Precalculus (6th Edition) Blitzer

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

Chapter 10 - Section 10.4 - Mathematical Induction - Exercise Set - Page 1086: 48

Answer

foci $(2\pm\sqrt {41},-3)$, asymptotes $y=\pm\frac{5}{4}(x-2)-3$; see graph.

Work Step by Step

Step 1. Rewrite the equation as $25(x^2-4x+4)-16(y^2+6y+9)=444+100-144$ or $25(x-2)^2-16(y+3)^2=400$ and in standard form as $\frac{(x-2)^2}{16}-\frac{(y+3)^2}{25}=1$ Step 2. We can identify $a=4, b=5, c=\sqrt {a^2+b^2}=\sqrt {41}$, center $(2,-3)$ and a horizontal transverse axis. Step 3. We can locate the foci at $(2\pm\sqrt {41},-3)$ and asymptotes as $y+3=\pm\frac{b}{a}(x-2)$ or $y=\pm\frac{5}{4}(x-2)-3$ Step 4. We can graph the equation as shown in the figure.
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