## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 9 - Section 9.4 - Rotation of Axes - Exercise Set - Page 1010: 28

#### Answer

a. $x'^2-9y'^2+36=0$ b. $\frac{y'^2}{4}-\frac{x'^2}{36}=1$ c. see graph #### Work Step by Step

a. Step 1. Given $3xy-4y^{2}+18=0$, we have $A=0, B=3, C=-4$, and the rotation angle $cot(2\theta)=\frac{A-C}{B}=\frac{0+4}{3}=\frac{4}{3}$, Step 2. We have the transformations (see also exercise 18): $x=\frac{\sqrt 10}{10}(3x'-y')$ and $y=\frac{\sqrt 10}{10}(x'+3y')$ Step 3. The equation becomes $3xy-4y^{2}+18=3(\frac{\sqrt 10}{10}(3x'-y'))(\frac{\sqrt 10}{10}(x'+3y'))-4(\frac{\sqrt 10}{10}(x'+3y'))^{2}+18=\frac{1}{2}x'^2-\frac{9}{2}y'^2+18=0$ or, $x'^2-9y'^2+36=0$ b. In standard form, we have $\frac{y'^2}{4}-\frac{x'^2}{36}=1$ which represents a hyperbola. c. We can graph the equation as shown in the figure.

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