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Andreza Costa Batista.pdf - mtc-m17:80 - Inpe

Andreza Costa Batista.pdf - mtc-m17:80 - Inpe

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Particularizando para nosso problema, temos:Da equação da elipse temos:2 2x y+ = 1(5.9)2 b2alogo:22 =2 xy b1−(5.10)2 a ou ainda,y2=b2b−a22x2(5.11)que também pode ser escrito como:ou ainda,2 2[ y ] = [ x 1. ] b −a2 b22(5.12) y1y2 y3 yi22 x1 x2 = x3 xi2222221 b1 −a1. 2 b 1 22(5.13)Y=X.A, onde:Y y1y2= y3 y i22222x11 2 x212, X = x 1e3 2 x i1A = b −a2 b22(5.14)25

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