## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 9 - Section 9.4 - Rotation of Axes - Exercise Set - Page 1010: 27

#### Answer

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

a. Step 1. Given $x^{2}+4xy-2y^{2}-1=0$ we have $A=1, B=4, C=-2$ The rotation angle is $cot(2\theta)=\frac{A-C}{B}=\frac{1+2}{4}=\frac{3}{4}$, Step 2. We have the transformations (see also exercise 23): $x=\frac{\sqrt 5}{5}(2x'-y')$ and $y=\frac{\sqrt 5}{5}(x'+2y')$ Step 3. The equation becomes $x^{2}+4xy-2y^{2}-1=(\frac{\sqrt 5}{5}(2x'-y'))^{2}+4(\frac{\sqrt 5}{5}(2x'-y'))(\frac{\sqrt 5}{5}(x'+2y'))-2(\frac{\sqrt 5}{5}(x'+2y'))^{2}-1=2x'^2-3y'^2-1=0$ or, $2x'^2-3y'^2=1$ b. In standard form, we have $\frac{x'^2}{1/2}-\frac{y'^2}{1/3}=1$, which represents a hyperbola. c. We can graph the equation as shown in the figure.

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