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:
- 4x^2+8x-16y+16=0
- x^2+y^2-2x+1=0
- 5x^2–6y^2–20x–36y–64=0
- 4x^2+4y^2-12x+4y+3=0
- Plot:
- z=-x^2-x*y+y+3*x+6*y
-
y^4-x^4=x*y
- y^2=4*(x-1)/x^2
- y^2=9*x
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Identical expressions
- 5y^ two +5x+ six = zero
- 5y squared plus 5x plus 6 equally 0
- 5y to the power of two plus 5x plus six equally zero
- 5y2+5x+6=0
- 5y²+5x+6=0
- 5y to the power of 2+5x+6=0
- 5y^2+5x+6=O
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Similar expressions
- 5y^2-5x+6=0
- 5y^2+5x-6=0
5y^2+5x+6=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$5 x + 5 y^{2} + 6 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{5}{2}$$
$$a_{22} = 5$$
$$a_{23} = 0$$
$$a_{33} = 6$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 5\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$5 \tilde y^{2} = - 5 \tilde x - 6$$
$$\tilde y^{2} = - \tilde x - \frac{6}{5}$$
$$\tilde y'^{2} = - \tilde x'$$
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
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$5 x + 5 y^{2} + 6 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{5}{2}$$
$$a_{22} = 5$$
$$a_{23} = 0$$
$$a_{33} = 6$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 5\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$5 \tilde y^{2} = - 5 \tilde x - 6$$
$$\tilde y^{2} = - \tilde x - \frac{6}{5}$$
$$\tilde y'^{2} = - \tilde x'$$
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:
$$5 x + 5 y^{2} + 6 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{5}{2}$$
$$a_{22} = 5$$
$$a_{23} = 0$$
$$a_{33} = 6$$
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
$$I_{1} = 5$$
$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{5}{2}\\0 & 5 & 0\\\frac{5}{2} & 0 & 6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 5 - \lambda\end{matrix}\right|$$
$$I_{1} = 5$$
$$I_{2} = 0$$
$$I_{3} = - \frac{125}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda$$
$$K_{2} = \frac{95}{4}$$
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
$$5 \tilde x + 5 \tilde y^{2} = 0$$
$$\tilde y^{2} = \tilde x$$
- reduced to canonical form
$$5 x + 5 y^{2} + 6 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{5}{2}$$
$$a_{22} = 5$$
$$a_{23} = 0$$
$$a_{33} = 6$$
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
$$I_{1} = 5$$
|0 0|
I2 = | |
|0 5|$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{5}{2}\\0 & 5 & 0\\\frac{5}{2} & 0 & 6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 5 - \lambda\end{matrix}\right|$$
| 0 5/2| |5 0|
K2 = | | + | |
|5/2 6 | |0 6|$$I_{1} = 5$$
$$I_{2} = 0$$
$$I_{3} = - \frac{125}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 5 \lambda$$
$$K_{2} = \frac{95}{4}$$
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
$$5 \tilde x + 5 \tilde y^{2} = 0$$
$$\tilde y^{2} = \tilde x$$
- reduced to canonical form