Chapter 10 - Analytic Geometry - 10.4 Summary of the Conic Sections - 10.4 Exercises - Page 996: 22

$${\text{hyperbola}}$$

$$\eqalign{ & \frac{{{x^2}}}{4} = 1 + \frac{{{y^2}}}{9} \cr & {\text{Subtract }}\frac{{{y^2}}}{9}{\text{ from both sides of the equation}} \cr & \frac{{{x^2}}}{4} - \frac{{{y^2}}}{9} = 1 \cr & {\text{The equation is written in the form }}\frac{{{x^2}}}{{{b^2}}} - \frac{{{y^2}}}{{{a^2}}} = 1\, \cr & {\text{Therefore,}} \cr & {\text{The graph of the equation }}\frac{{{x^2}}}{4} = 1 + \frac{{{y^2}}}{9}{\text{ a hyperbola}} \cr} $$