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: 37



Work Step by Step

Step 1. Given $5x^{2}-6xy+5y^{2}-8=0$ we have $A=5, B=-6, C=5$ and the rotation angle is $cot(2\theta)=\frac{A-C}{B}=\frac{5-5}{-6}=0, 2\theta=\frac{\pi}{2}, \theta=\frac{\pi}{4}$ Step 2. We have the transformations (see also exercise 7): $x=\frac{\sqrt 2}{2}(x'-y')$ and $y=\frac{\sqrt 2}{2}(x'+y')$ Step 3. The equation becomes $5x^{2}-6xy+5y^{2}-8=5(\frac{\sqrt 2}{2}(x'-y'))^{2}-6(\frac{\sqrt 2}{2}(x'-y'))(\frac{\sqrt 2}{2}(x'+y'))+5(\frac{\sqrt 2}{2}(x'+y'))^{2}-8=2x'^2+8y'^2-8=0$ or, $x'^2+4y'^2=4$ Step 4. In standard form, we have $\frac{x'^2}{4}+\frac{y'^2}{1}=1$, which represents an ellipse. Step 5. We can find the endpoints of the minor axis in the $x'-y'$ system as $(0,\pm1)$
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