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: 24

Answer

a. $x'^2+4y'^2=4$ b. $\frac{x'^2}{4}+\frac{y'^2}{1}=1$ c. see graph
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Work Step by Step

a. Step 1. Given $7x^{2}-6\sqrt{3}xy+13y^{2}-16=0$ we have $A=7, B=-6\sqrt 3, C=13$ and the rotation angle is $cot(2\theta)=\frac{A-C}{B}=\frac{7-13}{-6\sqrt 3}=\frac{\sqrt 3}{3}, 2\theta=\frac{\pi}{3}, \theta=\frac{\pi}{6}$ Step 2. We have the transformations (see also exercise 12): $x=\frac{1}{2}(\sqrt 3x'-y')$ and $y=\frac{1}{2}(x'+\sqrt 3y')$ Step 3. The equation becomes $7x^{2}-6\sqrt{3}xy+13y^{2}-16=7(\frac{1}{2}(\sqrt 3x'-y'))^{2}-6\sqrt{3}(\frac{1}{2}(\sqrt 3x'-y'))(\frac{1}{2}(x'+\sqrt 3y'))+13(\frac{1}{2}(x'+\sqrt 3y'))^{2}-16=4x'^2+16y'^2-16=0$ or, $x'^2+4y'^2=4$ b. In standard form, we have $\frac{x'^2}{4}+\frac{y'^2}{1}=1$, which represents an ellipse. c. We can graph the equation as shown in the figure.
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