Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 11 - Section 11.3 - Hyperbolas - 11.3 Exercises - Page 806: 31

Answer

$\frac{y^2}{9}-x^2=1$

Work Step by Step

Vertices: $(0,±a)=(0,±3)$ $a=3$ Asymptotes: $y=±\frac{a}{b}x$ $y=±3x$ $\frac{a}{b}=3$ $\frac{3}{b}=3$ $b=1$ Hyperbola with vertical transverse axis: $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ $\frac{y^2}{3^2}-\frac{x^2}{1^2}=1$ $\frac{y^2}{9}-x^2=1$
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