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:
- 4𝑥1𝑥3−6𝑥2𝑥4
-
7y^2+3x-28y
- x*x
- 4x-y-z+12=0
- Plot:
-
e^y/x=x*y+1
- log(x)=y
-
3*x*y^2+9*x*y+y^2+6*x+3*y=2022
- x^2+y^2+10z^2=100
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Identical expressions
- 7y^ two +3x-28y
- 7y squared plus 3x minus 28y
- 7y to the power of two plus 3x minus 28y
- 7y2+3x-28y
- 7y²+3x-28y
- 7y to the power of 2+3x-28y
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Similar expressions
- 7y^2-3x-28y
- 7y^2+3x+28y
7y^2+3x-28y canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$7 y^{2} + 3 x - 28 y = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{3}{2}$$
$$a_{22} = 7$$
$$a_{23} = -14$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 7\end{matrix}\right|$$
$$\Delta = 0$$
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
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 1 + 0 \cdot 0$$
$$y_{0} = 0 \cdot 0 + 0 \cdot 1$$
$$x_{0} = 0$$
$$y_{0} = 0$$
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( 1, \ 0\right)$$
$$\vec e_{2} = \left( 0, \ 1\right)$$
$$7 y^{2} + 3 x - 28 y = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{3}{2}$$
$$a_{22} = 7$$
$$a_{23} = -14$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & 0\\0 & 7\end{matrix}\right|$$
$$\Delta = 0$$
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
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 1 + 0 \cdot 0$$
$$y_{0} = 0 \cdot 0 + 0 \cdot 1$$
$$x_{0} = 0$$
$$y_{0} = 0$$
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:
$$7 y^{2} + 3 x - 28 y = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{3}{2}$$
$$a_{22} = 7$$
$$a_{23} = -14$$
$$a_{33} = 0$$
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
$$I_{1} = 7$$
$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{3}{2}\\0 & 7 & -14\\\frac{3}{2} & -14 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & - \lambda + 7\end{matrix}\right|$$
$$I_{1} = 7$$
$$I_{2} = 0$$
$$I_{3} = - \frac{63}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 7 \lambda$$
$$K_{2} = - \frac{793}{4}$$
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
$$7 \tilde y^{2} + 3 \tilde x = 0$$
$$\tilde y^{2} = \frac{3 \tilde x}{7}$$
- reduced to canonical form
$$7 y^{2} + 3 x - 28 y = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{3}{2}$$
$$a_{22} = 7$$
$$a_{23} = -14$$
$$a_{33} = 0$$
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
$$I_{1} = 7$$
|0 0|
I2 = | |
|0 7|$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{3}{2}\\0 & 7 & -14\\\frac{3}{2} & -14 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & - \lambda + 7\end{matrix}\right|$$
| 0 3/2| | 7 -14|
K2 = | | + | |
|3/2 0 | |-14 0 |$$I_{1} = 7$$
$$I_{2} = 0$$
$$I_{3} = - \frac{63}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 7 \lambda$$
$$K_{2} = - \frac{793}{4}$$
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
$$7 \tilde y^{2} + 3 \tilde x = 0$$
$$\tilde y^{2} = \frac{3 \tilde x}{7}$$
- reduced to canonical form
The graph