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How to use it?
- Canonical form:
-
14x^2+24xy+21y^2-4x-18y-139=0
-
9x^2+4y^2-18x+24y+9=0
- x1x2+x1x3+x1x4+x2x3+x2x4+x3x4
- -2c^3(3+16c^2)-(c^2+4)×3c^3+6(3c^3+1)
- Plot:
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y=x^4-26x^2+25
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xy=0
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xy=6
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4x^2+4y^2=25
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Identical expressions
- 14x^ two + two 4xy+21y^2-4x-18y- one hundred and thirty-nine = zero
- 14x squared plus 24xy plus 21y squared minus 4x minus 18y minus 139 equally 0
- 14x to the power of two plus two 4xy plus 21y squared minus 4x minus 18y minus one hundred and thirty minus nine equally zero
- 14x2+24xy+21y2-4x-18y-139=0
- 14x²+24xy+21y²-4x-18y-139=0
- 14x to the power of 2+24xy+21y to the power of 2-4x-18y-139=0
- 14x^2+24xy+21y^2-4x-18y-139=O
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Similar expressions
- 14x^2+24xy-21y^2-4x-18y-139=0
- 14x^2+24xy+21y^2-4x+18y-139=0
- 14x^2+24xy+21y^2-4x-18y+139=0
- 14x^2+24xy+21y^2+4x-18y-139=0
- 14x^2-24xy+21y^2-4x-18y-139=0
14x^2+24xy+21y^2-4x-18y-139=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
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2 2 -139 - 18*y - 4*x + 14*x + 21*y + 24*x*y = 0
$$14 x^{2} + 24 x y + 21 y^{2} - 4 x - 18 y - 139 = 0$$
14*x^2 + 24*x*y - 4*x + 21*y^2 - 18*y - 139 = 0
Detail solution
Given line equation of 2-order:
$$14 x^{2} + 24 x y + 21 y^{2} - 4 x - 18 y - 139 = 0$$
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
$$a_{11} = 14$$
$$a_{12} = 12$$
$$a_{13} = -2$$
$$a_{22} = 21$$
$$a_{23} = -9$$
$$a_{33} = -139$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}14 & 12\\12 & 21\end{matrix}\right|$$
$$\Delta = 150$$
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
$$14 x_{0} + 12 y_{0} - 2 = 0$$
$$12 x_{0} + 21 y_{0} - 9 = 0$$
then
$$x_{0} = - \frac{11}{25}$$
$$y_{0} = \frac{17}{25}$$
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
$$a'_{33} = - 2 x_{0} - 9 y_{0} - 139$$
$$a'_{33} = - \frac{3606}{25}$$
then The equation is transformed to
$$14 x'^{2} + 24 x' y' + 21 y'^{2} - \frac{3606}{25} = 0$$
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
$$\phi = - \frac{\operatorname{acot}{\left(\frac{7}{24} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{24}{25}$$
$$\cos{\left(2 \phi \right)} = \frac{7}{25}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{- \cos^{2}{\left(\phi \right)} + 1}$$
$$\cos{\left(\phi \right)} = \frac{4}{5}$$
$$\sin{\left(\phi \right)} = - \frac{3}{5}$$
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
$$14 x'^{2} + 24 x' y' + 21 y'^{2} - \frac{3606}{25} = 0$$
to
$$21 \left(- \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right)^{2} + 24 \left(- \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) \left(\frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}\right) + 14 \left(\frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}\right)^{2} - \frac{3606}{25} = 0$$
simplify
$$5 \tilde x^{2} + 30 \tilde y^{2} - \frac{3606}{25} = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{18030}}{25}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{3005}}{25}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \frac{4}{5}, \ - \frac{3}{5}\right)$$
$$\vec e_{2} = \left( \frac{3}{5}, \ \frac{4}{5}\right)$$
$$14 x^{2} + 24 x y + 21 y^{2} - 4 x - 18 y - 139 = 0$$
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
$$a_{11} = 14$$
$$a_{12} = 12$$
$$a_{13} = -2$$
$$a_{22} = 21$$
$$a_{23} = -9$$
$$a_{33} = -139$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}14 & 12\\12 & 21\end{matrix}\right|$$
$$\Delta = 150$$
