Chapter 10 - Analytic Geometry - 10.3 Hyperbolas - 10.3 Exercises - Page 988: 19

$$\eqalign{ & {\text{Center }}\left( {0,0} \right) \cr & {\text{Vertices}}:\left( { \pm \frac{1}{3},0} \right) \cr & {\text{Foci: }}\left( { \pm \frac{{\sqrt {13} }}{6},0} \right) \cr & {\text{Asymptotes: }}y = \pm \frac{3}{2}x \cr & {\text{domain: }}\left( { - \infty , - \frac{1}{3}} \right] \cup \left[ {\frac{1}{3},\infty } \right) \cr & {\text{range: }}\left( { - \infty , + \infty } \right) \cr} $$

$$\eqalign{ & 9{x^2} - 4{y^2} = 1 \cr & \frac{{{x^2}}}{{\left( {1/9} \right)}} - \frac{{{y^2}}}{{\left( {1/4} \right)}} = 1 \cr & {\text{The equation is written in the form }}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr & \frac{{{x^2}}}{{\left( {1/9} \right)}} - \frac{{{y^2}}}{{\left( {1/4} \right)}} = 1,{\text{ then }}a = 1/3,\,\,b = 1/2 \cr & c = \sqrt {{a^2} + {b^2}} = \sqrt {1/9 + 1/4} = \frac{{\sqrt {13} }}{6} \cr & \cr & {\text{Therefore,}} \cr & {\text{Center }}\left( {0,0} \right) \cr & {\text{Vertices: }}\left( { \pm a,0} \right):\left( { \pm \frac{1}{3},0} \right) \cr & {\text{Foci: }}\left( { \pm c,0} \right):\left( { \pm \frac{{\sqrt {13} }}{6},0} \right) \cr & {\text{Asymptotes: }}y = \pm \frac{b}{a}x \cr & {\text{Asymptotes: }}y = \pm \frac{3}{2}x \cr & {\text{The domain of the hyperbola is }}\left( { - \infty ,a} \right] \cup \left[ {a,\infty } \right) \cr & {\text{domain: }}\left( { - \infty , - 4} \right] \cup \left[ {4,\infty } \right) \cr & {\text{The range of the hyperbola is }}\left( { - \infty , + \infty } \right) \cr & \cr & {\text{Graph}} \cr} $$