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- 21x^2+18y^2-14z^2-294x-144y-56z+127=0
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sqrtx+sqrty=1
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Identical expressions
- -30x^ two -1 two xy-35y^2
- minus 30x squared minus 12xy minus 35y squared
- minus 30x to the power of two minus 1 two xy minus 35y squared
- -30x2-12xy-35y2
- -30x²-12xy-35y²
- -30x to the power of 2-12xy-35y to the power of 2
-
Similar expressions
- -30x^2+12xy-35y^2
- 30x^2-12xy-35y^2
- -30x^2-12xy+35y^2
-30x^2-12xy-35y^2 canonical form
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2 2 - 35*y - 30*x - 12*x*y = 0
$$- 30 x^{2} - 12 x y - 35 y^{2} = 0$$
-30*x^2 - 12*x*y - 35*y^2 = 0
Detail solution
Given line equation of 2-order:
$$- 30 x^{2} - 12 x y - 35 y^{2} = 0$$
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
$$a_{11} = -30$$
$$a_{12} = -6$$
$$a_{13} = 0$$
$$a_{22} = -35$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-30 & -6\\-6 & -35\end{matrix}\right|$$
$$\Delta = 1014$$
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
$$- 30 x_{0} - 6 y_{0} = 0$$
$$- 6 x_{0} - 35 y_{0} = 0$$
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
$$- 30 x'^{2} - 12 x' y' - 35 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{5}{12}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{5}{12} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{12}{13}$$
$$\cos{\left(2 \phi \right)} = \frac{5}{13}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{3 \sqrt{13}}{13}$$
$$\sin{\left(\phi \right)} = - \frac{2 \sqrt{13}}{13}$$
substitute coefficients
$$x' = \frac{3 \sqrt{13} \tilde x}{13} + \frac{2 \sqrt{13} \tilde y}{13}$$
$$y' = - \frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}$$
then the equation turns from
$$- 30 x'^{2} - 12 x' y' - 35 y'^{2} = 0$$
to
$$- 35 \left(- \frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}\right)^{2} - 12 \left(- \frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}\right) \left(\frac{3 \sqrt{13} \tilde x}{13} + \frac{2 \sqrt{13} \tilde y}{13}\right) - 30 \left(\frac{3 \sqrt{13} \tilde x}{13} + \frac{2 \sqrt{13} \tilde y}{13}\right)^{2} = 0$$
simplify
$$- 26 \tilde x^{2} - 39 \tilde y^{2} = 0$$
$$26 \tilde x^{2} + 39 \tilde y^{2} = 0$$
Given equation is degenerate ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{26}}{26}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{39}}{39}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{3 \sqrt{13}}{13}, \ - \frac{2 \sqrt{13}}{13}\right)$$
$$\vec e_2 = \left( \frac{2 \sqrt{13}}{13}, \ \frac{3 \sqrt{13}}{13}\right)$$
$$- 30 x^{2} - 12 x y - 35 y^{2} = 0$$
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
$$a_{11} = -30$$
$$a_{12} = -6$$
$$a_{13} = 0$$
$$a_{22} = -35$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-30 & -6\\-6 & -35\end{matrix}\right|$$
$$\Delta = 1014$$
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
$$- 30 x_{0} - 6 y_{0} = 0$$
$$- 6 x_{0} - 35 y_{0} = 0$$
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
$$- 30 x'^{2} - 12 x' y' - 35 y'^{2} = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{5}{12}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{5}{12} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{12}{13}$$
$$\cos{\left(2 \phi \right)} = \frac{5}{13}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{3 \sqrt{13}}{13}$$
$$\sin{\left(\phi \right)} = - \frac{2 \sqrt{13}}{13}$$
substitute coefficients
$$x' = \frac{3 \sqrt{13} \tilde x}{13} + \frac{2 \sqrt{13} \tilde y}{13}$$
$$y' = - \frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}$$
then the equation turns from
$$- 30 x'^{2} - 12 x' y' - 35 y'^{2} = 0$$
to
$$- 35 \left(- \frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}\right)^{2} - 12 \left(- \frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}\right) \left(\frac{3 \sqrt{13} \tilde x}{13} + \frac{2 \sqrt{13} \tilde y}{13}\right) - 30 \left(\frac{3 \sqrt{13} \tilde x}{13} + \frac{2 \sqrt{13} \tilde y}{13}\right)^{2} = 0$$
simplify
$$- 26 \tilde x^{2} - 39 \tilde y^{2} = 0$$
$$26 \tilde x^{2} + 39 \tilde y^{2} = 0$$
Given equation is degenerate ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{26}}{26}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{39}}{39}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)
Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{3 \sqrt{13}}{13}, \ - \frac{2 \sqrt{13}}{13}\right)$$
$$\vec e_2 = \left( \frac{2 \sqrt{13}}{13}, \ \frac{3 \sqrt{13}}{13}\right)$$
Invariants method
Given line equation of 2-order:
$$- 30 x^{2} - 12 x y - 35 y^{2} = 0$$
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
$$a_{11} = -30$$
$$a_{12} = -6$$
$$a_{13} = 0$$
$$a_{22} = -35$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = -65$$
$$I_{3} = \left|\begin{matrix}-30 & -6 & 0\\-6 & -35 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 30 & -6\\-6 & - \lambda - 35\end{matrix}\right|$$
$$I_{1} = -65$$
$$I_{2} = 1014$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + 65 \lambda + 1014$$
$$K_{2} = 0$$
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
$$\lambda^{2} + 65 \lambda + 1014 = 0$$
$$\lambda_{1} = -26$$
$$\lambda_{2} = -39$$
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
$$- 26 \tilde x^{2} - 39 \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{26}}{26}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{39}}{39}\right)^{2}} = 0$$
- reduced to canonical form
$$- 30 x^{2} - 12 x y - 35 y^{2} = 0$$
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
$$a_{11} = -30$$
$$a_{12} = -6$$
$$a_{13} = 0$$
$$a_{22} = -35$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = -65$$
|-30 -6 |
I2 = | |
|-6 -35|$$I_{3} = \left|\begin{matrix}-30 & -6 & 0\\-6 & -35 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 30 & -6\\-6 & - \lambda - 35\end{matrix}\right|$$
|-30 0| |-35 0|
K2 = | | + | |
| 0 0| | 0 0|$$I_{1} = -65$$
$$I_{2} = 1014$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + 65 \lambda + 1014$$
$$K_{2} = 0$$
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
$$\lambda^{2} + 65 \lambda + 1014 = 0$$
$$\lambda_{1} = -26$$
$$\lambda_{2} = -39$$
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
$$- 26 \tilde x^{2} - 39 \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{26}}{26}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{39}}{39}\right)^{2}} = 0$$
- reduced to canonical form