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: 2

Answer

$\frac{(y')^2}{8}-\frac{(x')^2}{8}=1$

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: $xy=(\frac{\sqrt 2}{2}(x'-y'))(\frac{\sqrt 2}{2}(x'+y'))=\frac{1}{2}((x')^2-(y')^2)=-4$ Step 3. We can rewrite the above result as $\frac{(y')^2}{8}-\frac{(x')^2}{8}=1$, which represents a hyperbola.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.