Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Section 9.4 - Rotation of Axes - Exercise Set - Page 1010: 3



Work Step by Step

Step 1. Using the transformation formulas $x=x'cos\theta-y'sin\theta$ $y=x'sin\theta+y'cos\theta$ we have $x=x'cos45^\circ-y'sin45^\circ=\frac{\sqrt 2}{2}(x'-y')$ $y=x'sin45^\circ+y'cos45^\circ=\frac{\sqrt 2}{2}(x'+y')$ Step 2. The original equation becomes: $x^2-4xy+y^2-3=(\frac{\sqrt 2}{2}(x'-y'))^2-4(\frac{\sqrt 2}{2}(x'-y'))(\frac{\sqrt 2}{2}(x'+y'))+(\frac{\sqrt 2}{2}(x'+y'))^2-3=\frac{1}{2}((x')^2-2x'y'+(y')^2)-2((x')^2-(y')^2)+\frac{1}{2}((x')^2+2x'y'+(y')^2)-3=-(x')^2+3(y')^2-3=0$ Step 3. We can rewrite the above result as $\frac{(y')^2}{1}-\frac{(x')^2}{3}=1$, which represents a hyperbola.
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