x2−2xy+y2−3x+2y+3=0 (1) We give a quadratic form:
x2−2xy+y2to the canonical form
B=(1−1−11)
Find the eigenvalues and eigenvectors of this matrix:
(1−λ)x1−1y1=0−2x1+(1−λ)y1=0
Characteristic equation:
∣∣1−λ−1−11−λ∣∣=λ2−2λ=0
λ2−2λ=0D=(−2)2−4∗1=4
Find the main axes of the quadratic form, that is, the eigenvectors of the matrix B.
λ1=0x1−y1=0−x1+y1=0 or
x1−y1=0
The eigenvector (λ1=0,forx1=1)
:
∣x1ˉ∣=12+12=2 So
The coordinates of the eigenvector (eigenvalue λ2=2):
−x1−y1=0−x1−y1=0or
−x1−y1=0
The eigenvector
(λ2=2,forx1=1)
∣x2ˉ∣=12+(−)12=2
Moving on to a new basis:
From (1) obtain
(21x1+21y1)2−2(21x1+21y1)(21x1−21y1)++(21x1−21y1)2−3(21x1+21y1)+2(21x1−21y1)+3=0
Now, completing the squares, we get
(y1−4521)2=21(21x1−1623)
Or
(y1−425)2=24121(x1−16232)
We received the equation of a parabola:
(y−y0)2=2p(x−x0),p=421
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