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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+4y^2-12x-12=0
- -11x1^2+12x2^2-x3^2-4sqrt(3)x1x2-10sqrt(3)x1x3-x2x3=0
- x1*x2+2x2*x3-3x3*x4
- 18x^2+24xy+11y^2-5=0
- Plot:
- 4*y^2=27*x+243
- 0=x^2+2*x*y+y^3
-
1+(36*x^2-36*y^2-444)/((x^2-y^2+5*x+6)^2+25*y^2)=0
- -7x-18
-
Identical expressions
- 18x^ two + two 4xy+11y^2- five = zero
- 18x squared plus 24xy plus 11y squared minus 5 equally 0
- 18x to the power of two plus two 4xy plus 11y squared minus five equally zero
- 18x2+24xy+11y2-5=0
- 18x²+24xy+11y²-5=0
- 18x to the power of 2+24xy+11y to the power of 2-5=0
- 18x^2+24xy+11y^2-5=O
-
Similar expressions
- 18x^2+24xy-11y^2-5=0
- 18x^2+24xy+11y^2+5=0
- 18x^2-24xy+11y^2-5=0
18x^2+24xy+11y^2-5=0 canonical form
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2 2 -5 + 11*y + 18*x + 24*x*y = 0
$$18 x^{2} + 24 x y + 11 y^{2} - 5 = 0$$
18*x^2 + 24*x*y + 11*y^2 - 5 = 0
Detail solution
Given line equation of 2-order:
$$18 x^{2} + 24 x y + 11 y^{2} - 5 = 0$$
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
$$a_{11} = 18$$
$$a_{12} = 12$$
$$a_{13} = 0$$
$$a_{22} = 11$$
$$a_{23} = 0$$
$$a_{33} = -5$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}18 & 12\\12 & 11\end{matrix}\right|$$
$$\Delta = 54$$
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
$$18 x_{0} + 12 y_{0} = 0$$
$$12 x_{0} + 11 y_{0} = 0$$
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
$$18 x'^{2} + 24 x' y' + 11 y'^{2} - 5 = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{7}{24}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{7}{24} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{24}{25}$$
$$\cos{\left(2 \phi \right)} = \frac{7}{25}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{4}{5}$$
$$\sin{\left(\phi \right)} = \frac{3}{5}$$
substitute coefficients
$$x' = \frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}$$
$$y' = \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}$$
then the equation turns from
$$18 x'^{2} + 24 x' y' + 11 y'^{2} - 5 = 0$$
to
$$11 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right)^{2} + 24 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right) + 18 \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right)^{2} - 5 = 0$$
simplify
$$27 \tilde x^{2} + 2 \tilde y^{2} - 5 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{9} \sqrt{3}}{\frac{1}{5} \sqrt{5}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{5} \sqrt{5}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{4}{5}, \ \frac{3}{5}\right)$$
$$\vec e_2 = \left( - \frac{3}{5}, \ \frac{4}{5}\right)$$
$$18 x^{2} + 24 x y + 11 y^{2} - 5 = 0$$
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
$$a_{11} = 18$$
$$a_{12} = 12$$
$$a_{13} = 0$$
$$a_{22} = 11$$
$$a_{23} = 0$$
$$a_{33} = -5$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}18 & 12\\12 & 11\end{matrix}\right|$$
$$\Delta = 54$$
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
$$18 x_{0} + 12 y_{0} = 0$$
$$12 x_{0} + 11 y_{0} = 0$$
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
$$18 x'^{2} + 24 x' y' + 11 y'^{2} - 5 = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{7}{24}$$
then
$$\phi = \frac{\operatorname{acot}{\left(\frac{7}{24} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{24}{25}$$
$$\cos{\left(2 \phi \right)} = \frac{7}{25}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{4}{5}$$
$$\sin{\left(\phi \right)} = \frac{3}{5}$$
substitute coefficients
$$x' = \frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}$$
$$y' = \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}$$
then the equation turns from
$$18 x'^{2} + 24 x' y' + 11 y'^{2} - 5 = 0$$
to
$$11 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right)^{2} + 24 \left(\frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right) + 18 \left(\frac{4 \tilde x}{5} - \frac{3 \tilde y}{5}\right)^{2} - 5 = 0$$
simplify
$$27 \tilde x^{2} + 2 \tilde y^{2} - 5 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{9} \sqrt{3}}{\frac{1}{5} \sqrt{5}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{5} \sqrt{5}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{4}{5}, \ \frac{3}{5}\right)$$
$$\vec e_2 = \left( - \frac{3}{5}, \ \frac{4}{5}\right)$$
Invariants method
Given line equation of 2-order:
$$18 x^{2} + 24 x y + 11 y^{2} - 5 = 0$$
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
$$a_{11} = 18$$
$$a_{12} = 12$$
$$a_{13} = 0$$
$$a_{22} = 11$$
$$a_{23} = 0$$
$$a_{33} = -5$$
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
$$I_{1} = 29$$
$$I_{3} = \left|\begin{matrix}18 & 12 & 0\\12 & 11 & 0\\0 & 0 & -5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}18 - \lambda & 12\\12 & 11 - \lambda\end{matrix}\right|$$
$$I_{1} = 29$$
$$I_{2} = 54$$
$$I_{3} = -270$$
$$I{\left(\lambda \right)} = \lambda^{2} - 29 \lambda + 54$$
$$K_{2} = -145$$
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
$$\lambda^{2} - 29 \lambda + 54 = 0$$
$$\lambda_{1} = 27$$
$$\lambda_{2} = 2$$
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
$$27 \tilde x^{2} + 2 \tilde y^{2} - 5 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{9} \sqrt{3}}{\frac{1}{5} \sqrt{5}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{5} \sqrt{5}}\right)^{2}} = 1$$
- reduced to canonical form
$$18 x^{2} + 24 x y + 11 y^{2} - 5 = 0$$
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
$$a_{11} = 18$$
$$a_{12} = 12$$
$$a_{13} = 0$$
$$a_{22} = 11$$
$$a_{23} = 0$$
$$a_{33} = -5$$
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
$$I_{1} = 29$$
|18 12|
I2 = | |
|12 11|$$I_{3} = \left|\begin{matrix}18 & 12 & 0\\12 & 11 & 0\\0 & 0 & -5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}18 - \lambda & 12\\12 & 11 - \lambda\end{matrix}\right|$$
|18 0 | |11 0 |
K2 = | | + | |
|0 -5| |0 -5|$$I_{1} = 29$$
$$I_{2} = 54$$
$$I_{3} = -270$$
$$I{\left(\lambda \right)} = \lambda^{2} - 29 \lambda + 54$$
$$K_{2} = -145$$
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
$$\lambda^{2} - 29 \lambda + 54 = 0$$
$$\lambda_{1} = 27$$
$$\lambda_{2} = 2$$
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
$$27 \tilde x^{2} + 2 \tilde y^{2} - 5 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{9} \sqrt{3}}{\frac{1}{5} \sqrt{5}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{5} \sqrt{5}}\right)^{2}} = 1$$
- reduced to canonical form