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: 29


a. $x'^2+2y'^2-1=0$ b. $\frac{x'^2}{1}+\frac{y'^2}{1/2}=1$ c. see graph

Work Step by Step

a. Step 1. Given $34x^{2}-24xy+41y^{2}-25=0$, we have $A=34, B=-24, C=41$, and the rotation angle $cot(2\theta)=\frac{A-C}{B}=\frac{34-41}{-24}=\frac{7}{24}$, Step 2. We have the transformations (see also exercise 17): $x=\frac{1}{5}(4x'-3y')$ and $y=\frac{1}{5}(3x'+4y')$ Step 3. The equation becomes $34x^{2}-24xy+41y^{2}-25=34(\frac{1}{5}(4x'-3y'))^{2}-24(\frac{1}{5}(4x'-3y'))(\frac{1}{5}(3x'+4y'))+41(\frac{1}{5}(3x'+4y'))^{2}-25=25x'^2+50y'^2-25=0$ or, $x'^2+2y'^2-1=0$ b. In standard form, we have $\frac{x'^2}{1}+\frac{y'^2}{1/2}=1$ which represents an ellipse. c. We can graph the equation as shown in the figure.
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