x_{0} = - \frac{1}{2}$$
$$y_{0} = - \frac{1}{2}$$
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
$$a'_{33} = 9 x_{0} + 11 y_{0} - 11$$
$$a'_{33} = -21$$
then equation turns into
$$9 x'^{2} + 18 x' y' + 13 y'^{2} - 21 = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = - \frac{2}{9}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{2}{9} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{9 \sqrt{85}}{85}$$
$$\cos{\left(2 \phi \right)} = \frac{2 \sqrt{85}}{85}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}}$$
$$y' = - \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}} + \tilde y \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}}$$
then the equation turns from
$$9 x'^{2} + 18 x' y' + 13 y'^{2} - 21 = 0$$
to
$$13 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}} + \tilde y \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}}\right)^{2} + 18 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}} + \tilde y \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}}\right) + 9 \left(\tilde x \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}}\right)^{2} - 21 = 0$$
simplify
$$- 18 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}} \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}} - \frac{4 \sqrt{85} \tilde x^{2}}{85} + 11 \tilde x^{2} - 8 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}} \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}} + \frac{36 \sqrt{85} \tilde x \tilde y}{85} + \frac{4 \sqrt{85} \tilde y^{2}}{85} + 18 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}} \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}} + 11 \tilde y^{2} - 21 = 0$$
$$- \sqrt{85} \tilde x^{2} + 11 \tilde x^{2} + \sqrt{85} \tilde y^{2} + 11 \tilde y^{2} - 21 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{\sqrt{21}}{21} \sqrt{11 - \sqrt{85}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{\sqrt{21}}{21} \sqrt{\sqrt{85} + 11}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}}\right)$$
$$\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}}, \ \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}}\right)$$
$$9 x^{2} + 18 x y + 18 x + 13 y^{2} + 22 y - 11 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 9$$
$$a_{13} = 9$$
$$a_{22} = 13$$
$$a_{23} = 11$$
$$a_{33} = -11$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}9 & 9\\9 & 13\end{matrix}\right|$$
$$\Delta = 36$$
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
$$9 x_{0} + 9 y_{0} + 9 = 0$$
$$9 x_{0} + 13 y_{0} + 11 = 0$$
then
$$x_{0} = - \frac{1}{2}$$
$$y_{0} = - \frac{1}{2}$$
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
$$a'_{33} = 9 x_{0} + 11 y_{0} - 11$$
$$a'_{33} = -21$$
then equation turns into
$$9 x'^{2} + 18 x' y' + 13 y'^{2} - 21 = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = - \frac{2}{9}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{2}{9} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{9 \sqrt{85}}{85}$$
$$\cos{\left(2 \phi \right)} = \frac{2 \sqrt{85}}{85}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}}$$
substitute coefficients
$$x' = \tilde x \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}}$$
$$y' = - \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}} + \tilde y \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}}$$
then the equation turns from
$$9 x'^{2} + 18 x' y' + 13 y'^{2} - 21 = 0$$
to
$$13 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}} + \tilde y \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}}\right)^{2} + 18 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}} + \tilde y \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}}\right) + 9 \left(\tilde x \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}}\right)^{2} - 21 = 0$$
simplify
$$- 18 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}} \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}} - \frac{4 \sqrt{85} \tilde x^{2}}{85} + 11 \tilde x^{2} - 8 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}} \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}} + \frac{36 \sqrt{85} \tilde x \tilde y}{85} + \frac{4 \sqrt{85} \tilde y^{2}}{85} + 18 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}} \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}} + 11 \tilde y^{2} - 21 = 0$$
$$- \sqrt{85} \tilde x^{2} + 11 \tilde x^{2} + \sqrt{85} \tilde y^{2} + 11 \tilde y^{2} - 21 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{\sqrt{21}}{21} \sqrt{11 - \sqrt{85}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{\sqrt{21}}{21} \sqrt{\sqrt{85} + 11}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(-1/2, -1/2)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}}\right)$$
$$\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{\sqrt{85}}{85}}, \ \sqrt{\frac{\sqrt{85}}{85} + \frac{1}{2}}\right)$$
Invariants method
Given line equation of 2-order:
$$9 x^{2} + 18 x y + 18 x + 13 y^{2} + 22 y - 11 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 9$$
$$a_{13} = 9$$
$$a_{22} = 13$$
$$a_{23} = 11$$
$$a_{33} = -11$$
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
$$I_{1} = 22$$
$$I_{3} = \left|\begin{matrix}9 & 9 & 9\\9 & 13 & 11\\9 & 11 & -11\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 9\\9 & 13 - \lambda\end{matrix}\right|$$
$$I_{1} = 22$$
$$I_{2} = 36$$
$$I_{3} = -756$$
$$I{\left(\lambda \right)} = \lambda^{2} - 22 \lambda + 36$$
$$K_{2} = -444$$
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
$$\lambda^{2} - 22 \lambda + 36 = 0$$
$$\lambda_{1} = 11 - \sqrt{85}$$
$$\lambda_{2} = \sqrt{85} + 11$$
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
$$\tilde x^{2} \left(11 - \sqrt{85}\right) + \tilde y^{2} \left(\sqrt{85} + 11\right) - 21 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{\sqrt{21}}{21} \sqrt{11 - \sqrt{85}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{\sqrt{21}}{21} \sqrt{\sqrt{85} + 11}}\right)^{2}} = 1$$
- reduced to canonical form
$$9 x^{2} + 18 x y + 18 x + 13 y^{2} + 22 y - 11 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 9$$
$$a_{13} = 9$$
$$a_{22} = 13$$
$$a_{23} = 11$$
$$a_{33} = -11$$
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
$$I_{1} = 22$$
|9 9 |
I2 = | |
|9 13|$$I_{3} = \left|\begin{matrix}9 & 9 & 9\\9 & 13 & 11\\9 & 11 & -11\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & 9\\9 & 13 - \lambda\end{matrix}\right|$$
|9 9 | |13 11 |
K2 = | | + | |
|9 -11| |11 -11|$$I_{1} = 22$$
$$I_{2} = 36$$
$$I_{3} = -756$$
$$I{\left(\lambda \right)} = \lambda^{2} - 22 \lambda + 36$$
$$K_{2} = -444$$
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
$$\lambda^{2} - 22 \lambda + 36 = 0$$
$$\lambda_{1} = 11 - \sqrt{85}$$
$$\lambda_{2} = \sqrt{85} + 11$$
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
$$\tilde x^{2} \left(11 - \sqrt{85}\right) + \tilde y^{2} \left(\sqrt{85} + 11\right) - 21 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{\sqrt{21}}{21} \sqrt{11 - \sqrt{85}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{\sqrt{21}}{21} \sqrt{\sqrt{85} + 11}}\right)^{2}} = 1$$
- reduced to canonical form