Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 13 - Conic Sections - Mid-Chapter Review - Mixed Review - Page 873: 15

Answer

Hyperbola

Work Step by Step

$16{{y}^{2}}-{{x}^{2}}=16$ Divide both sides of the equation by $16$, $\begin{align} & \frac{16{{y}^{2}}-{{x}^{2}}}{16}=\frac{16}{16} \\ & \frac{{{y}^{2}}}{{{1}^{2}}}-\frac{{{x}^{2}}}{{{4}^{2}}}=1 \end{align}$ Consider the standard form of the hyperbola $\frac{{{y}^{2}}}{{{b}^{2}}}-\frac{{{x}^{2}}}{{{a}^{2}}}=1$ The simplified equation $\frac{{{y}^{2}}}{{{1}^{2}}}-\frac{{{x}^{2}}}{{{4}^{2}}}=1$ resembles the standard equation of a hyperbola. Now, find points on the graph: $\begin{matrix} x & y \\ 0 & -1 \\ 0& 1 \\ -1 & 1 \\ 1 & 1 \\ \end{matrix}$ Plot the points and connect them.
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