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Other calculators
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- Canonical form of a double hyperboloid
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- Canonical form of a degenerate ellipse
- Canonical form of a parabola
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How to use it?
- Canonical form:
-
x^2+8y^2+8x+14y+5=0
- 14x^2-7y^2+20xy-8x-34y-35
- (y+3)^2=4x-2
- 5x^2+y^2+10x-6y-10z+14=0
- Plot:
-
x^2*y=x^2-y^2
-
-0.47y-1.02y^3+0.5ln((1+y)/(1-y))=x(1-exp(2.2*x/2))
- x^2+y^2=16
-
ln((y+x)/(x+3))=1+(1/(x+y))
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Identical expressions
- x^ two +8y^ two +8x+14y+ five = zero
- x squared plus 8y squared plus 8x plus 14y plus 5 equally 0
- x to the power of two plus 8y to the power of two plus 8x plus 14y plus five equally zero
- x2+8y2+8x+14y+5=0
- x²+8y²+8x+14y+5=0
- x to the power of 2+8y to the power of 2+8x+14y+5=0
- x^2+8y^2+8x+14y+5=O
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Similar expressions
- x^2+8y^2+8x+14y-5=0
- x^2+8y^2+8x-14y+5=0
- x^2-8y^2+8x+14y+5=0
- x^2+8y^2-8x+14y+5=0
x^2+8y^2+8x+14y+5=0 canonical form
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2 2 5 + x + 8*x + 8*y + 14*y = 0
$$x^{2} + 8 y^{2} + 8 x + 14 y + 5 = 0$$
x^2 + 8*x + 8*y^2 + 14*y + 5 = 0
Detail solution
Given line equation of 2-order:
$$x^{2} + 8 y^{2} + 8 x + 14 y + 5 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 4$$
$$a_{22} = 8$$
$$a_{23} = 7$$
$$a_{33} = 5$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & 8\end{matrix}\right|$$
$$\Delta = 8$$
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
$$x_{0} + 4 = 0$$
$$8 y_{0} + 7 = 0$$
then
$$x_{0} = -4$$
$$y_{0} = - \frac{7}{8}$$
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
$$a'_{33} = 4 x_{0} + 7 y_{0} + 5$$
$$a'_{33} = - \frac{137}{8}$$
then The equation is transformed to
$$x'^{2} + 8 y'^{2} - \frac{137}{8} = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{274}}{4}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{137}}{8}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$x^{2} + 8 y^{2} + 8 x + 14 y + 5 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 4$$
$$a_{22} = 8$$
$$a_{23} = 7$$
$$a_{33} = 5$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & 8\end{matrix}\right|$$
$$\Delta = 8$$
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
$$x_{0} + 4 = 0$$
$$8 y_{0} + 7 = 0$$
then
$$x_{0} = -4$$
$$y_{0} = - \frac{7}{8}$$
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
$$a'_{33} = 4 x_{0} + 7 y_{0} + 5$$
$$a'_{33} = - \frac{137}{8}$$
then The equation is transformed to
$$x'^{2} + 8 y'^{2} - \frac{137}{8} = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{274}}{4}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{137}}{8}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(-4, -7/8)
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:
$$x^{2} + 8 y^{2} + 8 x + 14 y + 5 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 4$$
$$a_{22} = 8$$
$$a_{23} = 7$$
$$a_{33} = 5$$
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
$$I_{1} = 9$$
$$I_{3} = \left|\begin{matrix}1 & 0 & 4\\0 & 8 & 7\\4 & 7 & 5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 1 & 0\\0 & - \lambda + 8\end{matrix}\right|$$
$$I_{1} = 9$$
$$I_{2} = 8$$
$$I_{3} = -137$$
$$I{\left(\lambda \right)} = \lambda^{2} - 9 \lambda + 8$$
$$K_{2} = -20$$
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
$$\lambda^{2} - 9 \lambda + 8 = 0$$
Solve this equation
$$\lambda_{1} = 8$$
$$\lambda_{2} = 1$$
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
$$8 \tilde x^{2} + \tilde y^{2} - \frac{137}{8} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{137}}{8}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{274}}{4}\right)^{2}} = 1$$
- reduced to canonical form
$$x^{2} + 8 y^{2} + 8 x + 14 y + 5 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 4$$
$$a_{22} = 8$$
$$a_{23} = 7$$
$$a_{33} = 5$$
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
$$I_{1} = 9$$
|1 0|
I2 = | |
|0 8|$$I_{3} = \left|\begin{matrix}1 & 0 & 4\\0 & 8 & 7\\4 & 7 & 5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 1 & 0\\0 & - \lambda + 8\end{matrix}\right|$$
|1 4| |8 7|
K2 = | | + | |
|4 5| |7 5|$$I_{1} = 9$$
$$I_{2} = 8$$
$$I_{3} = -137$$
$$I{\left(\lambda \right)} = \lambda^{2} - 9 \lambda + 8$$
$$K_{2} = -20$$
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
$$\lambda^{2} - 9 \lambda + 8 = 0$$
Solve this equation
$$\lambda_{1} = 8$$
$$\lambda_{2} = 1$$
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
$$8 \tilde x^{2} + \tilde y^{2} - \frac{137}{8} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{137}}{8}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{274}}{4}\right)^{2}} = 1$$
- reduced to canonical form
The graph