## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 9 - Section 9.4 - Rotation of Axes - Exercise Set - Page 1010: 25

#### Answer

a. $26x'^2+y'^2=9$ b. $\frac{x'^2}{9/26}+\frac{y'^2}{9}=1$ c. see graph #### Work Step by Step

a. Step 1. Given $10x^{2}+24xy+17y^{2}-9=0$ we have $A=10, B=24, C=17$ The rotation angle $cot(2\theta)=\frac{A-C}{B}=\frac{10-17}{24}=-\frac{7}{24}$ 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 $10x^{2}+24xy+17y^{2}-9=10(\frac{1}{5}(3x'-4y'))^{2}+24(\frac{1}{5}(3x'-4y'))(\frac{1}{5}(4x'+3y'))+17(\frac{1}{5}(4x'+3y'))^{2}-9=26x'^2+y'^2-9=0$ or $26x'^2+y'^2=9$ b. In standard form, we have $\frac{x'^2}{9/26}+\frac{y'^2}{9}=1$ which represents an ellipse. c. We can graph the equation as shown in the figure.

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