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
-
How to use it?
- Canonical form:
- 9x^2+4xy+6y^2+40x+20y-50=0
- 2x^2-5xy+3y^2=0
- 2x_1^2+x_2^2-4x_1x_2-4x_2x_3
- x^2-8x+15
- Plot:
- x^2=y
-
x^3(y-x)=x^3+1
- x^2/9-y^2/4=1
-
x^6+3*x^4*y^2-3*x^4+3*x^2*y^4-x^2*y^3-6*x^2*y^2+3*x^2+y^6-3*y^4+3*y^2-1=0
- Graphing y =:
- -2x^2+3x+5
-
Identical expressions
- - two x^2+3x+ five
- minus 2x squared plus 3x plus 5
- minus two x squared plus 3x plus five
- -2x2+3x+5
- -2x²+3x+5
- -2x to the power of 2+3x+5
-
Similar expressions
- -2x^2+3x-5
- -2x^2-3x+5
- 2x^2+3x+5
-2x^2+3x+5 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given line equation of 2-order:
$$- 2 x^{2} + 3 x + 5 = 0$$
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
$$a_{11} = -2$$
$$a_{12} = 0$$
$$a_{13} = \frac{3}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 5$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-2 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
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
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$- 2 x^{2} + 3 x + 5 = 0$$
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
$$a_{11} = -2$$
$$a_{12} = 0$$
$$a_{13} = \frac{3}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 5$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}-2 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
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:
$$- 2 x^{2} + 3 x + 5 = 0$$
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
$$a_{11} = -2$$
$$a_{12} = 0$$
$$a_{13} = \frac{3}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 5$$
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
$$I_{1} = -2$$
$$I_{3} = \left|\begin{matrix}-2 & 0 & \frac{3}{2}\\0 & 0 & 0\\\frac{3}{2} & 0 & 5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 2 & 0\\0 & - \lambda\end{matrix}\right|$$
$$I_{1} = -2$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + 2 \lambda$$
$$K_{2} = - \frac{49}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge K_{2} < 0 \wedge I_{1} \neq 0$$
then by line type:
this equation is of type : two parallel lines
$$I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
$$\frac{49}{8} - 2 \tilde y^{2} = 0$$
- reduced to canonical form
$$- 2 x^{2} + 3 x + 5 = 0$$
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
$$a_{11} = -2$$
$$a_{12} = 0$$
$$a_{13} = \frac{3}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 5$$
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
$$I_{1} = -2$$
|-2 0|
I2 = | |
|0 0|$$I_{3} = \left|\begin{matrix}-2 & 0 & \frac{3}{2}\\0 & 0 & 0\\\frac{3}{2} & 0 & 5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 2 & 0\\0 & - \lambda\end{matrix}\right|$$
|-2 3/2| |0 0|
K2 = | | + | |
|3/2 5 | |0 5|$$I_{1} = -2$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + 2 \lambda$$
$$K_{2} = - \frac{49}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge K_{2} < 0 \wedge I_{1} \neq 0$$
then by line type:
this equation is of type : two parallel lines
$$I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0$$
or
$$\frac{49}{8} - 2 \tilde y^{2} = 0$$
None
- reduced to canonical form