This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ( ) = vextendsingle vextendsingle vextendsingle vextendsingle 5 3 3 3 vextendsingle vextendsingle vextendsingle vextendsingle = 2 2 24 = ( 6)( + 4) . So, we have eigenvalues 1 = 5 and 2 = 4. For 1 = 6 we get A 1 I = bracketleftbigg 1 3 3 9 bracketrightbigg bracketleftbigg 1 3 bracketrightbigg . Thus, a corresponding eigenvector is vectorv 1 = bracketleftbigg 3 1 bracketrightbigg . For 2 we get A 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 ....
View
Full
Document
This note was uploaded on 05/19/2011 for the course MATH 235 taught by Professor Celmin during the Fall '08 term at Waterloo.
 Fall '08
 CELMIN
 Math, Linear Algebra, Algebra

Click to edit the document details