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:
- 2x^2-3y^2+z^2+4x+6y-4z-3=0
- y=4x^2-16x
- y=(2x)/(3x+1)
- x+z=6
- Plot:
-
tg(x*y+0.4)-x^2
- 4*x^2+y-3*x=10
- (x-1)^2+y^2=9
- -1+y^2/b^2+x^2/a^2=0
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Identical expressions
- y=4x^ two -16x
- y equally 4x squared minus 16x
- y equally 4x to the power of two minus 16x
- y=4x2-16x
- y=4x²-16x
- y=4x to the power of 2-16x
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Similar expressions
- y=4x^2+16x
y=4x^2-16x canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$- 4 x^{2} + 16 x + y = 0$$
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
$$a_{11} = -4$$
$$a_{12} = 0$$
$$a_{13} = 8$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-4 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\left(2 \tilde x + 4\right)^{2} = \tilde y + 16$$
$$\left(\tilde x + 2\right)^{2} = \frac{\tilde y}{4} + 4$$
$$\tilde x'^{2} = \frac{\tilde y}{4} + 4$$
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
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$- 4 x^{2} + 16 x + y = 0$$
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
$$a_{11} = -4$$
$$a_{12} = 0$$
$$a_{13} = 8$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-4 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\left(2 \tilde x + 4\right)^{2} = \tilde y + 16$$
$$\left(\tilde x + 2\right)^{2} = \frac{\tilde y}{4} + 4$$
$$\tilde x'^{2} = \frac{\tilde y}{4} + 4$$
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:
$$- 4 x^{2} + 16 x + y = 0$$
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
$$a_{11} = -4$$
$$a_{12} = 0$$
$$a_{13} = 8$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 0$$
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
$$I_{1} = -4$$
$$I_{3} = \left|\begin{matrix}-4 & 0 & 8\\0 & 0 & \frac{1}{2}\\8 & \frac{1}{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 4 & 0\\0 & - \lambda\end{matrix}\right|$$
$$I_{1} = -4$$
$$I_{2} = 0$$
$$I_{3} = 1$$
$$I{\left(\lambda \right)} = \lambda^{2} + 4 \lambda$$
$$K_{2} = - \frac{257}{4}$$
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
$$\tilde x - 4 \tilde y^{2} = 0$$
$$\tilde y^{2} = - \frac{\tilde x}{4}$$
- reduced to canonical form
$$- 4 x^{2} + 16 x + y = 0$$
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
$$a_{11} = -4$$
$$a_{12} = 0$$
$$a_{13} = 8$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 0$$
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
$$I_{1} = -4$$
|-4 0|
I2 = | |
|0 0|$$I_{3} = \left|\begin{matrix}-4 & 0 & 8\\0 & 0 & \frac{1}{2}\\8 & \frac{1}{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 4 & 0\\0 & - \lambda\end{matrix}\right|$$
|-4 8| | 0 1/2|
K2 = | | + | |
|8 0| |1/2 0 |$$I_{1} = -4$$
$$I_{2} = 0$$
$$I_{3} = 1$$
$$I{\left(\lambda \right)} = \lambda^{2} + 4 \lambda$$
$$K_{2} = - \frac{257}{4}$$
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
$$\tilde x - 4 \tilde y^{2} = 0$$
$$\tilde y^{2} = - \frac{\tilde x}{4}$$
- reduced to canonical form