Mister Exam

Canonical form of a parabola

For example, you have entered:
x2−2xy−10x+y2−6y+25=0x^{2} - 2 x y - 10 x + y^{2} - 6 y + 25 = 0

Detail solution

Given line equation of 2-order:
x2−2xy−10x+y2−6y+25=0x^{2} - 2 x y - 10 x + y^{2} - 6 y + 25 = 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=1a_{11} = 1
a12=−1a_{12} = -1
a13=−5a_{13} = -5
a22=1a_{22} = 1
a23=−3a_{23} = -3
a33=25a_{33} = 25
To calculate the determinant
Δ=∣a11a12a12a22∣\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|
or, substitute
Δ=∣1−1−11∣\Delta = \left|\begin{matrix}1 & -1\\-1 & 1\end{matrix}\right|
Δ=0\Delta = 0
Because
Δ\Delta
is equal to 0, then
Rotate the resulting coordinate system by an angle φ
x′=x~cos⁡(ϕ)−y~sin⁡(ϕ)x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}
y′=x~sin⁡(ϕ)+y~cos⁡(ϕ)y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}
φ - determined from the formula
cot⁡(2ϕ)=a11−a222a12\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}
substitute coefficients
cot⁡(2ϕ)=0\cot{\left(2 \phi \right)} = 0
then
ϕ=π4\phi = \frac{\pi}{4}
sin⁡(2ϕ)=1\sin{\left(2 \phi \right)} = 1
cos⁡(2ϕ)=0\cos{\left(2 \phi \right)} = 0
cos⁡(ϕ)=cos⁡(2ϕ)2+12\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}
sin⁡(ϕ)=1−cos⁡2(ϕ)\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}
cos⁡(ϕ)=22\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}
sin⁡(ϕ)=22\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}
substitute coefficients
x′=2x~2−2y~2x' = \frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}
y′=2x~2+2y~2y' = \frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}
then the equation turns from
x′2−2x′y′−10x′+y′2−6y′+25=0x'^{2} - 2 x' y' - 10 x' + y'^{2} - 6 y' + 25 = 0
to
(2x~2−2y~2)2−2(2x~2−2y~2)(2x~2+2y~2)−10(2x~2−2y~2)+(2x~2+2y~2)2−6(2x~2+2y~2)+25=0\left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 2 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) - 10 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) + \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 6 \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) + 25 = 0
simplify
−82x~+2y~2+22y~+25=0- 8 \sqrt{2} \tilde x + 2 \tilde y^{2} + 2 \sqrt{2} \tilde y + 25 = 0
(2y~+1)2=82x~−24\left(\sqrt{2} \tilde y + 1\right)^{2} = 8 \sqrt{2} \tilde x - 24
(y~+22)2=42(x~−322)\left(\tilde y + \frac{\sqrt{2}}{2}\right)^{2} = 4 \sqrt{2} \left(\tilde x - \frac{3 \sqrt{2}}{2}\right)
y~′2=42x~′\tilde y'^{2} = 4 \sqrt{2} \tilde x'
Given equation is by parabola
- 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=022+022x_{0} = 0 \frac{\sqrt{2}}{2} + 0 \frac{\sqrt{2}}{2}
y0=022+022y_{0} = 0 \frac{\sqrt{2}}{2} + 0 \frac{\sqrt{2}}{2}
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=(22, 22)\vec e_1 = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)
e⃗2=(−22, 22)\vec e_2 = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)

Invariants method

Given line equation of 2-order:
x2−2xy−10x+y2−6y+25=0x^{2} - 2 x y - 10 x + y^{2} - 6 y + 25 = 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=1a_{11} = 1
a12=−1a_{12} = -1
a13=−5a_{13} = -5
a22=1a_{22} = 1
a23=−3a_{23} = -3
a33=25a_{33} = 25
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=2I_{1} = 2
     |1   -1|
I2 = |      |
     |-1  1 |

I3=∣1−1−5−11−3−5−325∣I_{3} = \left|\begin{matrix}1 & -1 & -5\\-1 & 1 & -3\\-5 & -3 & 25\end{matrix}\right|
I(λ)=∣1−λ−1−11−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & -1\\-1 & 1 - \lambda\end{matrix}\right|
     |1   -5|   |1   -3|
K2 = |      | + |      |
     |-5  25|   |-3  25|

I1=2I_{1} = 2
I2=0I_{2} = 0
I3=−64I_{3} = -64
I(λ)=λ2−2λI{\left(\lambda \right)} = \lambda^{2} - 2 \lambda
K2=16K_{2} = 16
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
82x~+2y~2=08 \sqrt{2} \tilde x + 2 \tilde y^{2} = 0
y~2=42x~\tilde y^{2} = 4 \sqrt{2} \tilde x
- reduced to canonical form
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