Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Section 9.4 - Rotation of Axes - Exercise Set - Page 1010: 6

Answer

$\frac{(x')^2}{4}+\frac{(y')^2}{1}=1$

Work Step by Step

Step 1. Using the transformation formulas $x=x'cos\theta-y'sin\theta$ $y=x'sin\theta+y'cos\theta$ we have $x=x'cos60^\circ-y'sin60^\circ=\frac{1}{2}(x'-\sqrt 3y')$ $y=x'sin60^\circ+y'cos60^\circ=\frac{1}{2}(\sqrt 3x'+y')$ Step 2. The original equation becomes: $13x^2-6\sqrt 3xy+7y^2-16=13(\frac{1}{2}(x'-\sqrt 3y'))^2-6\sqrt 3(\frac{1}{2}(x'-\sqrt 3y'))(\frac{1}{2}(\sqrt 3x'+y'))+7(\frac{1}{2}(\sqrt 3x'+y'))^2-16=\frac{13}{4}((x')^2-2\sqrt 3x'y'+3(y')^2)-\frac{6\sqrt 3}{4}(\sqrt 3(x')^2-2x'y'-\sqrt 3(y')^2)+\frac{7}{4}(3(x')^2+2\sqrt 3x'y'+(y')^2)-16=4(x')^2+16(y')^2-16=0$ Step 3. We can rewrite the above result as $\frac{(x')^2}{4}+\frac{(y')^2}{1}=1$, which represents an ellipse.
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