Precalculus (6th Edition) Blitzer

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

Chapter 9 - Section 9.2 - The Hyperbola - Exercise Set - Page 982: 47

Answer

asymptotes: $y=\pm\frac{2}{3}(x-2)+3$ foci: $(2\pm\sqrt {13},3)$
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Work Step by Step

Step 1. Rewriting the equation as $4(x^2-4x+4)-9(y^2-6y+9)=101+16-81=36$ or $\frac{(x-2)^2}{9}-\frac{(y-3)^2}{4}=1$ we have $a=3,b=2, c=\sqrt {a^2+b^2}=\sqrt {13}$ centered at $(2,3)$ with a horizontal transverse axis. Step 2. We can find the vertices as $(-1,3),(5,3)$ and asymptotes as $y-3=\pm\frac{b}{a}(x-2)$ or $y=\pm\frac{2}{3}(x-2)+3$ Step 3. We can graph the equation as shown in the figure with foci at $(2\pm\sqrt {13},3)$
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