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
$$14 x_{0} + 12 y_{0} - 2 = 0$$
$$12 x_{0} + 21 y_{0} - 9 = 0$$
then
$$x_{0} = - \frac{11}{25}$$
$$y_{0} = \frac{17}{25}$$
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
$$a'_{33} = - 2 x_{0} - 9 y_{0} - 139$$
$$a'_{33} = - \frac{3606}{25}$$
then The equation is transformed to
$$14 x'^{2} + 24 x' y' + 21 y'^{2} - \frac{3606}{25} = 0$$
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
$$\phi = - \frac{\operatorname{acot}{\left(\frac{7}{24} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{24}{25}$$
$$\cos{\left(2 \phi \right)} = \frac{7}{25}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{- \cos^{2}{\left(\phi \right)} + 1}$$
$$\cos{\left(\phi \right)} = \frac{4}{5}$$
$$\sin{\left(\phi \right)} = - \frac{3}{5}$$
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
$$14 x'^{2} + 24 x' y' + 21 y'^{2} - \frac{3606}{25} = 0$$
to
$$21 \left(- \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right)^{2} + 24 \left(- \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) \left(\frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}\right) + 14 \left(\frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}\right)^{2} - \frac{3606}{25} = 0$$
simplify
$$5 \tilde x^{2} + 30 \tilde y^{2} - \frac{3606}{25} = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{18030}}{25}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{3005}}{25}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
-11 17 (----, --) 25 25
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:
$$14 x^{2} + 24 x y + 21 y^{2} - 4 x - 18 y - 139 = 0$$
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
$$a_{11} = 14$$
$$a_{12} = 12$$
$$a_{13} = -2$$
$$a_{22} = 21$$
$$a_{23} = -9$$
$$a_{33} = -139$$
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
$$I_{1} = 35$$
$$I_{3} = \left|\begin{matrix}14 & 12 & -2\\12 & 21 & -9\\-2 & -9 & -139\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 14 & 12\\12 & - \lambda + 21\end{matrix}\right|$$
$$I_{1} = 35$$
$$I_{2} = 150$$
$$I_{3} = -21636$$
$$I{\left(\lambda \right)} = \lambda^{2} - 35 \lambda + 150$$
$$K_{2} = -4950$$
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:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - 35 \lambda + 150 = 0$$
Solve this equation
$$\lambda_{1} = 30$$
$$\lambda_{2} = 5$$
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
$$30 \tilde x^{2} + 5 \tilde y^{2} - \frac{3606}{25} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{3005}}{25}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{18030}}{25}\right)^{2}} = 1$$
- reduced to canonical form
$$14 x^{2} + 24 x y + 21 y^{2} - 4 x - 18 y - 139 = 0$$
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
$$a_{11} = 14$$
$$a_{12} = 12$$
$$a_{13} = -2$$
$$a_{22} = 21$$
$$a_{23} = -9$$
$$a_{33} = -139$$
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
$$I_{1} = 35$$
|14 12|
I2 = | |
|12 21|$$I_{3} = \left|\begin{matrix}14 & 12 & -2\\12 & 21 & -9\\-2 & -9 & -139\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 14 & 12\\12 & - \lambda + 21\end{matrix}\right|$$
|14 -2 | |21 -9 |
K2 = | | + | |
|-2 -139| |-9 -139|$$I_{1} = 35$$
$$I_{2} = 150$$
$$I_{3} = -21636$$
$$I{\left(\lambda \right)} = \lambda^{2} - 35 \lambda + 150$$
$$K_{2} = -4950$$
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:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - 35 \lambda + 150 = 0$$
Solve this equation
$$\lambda_{1} = 30$$
$$\lambda_{2} = 5$$
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
$$30 \tilde x^{2} + 5 \tilde y^{2} - \frac{3606}{25} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{3005}}{25}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{18030}}{25}\right)^{2}} = 1$$
- reduced to canonical form
The graph