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A8_soln

# A8_soln - Math 235 Assignment 8 Solutions 1 Sketch the...

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Unformatted text preview: Math 235 Assignment 8 Solutions 1. Sketch the graph of 5 x 2 + 6 xy − 3 y 2 = 15 showing both the original and new axes. Solution: The corresponding symmetric matrix is bracketleftbigg 5 3 3 − 3 bracketrightbigg . We find that the characteristic polynomial is C ( λ ) = vextendsingle vextendsingle vextendsingle vextendsingle 5 − λ 3 3 − 3 − λ vextendsingle vextendsingle vextendsingle vextendsingle = λ 2 − 2 λ − 24 = ( λ − 6)( λ + 4) . So, we have eigenvalues λ 1 = 5 and λ 2 = − 4. For λ 1 = 6 we get A − λ 1 I = bracketleftbigg − 1 3 3 − 9 bracketrightbigg ∼ bracketleftbigg 1 − 3 bracketrightbigg . Thus, a corresponding eigenvector is vectorv 1 = bracketleftbigg 3 1 bracketrightbigg . For λ 2 we get A − λ 2 I = bracketleftbigg 9 3 3 1 bracketrightbigg ∼ bracketleftbigg 1 1 / 3 bracketrightbigg . A corresponding eigenvector is vectorv 2 = bracketleftbigg − 1 3 bracketrightbigg . Thus, we have the hyperbola 6 x 2 1 − 4 y 2 1 = 15 with principal axis vectorv 1 for x 1 and vectorv 2 for y 1 . The asymptotes of the hyperbola are when 0 = 6 x 2 1 − 4 y 2 1 = ( √ 6 x 1 + 2 y 1 )( √ 6 x 1 − 2 y 1 ) We have (finding the inverse of the change of basis matrix by taking the transpose) bracketleftbigg x 1 y 1 bracketrightbigg = 1 √ 10 bracketleftbigg 3 1 − 1 3 bracketrightbiggbracketleftbigg x y bracketrightbigg = 1 √ 10 bracketleftbigg 3 x + y x − 3 y bracketrightbigg ....
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A8_soln - Math 235 Assignment 8 Solutions 1 Sketch the...

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