Mister Exam

Other calculators

Canonical form/ 3y^2+5x+6y+13=0

3y^2+5x+6y+13=0 canonical form

The teacher will be very surprised to see your correct solution 😉

v

The graph:

x: [, ]
y: [, ]
z: [, ]

Quality:

 (Number of points on the axis)

Plot type:

The solution

You have entered [src] 
        2                
13 + 3*y  + 5*x + 6*y = 0
5x+3y2+6y+13=05 x + 3 y^{2} + 6 y + 13 = 0 
5*x + 3*y^2 + 6*y + 13 = 0
Detail solution
Given line equation of 2-order:
5x+3y2+6y+13=05 x + 3 y^{2} + 6 y + 13 = 0
This equation looks like:
a11x2+2a12xy+2a13x+a22y2+2a23y+a33=0a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0
where
a11=0a_{11} = 0
a12=0a_{12} = 0
a13=52a_{13} = \frac{5}{2}
a22=3a_{22} = 3
a23=3a_{23} = 3
a33=13a_{33} = 13
To calculate the determinant
Δ=∣a11a12a12a22∣\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|
or, substitute
Δ=∣0003∣\Delta = \left|\begin{matrix}0 & 0\\0 & 3\end{matrix}\right|
Δ=0\Delta = 0
Because
Δ\Delta
is equal to 0, then
Given equation is straight line
- reduced to canonical form
The center of the canonical coordinate system in OXY
x0=x~cos⁡(ϕ)−y~sin⁡(ϕ)x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}
y0=x~sin⁡(ϕ)+y~cos⁡(ϕ)y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}
x0=0⋅0x_{0} = 0 \cdot 0
y0=0⋅0y_{0} = 0 \cdot 0
x0=0x_{0} = 0
y0=0y_{0} = 0
The center of canonical coordinate system at point O
(0, 0)

Basis of the canonical coordinate system
e⃗1=(1, 0)\vec e_1 = \left( 1, \ 0\right)
e⃗2=(0, 1)\vec e_2 = \left( 0, \ 1\right)
Invariants method
Given line equation of 2-order:
5x+3y2+6y+13=05 x + 3 y^{2} + 6 y + 13 = 0
This equation looks like:
a11x2+2a12xy+2a13x+a22y2+2a23y+a33=0a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0
where
a11=0a_{11} = 0
a12=0a_{12} = 0
a13=52a_{13} = \frac{5}{2}
a22=3a_{22} = 3
a23=3a_{23} = 3
a33=13a_{33} = 13
The invariants of the equation when converting coordinates are determinants:
I1=a11+a22I_{1} = a_{11} + a_{22}
     |a11  a12|
I2 = |        |
     |a12  a22|

I3=∣a11a12a13a12a22a23a13a23a33∣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(λ)=∣a11−λa12a12a22−λ∣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
I1=3I_{1} = 3
     |0  0|
I2 = |    |
     |0  3|

I3=∣005203352313∣I_{3} = \left|\begin{matrix}0 & 0 & \frac{5}{2}\\0 & 3 & 3\\\frac{5}{2} & 3 & 13\end{matrix}\right|
I(λ)=∣−λ003−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 3 - \lambda\end{matrix}\right|
     | 0   5/2|   |3  3 |
K2 = |        | + |     |
     |5/2  13 |   |3  13|

I1=3I_{1} = 3
I2=0I_{2} = 0
I3=−754I_{3} = - \frac{75}{4}
I(λ)=λ2−3λI{\left(\lambda \right)} = \lambda^{2} - 3 \lambda
K2=954K_{2} = \frac{95}{4}
Because
I2=0∧I3≠0I_{2} = 0 \wedge I_{3} \neq 0
then by line type:
this equation is of type : parabola
I1y~2+2x~−I3I1=0I_{1} \tilde y^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0
or
5x~+3y~2=05 \tilde x + 3 \tilde y^{2} = 0
y~2=5x~3\tilde y^{2} = \frac{5 \tilde x}{3}
- reduced to canonical form
    © Mister Exam – Calculator