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: 46


asymptotes: $y=\pm\frac{2}{3}(x-3)+1$ foci: $(3,1\pm\sqrt {13})$

Work Step by Step

Step 1. Rewriting the equation as $9(y^2-2y+1)-4(x^2-6x+9)=63+9-36=36$ or $\frac{(y-1)^2}{4}-\frac{(x-3)^2}{9}=1$ we have $a=2,b=3, c=\sqrt {a^2+b^2}=\sqrt {13}$ centered at $(3,1)$ with a vertical transverse axis. Step 2. We can find the vertices as $(3,-1),(3,3)$ and asymptotes as $y-1=\pm\frac{a}{b}(x-3)$ or $y=\pm\frac{2}{3}(x-3)+1$ Step 3. We can graph the equation as shown in the figure with foci at $(3,1\pm\sqrt {13})$
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