## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 9 - Section 9.4 - Rotation of Axes - Exercise Set - Page 1010: 19

#### Answer

a. $3x'^2+y'^2=20$ b. $\frac{3x'^2}{20}+\frac{y'^2}{20}=1$ c. see graph #### Work Step by Step

a. Step 1. Given $x^2+xy+y^2-10=0$, we have $A=B=C=1$. 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 $x^2+xy+y^2-10=(\frac{\sqrt 2}{2}(x'-y'))^2+(\frac{\sqrt 2}{2}(x'-y'))(\frac{\sqrt 2}{2}(x'+y'))+(\frac{\sqrt 2}{2}(x'+y'))^2-10=\frac{3}{2}x'^2+\frac{1}{2}y'^210=0$ or $3x'^2+y'^2=20$ b. In standard form, we have $\frac{3x'^2}{20}+\frac{y'^2}{20}=1$ which represents an ellipse. c. We can graph the equation as shown in the figure.

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