Chapter 10 - Analytic Geometry - 10.3 Hyperbolas - 10.3 Exercises - Page 988: 15

$$\eqalign{ & {\text{Center }}\left( {0,0} \right) \cr & {\text{Vertices}}:\left( { \pm 5,0} \right) \cr & {\text{Foci: }}\left( { \pm \sqrt {34} ,0} \right) \cr & {\text{Asymptotes: }}y = \pm \frac{3}{5}x \cr & {\text{domain: }}\left( { - \infty , - 5} \right] \cup \left[ {5,\infty } \right) \cr & {\text{range: }}\left( { - \infty , + \infty } \right) \cr} $$

$$\eqalign{ & 9{x^2} - 25{y^2} = 225 \cr & {\text{Divide both sides by }}225 \cr & \frac{{9{x^2}}}{{225}} - \frac{{25{y^2}}}{{225}} = \frac{{225}}{{225}} \cr & \frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{9} = 1 \cr & {\text{The equation is written in the form }}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr & \frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{9} = 1,{\text{ then }}a = 5,\,\,b = 3 \cr & c = \sqrt {{a^2} + {b^2}} = \sqrt {25 + 9} = \sqrt {34} \cr & \cr & {\text{Therefore,}} \cr & {\text{Center }}\left( {0,0} \right) \cr & {\text{Vertices: }}\left( { \pm a,0} \right):\left( { \pm 5,0} \right) \cr & {\text{Foci: }}\left( { \pm c,0} \right):\left( { \pm \sqrt {34} ,0} \right) \cr & {\text{Asymptotes: }}y = \pm \frac{b}{a}x \cr & {\text{Asymptotes: }}y = \pm \frac{3}{5}x \cr & {\text{The domain of the hyperbola is }}\left( { - \infty ,a} \right] \cup \left[ {a,\infty } \right) \cr & {\text{domain: }}\left( { - \infty , - 5} \right] \cup \left[ {5,\infty } \right) \cr & {\text{The range of the hyperbola is }}\left( { - \infty , + \infty } \right) \cr & \cr & {\text{Graph}} \cr} $$