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: 26

Answer

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

Work Step by Step

a. Step 1. Given $32x^{2}-48xy+18y^{2}-15x-20y=0$ we have $A=32, B=-48, C=18$ The rotation angle is $cot(2\theta)=\frac{A-C}{B}=\frac{32-18}{-48}=-\frac{7}{24}$, Step 2. We have the transformations (see also exercise 21): $x=\frac{1}{5}(3x'-4y')$ and $y=\frac{1}{5}(4x'+3y')$ Step 3. The equation becomes $32x^{2}-48xy+18y^{2}-15x-20y=32(\frac{1}{5}(3x'-4y'))^{2}-48(\frac{1}{5}(3x'-4y'))(\frac{1}{5}(4x'+3y'))+18(\frac{1}{5}(4x'+3y'))^{2}-15(\frac{1}{5}(3x'-4y'))-20(\frac{1}{5}(4x'+3y'))=-25x'+50y'^2=0$ or, $x'-2y'^2=0$ b. In standard form, we have $y'^2=\frac{1}{2}x'$ which represents a parabola. c. We can graph the equation as shown in the figure.
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