## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 9 - Section 9.4 - Rotation of Axes - Exercise Set - Page 1010: 22

#### Answer

a. $x'^2+9y'^2=9$ b. $\frac{x'^2}{9}+\frac{y'^2}{1}=1$ c. see graph

#### Work Step by Step

a. Step 1. Given $5x^{2}-8xy+5y^{2}-9=0$ we have $A=C=5, B=-8$ The rotation angle is $cot(2\theta)=\frac{A-C}{B}=0, 2\theta=\frac{\pi}{2}, \theta=\frac{\pi}{4}$ Step 2. We have the transformations: $x=\frac{\sqrt 2}{2}(x'-y')$ and $y=\frac{\sqrt 2}{2}(x'+y')$ Step 3. The equation becomes $5x^{2}-8xy+5y^{2}-9=5(\frac{\sqrt 2}{2}(x'-y'))^{2}-8(\frac{\sqrt 2}{2}(x'-y'))(\frac{\sqrt 2}{2}(x'+y'))+5(\frac{\sqrt 2}{2}(x'+y'))^{2}-9=x'^2+9y'^2-9=0$ or $x'^2+9y'^2=9$ b. In standard form, we have $\frac{x'^2}{9}+\frac{y'^2}{1}=1$ which represents an ellipse. c. We can graph the equation as shown in the figure.

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