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How to use it?
- Canonical form:
- y²-y=0
- 16x^2+25y^2-64x+50y-311=0
-
x1^2+4x2^2
- 9x²-81y²=729
- Plot:
-
(-x^2)*y^3+(x^2+y^2-1)^3=0
-
x^2+(y+cbrt(x^2))^2=1
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sin(x^e)+cos(y^e)=1
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Identical expressions
- 16x^ two + two 5y^2-64x+5 zero y- three hundred and eleven =0
- 16x squared plus 25y squared minus 64x plus 50y minus 311 equally 0
- 16x to the power of two plus two 5y squared minus 64x plus 5 zero y minus three hundred and eleven equally 0
- 16x2+25y2-64x+50y-311=0
- 16x²+25y²-64x+50y-311=0
- 16x to the power of 2+25y to the power of 2-64x+50y-311=0
- 16x^2+25y^2-64x+50y-311=O
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Similar expressions
- 16x^2-25y^2-64x+50y-311=0
- 16x^2+25y^2+64x+50y-311=0
- 16x^2+25y^2-64x+50y+311=0
- 16x^2+25y^2-64x-50y-311=0
16x^2+25y^2-64x+50y-311=0 canonical form
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The solution
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2 2 -311 - 64*x + 16*x + 25*y + 50*y = 0
$$16 x^{2} - 64 x + 25 y^{2} + 50 y - 311 = 0$$
16*x^2 - 64*x + 25*y^2 + 50*y - 311 = 0
Detail solution
Given line equation of 2-order:
$$16 x^{2} - 64 x + 25 y^{2} + 50 y - 311 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = 0$$
$$a_{13} = -32$$
$$a_{22} = 25$$
$$a_{23} = 25$$
$$a_{33} = -311$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}16 & 0\\0 & 25\end{matrix}\right|$$
$$\Delta = 400$$
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
$$16 x_{0} - 32 = 0$$
$$25 y_{0} + 25 = 0$$
then
$$x_{0} = 2$$
$$y_{0} = -1$$
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
$$a'_{33} = - 32 x_{0} + 25 y_{0} - 311$$
$$a'_{33} = -400$$
then equation turns into
$$16 x'^{2} + 25 y'^{2} - 400 = 0$$
Given equation is ellipse
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$16 x^{2} - 64 x + 25 y^{2} + 50 y - 311 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = 0$$
$$a_{13} = -32$$
$$a_{22} = 25$$
$$a_{23} = 25$$
$$a_{33} = -311$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}16 & 0\\0 & 25\end{matrix}\right|$$
$$\Delta = 400$$
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
$$16 x_{0} - 32 = 0$$
$$25 y_{0} + 25 = 0$$
then
$$x_{0} = 2$$
$$y_{0} = -1$$
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
$$a'_{33} = - 32 x_{0} + 25 y_{0} - 311$$
$$a'_{33} = -400$$
then equation turns into
$$16 x'^{2} + 25 y'^{2} - 400 = 0$$
Given equation is ellipse
2 2
\tilde x \tilde y
--------- + --------- = 1
2 2
/ 1 \ / 1 \
|------| |------|
\4*1/20/ \5*1/20/ - reduced to canonical form
The center of canonical coordinate system at point O
(2, -1)
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
Invariants method
Given line equation of 2-order:
$$16 x^{2} - 64 x + 25 y^{2} + 50 y - 311 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = 0$$
$$a_{13} = -32$$
$$a_{22} = 25$$
$$a_{23} = 25$$
$$a_{33} = -311$$
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
$$I_{1} = 41$$
$$I_{3} = \left|\begin{matrix}16 & 0 & -32\\0 & 25 & 25\\-32 & 25 & -311\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}16 - \lambda & 0\\0 & 25 - \lambda\end{matrix}\right|$$
$$I_{1} = 41$$
$$I_{2} = 400$$
$$I_{3} = -160000$$
$$I{\left(\lambda \right)} = \lambda^{2} - 41 \lambda + 400$$
$$K_{2} = -14400$$
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
$$\lambda^{2} - 41 \lambda + 400 = 0$$
$$\lambda_{1} = 25$$
$$\lambda_{2} = 16$$
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
$$25 \tilde x^{2} + 16 \tilde y^{2} - 400 = 0$$
- reduced to canonical form
$$16 x^{2} - 64 x + 25 y^{2} + 50 y - 311 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = 0$$
$$a_{13} = -32$$
$$a_{22} = 25$$
$$a_{23} = 25$$
$$a_{33} = -311$$
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
$$I_{1} = 41$$
|16 0 |
I2 = | |
|0 25|$$I_{3} = \left|\begin{matrix}16 & 0 & -32\\0 & 25 & 25\\-32 & 25 & -311\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}16 - \lambda & 0\\0 & 25 - \lambda\end{matrix}\right|$$
|16 -32 | |25 25 |
K2 = | | + | |
|-32 -311| |25 -311|$$I_{1} = 41$$
$$I_{2} = 400$$
$$I_{3} = -160000$$
$$I{\left(\lambda \right)} = \lambda^{2} - 41 \lambda + 400$$
$$K_{2} = -14400$$
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
$$\lambda^{2} - 41 \lambda + 400 = 0$$
$$\lambda_{1} = 25$$
$$\lambda_{2} = 16$$
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
$$25 \tilde x^{2} + 16 \tilde y^{2} - 400 = 0$$
2 2
\tilde x \tilde y
--------- + --------- = 1
2 2
/ 1 \ / 1 \
|------| |------|
\5*1/20/ \4*1/20/ - reduced to canonical form