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: 51

Answer

a. $x'^2+6y'=0$ b. $x'^2=-6y'$ . c. see graph
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Work Step by Step

a. Step 1. Given $x^{2}+2\sqrt 3xy+3y^{2}-12\sqrt 3x+12y=0$ we have $A=1, B=2\sqrt 3, C=3$ and the rotation angle is $cot(2\theta)=\frac{A-C}{B}=\frac{1-3}{2\sqrt 3}=-\frac{\sqrt 3}{3}, 2\theta=\frac{2\pi}{3}, \theta=\frac{\pi}{3}$, Step 2. We have the transformations (see also section 9.4 exercise 34): $x=\frac{1}{2}(x'-\sqrt 3y')$ and $y=\frac{1}{2}(\sqrt 3x'+y')$ Step 3. The equation becomes $x^{2}+2\sqrt 3xy+3y^{2}-12\sqrt 3x+12y=(\frac{1}{2}(x'-\sqrt 3y'))^{2}+2\sqrt 3(\frac{1}{2}(x'-\sqrt 3y'))(\frac{1}{2}(\sqrt 3x'+y'))+3(\frac{1}{2}(\sqrt 3x'+y'))^{2}-12\sqrt 3(\frac{1}{2}(x'-\sqrt 3y'))+12(\frac{1}{2}(\sqrt 3x'+y'))=4x'^2+24y'=0$ or, $x'^2+6y'=0$ b. In standard form, we have $x'^2=-6y'$, which represents a parabola. c. We can graph the equation as shown in the figure
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