Mister Exam
Other calculators
- Canonical form of a elliptical paraboloid
- Canonical form of a double hyperboloid
- Canonical form of a imaginary ellipsoid
- Canonical form of a degenerate ellipse
- Canonical form of a parabola
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How to use it?
- Canonical form:
- 5*x^2+4*y^2-4*z^2-2*x+2*y-4*z-3*y*z-3*x*z-x*y=0
- 8x21+16x1x2+14x1x3−28x2+32x2x3+14x23
- x^2-4*x+9*y^2+18*y-4*z^2-24*z-23=0
- 6y^2-12x+24y-5=0
- Plot:
- 9*(x*cos(a)-y*sin(a))^2+24*(x*cos(a)-y*sin(a))*(y*sin(a)+y*cos(a))+16*(y*sin(a)+y*cos(a))^2+50*(x*cos(a)-y*sin(a))-100*(y*sin(a)+y*cos(a))+25=0
-
y=sin(60*x)
- y*(-x)=x2*e^x
-
x^5+y^5=1
-
Identical expressions
- 6y^ two -12x+24y- five = zero
- 6y squared minus 12x plus 24y minus 5 equally 0
- 6y to the power of two minus 12x plus 24y minus five equally zero
- 6y2-12x+24y-5=0
- 6y²-12x+24y-5=0
- 6y to the power of 2-12x+24y-5=0
- 6y^2-12x+24y-5=O
-
Similar expressions
- 6y^2-12x-24y-5=0
- 6y^2-12x+24y+5=0
- 6y^2+12x+24y-5=0
6y^2-12x+24y-5=0 canonical form
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The solution
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2 -5 - 12*x + 6*y + 24*y = 0
$$- 12 x + 6 y^{2} + 24 y - 5 = 0$$
-12*x + 6*y^2 + 24*y - 5 = 0
Detail solution
Given line equation of 2-order:
$$- 12 x + 6 y^{2} + 24 y - 5 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = -6$$
$$a_{22} = 6$$
$$a_{23} = 12$$
$$a_{33} = -5$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 6\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
Given equation is straight line
- 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
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$- 12 x + 6 y^{2} + 24 y - 5 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = -6$$
$$a_{22} = 6$$
$$a_{23} = 12$$
$$a_{33} = -5$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 6\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
Given equation is straight line
- 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:
$$- 12 x + 6 y^{2} + 24 y - 5 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = -6$$
$$a_{22} = 6$$
$$a_{23} = 12$$
$$a_{33} = -5$$
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
$$I_{1} = 6$$
$$I_{3} = \left|\begin{matrix}0 & 0 & -6\\0 & 6 & 12\\-6 & 12 & -5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 6 - \lambda\end{matrix}\right|$$
$$I_{1} = 6$$
$$I_{2} = 0$$
$$I_{3} = -216$$
$$I{\left(\lambda \right)} = \lambda^{2} - 6 \lambda$$
$$K_{2} = -210$$
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
$$12 \tilde x + 6 \tilde y^{2} = 0$$
$$\tilde y^{2} = 2 \tilde x$$
- reduced to canonical form
$$- 12 x + 6 y^{2} + 24 y - 5 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = -6$$
$$a_{22} = 6$$
$$a_{23} = 12$$
$$a_{33} = -5$$
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
$$I_{1} = 6$$
|0 0|
I2 = | |
|0 6|$$I_{3} = \left|\begin{matrix}0 & 0 & -6\\0 & 6 & 12\\-6 & 12 & -5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 6 - \lambda\end{matrix}\right|$$
|0 -6| |6 12|
K2 = | | + | |
|-6 -5| |12 -5|$$I_{1} = 6$$
$$I_{2} = 0$$
$$I_{3} = -216$$
$$I{\left(\lambda \right)} = \lambda^{2} - 6 \lambda$$
$$K_{2} = -210$$
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
$$12 \tilde x + 6 \tilde y^{2} = 0$$
$$\tilde y^{2} = 2 \tilde x$$
- reduced to canonical form