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: 39


$(y'+1)^2=-(x'-3)$, vertex $(3,-1)$

Work Step by Step

Step 1. Given $x^{2}-4xy+4y^{2}+5\sqrt 5y-10=0$ we have $A=1, B=-4, C=4$ and the rotation angle is $cot(2\theta)=\frac{A-C}{B}=\frac{1-4}{-4}=\frac{3}{4}$ 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 $x^{2}-4xy+4y^{2}+5\sqrt 5y-10=(\frac{\sqrt 5}{5}(2x'-y'))^{2}-4(\frac{\sqrt 5}{5}(2x'-y'))(\frac{\sqrt 5}{5}(x'+2y'))+4(\frac{\sqrt 5}{5}(x'+2y'))^{2}+5\sqrt 5(\frac{\sqrt 5}{5}(x'+2y'))-10=5x;+5y'^2+10y'-10$ or, $x'+y'^2+2y'-2=0$ and $x'+(y'+1)^2-3=0$ Step 4. In standard form, we have $(y'+1)^2=-(x'-3)$, which represents a parabola. Step 5. We can find the vertex in the $x'-y'$ system as $(3,-1)$
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