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How to use it?
- Canonical form:
-
3x^2+10xy+3y^2-2x-14y-13
- 377
- x^2+y^2+z^2-xy+xz+yz+3x+3y-3z=0
- -x^2-10x^2-6x1x2
- Plot:
- x^2+(y-sqrt(|x|))^2=1
- 3(5a-b+6c)=3(5a−b+6c)
- xy^2
- (x2+y2–1)^3-x2y3=0
-
Identical expressions
- 3x^ two +10xy+3y^ two -2x-14y- thirteen
- 3x squared plus 10xy plus 3y squared minus 2x minus 14y minus 13
- 3x to the power of two plus 10xy plus 3y to the power of two minus 2x minus 14y minus thirteen
- 3x2+10xy+3y2-2x-14y-13
- 3x²+10xy+3y²-2x-14y-13
- 3x to the power of 2+10xy+3y to the power of 2-2x-14y-13
-
Similar expressions
- 3x^2+10xy+3y^2-2x-14y+13
- 3x^2+10xy+3y^2+2x-14y-13
- 3x^2+10xy-3y^2-2x-14y-13
- 3x^2-10xy+3y^2-2x-14y-13
- 3x^2+10xy+3y^2-2x+14y-13
3x^2+10xy+3y^2-2x-14y-13 canonical form
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The solution
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2 2 -13 - 14*y - 2*x + 3*x + 3*y + 10*x*y = 0
$$3 x^{2} + 10 x y + 3 y^{2} - 2 x - 14 y - 13 = 0$$
3*x^2 + 10*x*y - 2*x + 3*y^2 - 14*y - 13 = 0
Detail solution
Given line equation of 2-order:
$$3 x^{2} + 10 x y + 3 y^{2} - 2 x - 14 y - 13 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 3$$
$$a_{12} = 5$$
$$a_{13} = -1$$
$$a_{22} = 3$$
$$a_{23} = -7$$
$$a_{33} = -13$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 5\\5 & 3\end{matrix}\right|$$
$$\Delta = -16$$
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
$$3 x_{0} + 5 y_{0} - 1 = 0$$
$$5 x_{0} + 3 y_{0} - 7 = 0$$
then
$$x_{0} = 2$$
$$y_{0} = -1$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - x_{0} - 7 y_{0} - 13$$
$$a'_{33} = -8$$
then The equation is transformed to
$$3 x'^{2} + 10 x' y' + 3 y'^{2} - 8 = 0$$
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
$$\cot{\left(2 \phi \right)} = 0$$
then
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{- \cos^{2}{\left(\phi \right)} + 1}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
substitute coefficients
$$x' = \frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}$$
$$y' = \frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}$$
then the equation turns from
$$3 x'^{2} + 10 x' y' + 3 y'^{2} - 8 = 0$$
to
$$3 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right)^{2} + 10 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) + 3 \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 8 = 0$$
simplify
$$8 \tilde x^{2} - 2 \tilde y^{2} - 8 = 0$$
$$- 8 \tilde x^{2} + 2 \tilde y^{2} + 8 = 0$$
Given equation is hyperbole
$$\tilde x^{2} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$\vec e_{2} = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$3 x^{2} + 10 x y + 3 y^{2} - 2 x - 14 y - 13 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 3$$
$$a_{12} = 5$$
$$a_{13} = -1$$
$$a_{22} = 3$$
$$a_{23} = -7$$
$$a_{33} = -13$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}3 & 5\\5 & 3\end{matrix}\right|$$
$$\Delta = -16$$
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
$$3 x_{0} + 5 y_{0} - 1 = 0$$
$$5 x_{0} + 3 y_{0} - 7 = 0$$
then
$$x_{0} = 2$$
$$y_{0} = -1$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - x_{0} - 7 y_{0} - 13$$
$$a'_{33} = -8$$
then The equation is transformed to
$$3 x'^{2} + 10 x' y' + 3 y'^{2} - 8 = 0$$
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
$$\cot{\left(2 \phi \right)} = 0$$
then
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{- \cos^{2}{\left(\phi \right)} + 1}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
substitute coefficients
$$x' = \frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}$$
$$y' = \frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}$$
then the equation turns from
$$3 x'^{2} + 10 x' y' + 3 y'^{2} - 8 = 0$$
to
$$3 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right)^{2} + 10 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) + 3 \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 8 = 0$$
simplify
$$8 \tilde x^{2} - 2 \tilde y^{2} - 8 = 0$$
$$- 8 \tilde x^{2} + 2 \tilde y^{2} + 8 = 0$$
Given equation is hyperbole
$$\tilde x^{2} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(2, -1)
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$\vec e_{2} = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
Invariants method
Given line equation of 2-order:
$$3 x^{2} + 10 x y + 3 y^{2} - 2 x - 14 y - 13 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 3$$
$$a_{12} = 5$$
$$a_{13} = -1$$
$$a_{22} = 3$$
$$a_{23} = -7$$
$$a_{33} = -13$$
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
$$I_{1} = 6$$
$$I_{3} = \left|\begin{matrix}3 & 5 & -1\\5 & 3 & -7\\-1 & -7 & -13\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 3 & 5\\5 & - \lambda + 3\end{matrix}\right|$$
$$I_{1} = 6$$
$$I_{2} = -16$$
$$I_{3} = 128$$
$$I{\left(\lambda \right)} = \lambda^{2} - 6 \lambda - 16$$
$$K_{2} = -128$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - 6 \lambda - 16 = 0$$
Solve this equation
$$\lambda_{1} = 8$$
$$\lambda_{2} = -2$$
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
$$8 \tilde x^{2} - 2 \tilde y^{2} - 8 = 0$$
$$\tilde x^{2} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
$$3 x^{2} + 10 x y + 3 y^{2} - 2 x - 14 y - 13 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 3$$
$$a_{12} = 5$$
$$a_{13} = -1$$
$$a_{22} = 3$$
$$a_{23} = -7$$
$$a_{33} = -13$$
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
$$I_{1} = 6$$
|3 5|
I2 = | |
|5 3|$$I_{3} = \left|\begin{matrix}3 & 5 & -1\\5 & 3 & -7\\-1 & -7 & -13\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 3 & 5\\5 & - \lambda + 3\end{matrix}\right|$$
|3 -1 | |3 -7 |
K2 = | | + | |
|-1 -13| |-7 -13|$$I_{1} = 6$$
$$I_{2} = -16$$
$$I_{3} = 128$$
$$I{\left(\lambda \right)} = \lambda^{2} - 6 \lambda - 16$$
$$K_{2} = -128$$
Because
$$I_{2} < 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - 6 \lambda - 16 = 0$$
Solve this equation
$$\lambda_{1} = 8$$
$$\lambda_{2} = -2$$
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
$$8 \tilde x^{2} - 2 \tilde y^{2} - 8 = 0$$
$$\tilde x^{2} - \frac{\tilde y^{2}}{4} = 1$$
- reduced to canonical form
The graph