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How to use it?
- Canonical form:
- x^2+2y^2-3z^2-8x+12y-6z-23=0
- x^2+8x-27y+70=0
- sqrt((x^2/16+y^2/25-1))
- 9(x²-10x)-4(y²+2y)+185=0
- Plot:
- y=sin(x+pi/4)^(2)-1
-
y=x*sin(x)^2+x*cos(x)^2
- y=(2*x+3)*e5*x/1
- y*(-x)=x2*e^x
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Identical expressions
- x^ two +8x-27y+ seventy = zero
- x squared plus 8x minus 27y plus 70 equally 0
- x to the power of two plus 8x minus 27y plus seventy equally zero
- x2+8x-27y+70=0
- x²+8x-27y+70=0
- x to the power of 2+8x-27y+70=0
- x^2+8x-27y+70=O
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Similar expressions
- x^2+8x+27y+70=0
- x^2+8x-27y-70=0
- x^2-8x-27y+70=0
x^2+8x-27y+70=0 canonical form
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2 70 + x - 27*y + 8*x = 0
$$x^{2} + 8 x - 27 y + 70 = 0$$
x^2 + 8*x - 27*y + 70 = 0
Detail solution
Given line equation of 2-order:
$$x^{2} + 8 x - 27 y + 70 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 4$$
$$a_{22} = 0$$
$$a_{23} = - \frac{27}{2}$$
$$a_{33} = 70$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\left(\tilde x + 4\right)^{2} = 27 \tilde y - 54$$
$$\tilde x'^{2} = 27 \tilde y - 54$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$y_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$x^{2} + 8 x - 27 y + 70 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 4$$
$$a_{22} = 0$$
$$a_{23} = - \frac{27}{2}$$
$$a_{33} = 70$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\left(\tilde x + 4\right)^{2} = 27 \tilde y - 54$$
$$\tilde x'^{2} = 27 \tilde y - 54$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$y_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
(0, 0)
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:
$$x^{2} + 8 x - 27 y + 70 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 4$$
$$a_{22} = 0$$
$$a_{23} = - \frac{27}{2}$$
$$a_{33} = 70$$
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
$$I_{1} = 1$$
$$I_{3} = \left|\begin{matrix}1 & 0 & 4\\0 & 0 & - \frac{27}{2}\\4 & - \frac{27}{2} & 70\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = - \frac{729}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda$$
$$K_{2} = - \frac{513}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : parabola
$$I_{1} \tilde y^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0$$
or
$$27 \tilde x + \tilde y^{2} = 0$$
$$\tilde y^{2} = 27 \tilde x$$
- reduced to canonical form
$$x^{2} + 8 x - 27 y + 70 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 4$$
$$a_{22} = 0$$
$$a_{23} = - \frac{27}{2}$$
$$a_{33} = 70$$
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
$$I_{1} = 1$$
|1 0|
I2 = | |
|0 0|$$I_{3} = \left|\begin{matrix}1 & 0 & 4\\0 & 0 & - \frac{27}{2}\\4 & - \frac{27}{2} & 70\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda\end{matrix}\right|$$
|1 4 | | 0 -27/2|
K2 = | | + | |
|4 70| |-27/2 70 |$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = - \frac{729}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda$$
$$K_{2} = - \frac{513}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : parabola
$$I_{1} \tilde y^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0$$
or
$$27 \tilde x + \tilde y^{2} = 0$$
$$\tilde y^{2} = 27 \tilde x$$
- reduced to canonical form