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: 50


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

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 is $cot(2\theta)=\frac{A-C}{B}=\frac{6-14}{-6}=\frac{4}{3}$, Step 2. We have the transformations (see also section 9.4 exercise 16): $x=\frac{\sqrt 10}{10}(3x'-y')$ 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
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