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: 20

Answer

a. $3x'^2-y'^2=3$ b. $\frac{x'^2}{1}-\frac{y'^2}{3}=1$ c. see graph
1585959185

Work Step by Step

a. Step 1. Given $x^2+4xy+y^2-3=0$, we have $A=C=1, B=4$. The rotation angle is $cot(2\theta)=\frac{A-C}{B}=0, 2\theta=\frac{\pi}{2}, \theta=\frac{\pi}{4}$ Step 2. We have the transformations: $x=\frac{\sqrt 2}{2}(x'-y')$ and $y=\frac{\sqrt 2}{2}(x'+y')$ Step 3. The equation becomes $x^2+4xy+y^2-3=(\frac{\sqrt 2}{2}(x'-y'))^2+4(\frac{\sqrt 2}{2}(x'-y'))(\frac{\sqrt 2}{2}(x'+y'))+(\frac{\sqrt 2}{2}(x'+y'))^2-3=3x'^2-y'^2-3=0$ or $3x'^2-y'^2=3$ b. In standard form, we have $\frac{x'^2}{1}-\frac{y'^2}{3}=1$ which represents a hyperbola. c. We can graph the equation as shown in the figure.
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.