## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 9 - Section 9.4 - Rotation of Axes - Exercise Set - Page 1010: 40

#### Answer

$\frac{x'^2}{3}-\frac{y'^2}{2}=1$, asymptotes $y' =\pm\frac{\sqrt 6}{3}x'$

#### Work Step by Step

Step 1. Given $x^{2}+4xy-2y^{2}-6=0$ we have $A=1, B=4, C=-2$ and 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 15): $x=\frac{\sqrt 5}{5}(2x'-y')$ and $y=\frac{\sqrt 5}{5}(x'+2y')$ Step 3. The equation becomes $x^{2}+4xy-2y^{2}-6=(\frac{\sqrt 5}{5}(2x'-y'))^{2}+4(\frac{\sqrt 5}{5}(2x'-y'))(\frac{\sqrt 5}{5}(x'+2y'))-2(\frac{\sqrt 5}{5}(x'+2y'))^{2}-6=2x'^2-3y'^2-6=0$ or, $2x'^2-3y'^2=6$ Step 4. In standard form, we have $\frac{x'^2}{3}-\frac{y'^2}{2}=1$ which represents a hyperbola. Step 5. We have $a=\sqrt 3, b=\sqrt 2$ and the asymptotes in the $x'-y'$ system as $y'=\pm\frac{b}{a}x'=\pm\frac{\sqrt 6}{3}x'$

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