Unformatted text preview: 1 Review Last lecture we learnt how to orthogonally diagonalise a real symmetric matrix A. Quadratic Forms: Conic Sections 2 The equation: ax 2 + bxy + cy 2 + dx + ey + f = 0 can always be
written in matrix form:
x y a
b/2 b/2
c x
+d
y e x
+f =0
y or as xt Ax + K x + f = 0
where A= a
b/2 b/2
c x= x
y and K= d e. xt Ax is called a quadratic form in two variables. More generally,
if A is a symmetric n × n (real) matrix and x ∈ Rn , xt Ax is a
quadratic form in n variables. General procedure for identifying conics. 3 Case 1: K = [0 0].
(a) Write equation as xt Ax + f = 0 where A is real and symmetric.
(b) Find P orthogonally diagonalising A and such that det(P ) = 1.
(c) Substitute x = P x to transform equation to
(x )t P t AP x + f = (x )t D x + f = λ1 (x )2 + λ2 (y )2 + f = 0
where D has diagonal entries λ1 and λ2 .
(d) Rearrange into standard form to identify conic.
Note: If P is orthogonal and det(P ) = 1 then P is a rotation:
P= cos θ
sin θ − sin θ
.
cos θ 4
Case 2: K ≠ [0 0].
Do (a), (b) and (c) as before to get:
λ1 (x )2 + λ2 (y )2 + KP x + f = 0
or
λ1 (x )2 + λ2 (y )2 + c x + d y + f = 0.
(d) Complete the square to get
λ1 (x − x0 )2 + λ2 (y − y0 )2 = f
(e) Substitute x = x − x0 and y
standard form to identify conic. = y = y0 and rearrange into ...
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 Spring '12
 ankit
 Algebra, Conic Sections, Conic section, Quadratic form, Symmetric matrix, xT Ax

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