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: 30

Answer

a. $x'^2+3y'^2=9$ b. $\frac{x'^2}{9}+\frac{y'^2}{3}=1$ c. see graph
1586011514

Work Step by Step

a. Step 1. Given $6x^{2}-6xy+14y^{2}-45=0$, we have $A=6, B=-6, C=14$, and the rotation angle $cot(2\theta)=\frac{A-C}{B}=\frac{6-14}{-6}=\frac{4}{3}$ Step 2. We have the transformations (see also exercise 18): $x=\frac{\sqrt{10}}{10}(3x'-y')$ and $y=\frac{\sqrt{10}}{10}(x'+3y')$ Step 3. The equation becomes $6x^{2}-6xy+14y^{2}-45=6(\frac{\sqrt{10}}{10}(3x'-y'))^{2}-6(\frac{\sqrt{10}}{10}(3x'-y'))(\frac{\sqrt{10}}{10}(x'+3y'))+14(\frac{\sqrt{10}}{10}(x'+3y'))^{2}-45=5x'^2+15y'^2-45=0$ or, $x'^2+3y'^2=9$ b. In standard form, we have $\frac{x'^2}{9}+\frac{y'^2}{3}=1$ which represents an ellipse. 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.