Precalculus (6th Edition) Blitzer

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

Chapter 9 - Review Exercises - Page 1036: 47


a. $x'^2-y'^2=8$ b. $\frac{x'^2}{8}-\frac{y'^2}{8}=1$ c. See graph.

Work Step by Step

a. From the given equation, we have $A=C=0, B=1$. Thus $cot\theta=\frac{A-C}{B}=0$, $2\theta=90^\circ$ and $\theta=45^\circ$. Use the transformation formulas: $x=x'cos\theta-y'sin\theta=x'cos45^\circ-y'sin45^\circ=\frac{\sqrt 2}{2}(x'-y')$ $y=x'sin\theta+y'cos\theta=x'sin45^\circ+y'cos45^\circ=\frac{\sqrt 2}{2}(x'+y')$ The original equation becomes $xy-4=(\frac{\sqrt 2}{2}(x'-y'))(\frac{\sqrt 2}{2}(x'+y'))-4=\frac{1}{2}(x'^2-y'^2)-4=0$ or $x'^2-y'^2=8$ b. We can express the equation involving x′ and y′ in the standard form of a conic section as $\frac{x'^2}{8}-\frac{y'^2}{8}=1$ c. See graph.
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