lecture 36-draft.pdf - 1 AM 1411a Fall 2014 Lectures 36 December 1 Quadratic forms(Sec 7.3 Conic sections Conics in standard position A general

lecture 36-draft.pdf - 1 AM 1411a Fall 2014 Lectures 36...

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1 AM 1411a, Fall 2014 Lectures 36, December 1 Quadratic forms (Sec. 7.3) Conic sections
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Conics in standard position
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A general quadratic equation in R 2 : ax 2 1 + 2 bx 1 x 2 + cx 2 2 + dx 1 + ex 2 + f = 0 , where a; b; c; d; e; f are constants. It can be rewritten as h x 1 x 2 i " a b b c # " x 1 x 2 # + h d e i " x 1 x 2 # + f = 0 , or x T A x + v T x + f = 0 : Note: a matrix A is symmetric. Ex 1. Express 2 x 2 1 +3 x 1 x 2 ° 4 x 2 2 + x 1 +5 = 0 in matrix notation. h x 1 x 2 i " 2 3 2 3 2 ° 4 # " x 1 x 2 # + h 1 0 i " x 1 x 2 # +5 = 0 :
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A conic in non-standard position To eliminate a cross product term (proportional to x 1 x 2 ) in x T A x + v T x + f = 0 ; use diagonalization of the matrix A ; to eliminate linear terms °shift a center of the coordinate system. Ex 2. Let 3 x 2 1 +2 x 1 x 2 +3 x 2 2 ° 8 = 0 : Identify the conic by eliminating the cross product term. h x 1 x 2 i " 3 1 1 1 # " x 1 x 2 # = 8
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Diagonalize A = " 3 1 1 1 # : 1) det ( °I ° A ) = det " ° ° 3 ° 1 ° 1 ° ° 3 # = 0 ° 2 ° 6 ° + 8 = 0 ° 1 = 2 ; ° 2 = 4 2) (a) ° 1 = 2 ) (2 ± I ° A ) x = 0 " ° 1 ° 1 0 ° 1 ° 1 0 # RREF °°°°! " 1 1 0 0 0 0 # Let x 2 = t , then x 1 = ° t: x = " ° t t # = ° t " 1 ° 1 # ; p 1 = " 1 ° 1 # (b) ° 2 = 4 ) (4 I ° A ) x = 0 )
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" 1 ° 1 0 ° 1 1 0 # RREF °°°°! " 1 ° 1 0 0 0 0 # Let x 2 = t , then x 1 = t: x = " t t # = t " 1 1 # ; p 2 = " 1 1 # p 1 ± p 2 = 0 ) orthogonal. Normalize them: q 1 = p 1 k p 1 k = 1 p 2 " 1 ° 1 # ; q 2 = p 2 k p 2 k = 1 p 2 " 1 1 # Q = 2 4 1 p 2 1 p 2 ° 1 p 2 1 p 2 3 5 is a transformation matrix that rep- resent rotation about the origin on 45
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  • Fall '19
  • #, Conic section, 0

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