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: 37

Answer

$$\Rightarrow \dfrac {x^{2}}{9}-\dfrac {y^{2}}{16}=1$$

Work Step by Step

Equation of hyperbola with a format of $$\dfrac {x^{2}}{a^{2}}-\dfrac {y^{2}}{b^{2}}=1\left( a > 0,b > 0\right) $$ Foci are $\left( \pm c,0\right) ;c^{2}=a^{2}+b^{2}$ Vertices are $\left( \pm a,0\right) $ Given vertices $(\pm3,0)$ we find $a=3$ and given foci $(\pm5)$ we find $c=5$ $\Rightarrow c^{2}=a^{2}+b^{2}\Rightarrow 5^{2}=3^{2}+b^{2}\Rightarrow b=4$ Then the equation will be $$\Rightarrow \dfrac {x^{2}}{9}-\dfrac {y^{2}}{16}=1$$
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