## College Algebra (11th Edition)

Published by Pearson

# Chapter 6 - Section 6.4 - Summary of the Conic Sections - 6.4 Exercises - Page 619: 14

Hyperbola

#### Work Step by Step

We can see, that both x and y are on the power of two,one with a positive the other with a negative coefficient. Therefore the equation will be a hyperbola. The given equation is equivalent to: $\frac{x^2}{4}-\frac{y^2}{9}=1$ Which can be easily transformed into the standard equation of a hyperbola: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ Here, it is: $\frac{x^2}{2^2}-\frac{y^2}{3^2}=1$

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