Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 4 - Review Exercises - Page 354: 62

Answer

$\dfrac{y^2}{4}-\dfrac{x^2}{5}=1$

Work Step by Step

The standard form of the equation of the hyperbola with a horizontal transverse axis can be expressed as: $\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$ The vertices and foci have the form $(\pm a, 0) $ and $(\pm c,0)$. The standard form of the equation of the hyperbola with a vertical transverse axis can be expressed as: $\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$ The vertices and foci have the form $(0, \pm, a) $ and $(0, \pm c)$. We have: $a=2$ $y=\dfrac{ax}{b} \implies \dfrac{2x}{\sqrt x}=\dfrac{ax}{b}$ or, $b=\sqrt 5$ $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$ or, $\dfrac{y^2}{4}-\dfrac{x^2}{5}=1$
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