Chapter 10 - Analytic Geometry - 10.3 Hyperbolas - 10.3 Exercises - Page 988: 30

$$\eqalign{ & {\text{Domain: }}\left( { - \infty , - 2} \right] \cr & {\text{Range: }}\left( { - \infty ,\infty } \right) \cr} $$

$$\eqalign{ & 3y = \sqrt {4{x^2} - 16} \cr & {\text{Square each side}} \cr & {\left( {3y} \right)^2} = {\left( {\sqrt {4{x^2} - 16} } \right)^2} \cr & 9{y^2} = 4{x^2} - 16 \cr & {\text{Write in standard form}}. \cr & 4{x^2} - 9{y^2} = 16 \cr & \frac{{{x^2}}}{4} - \frac{{{y^2}}}{{16/9}} = 1 \cr & \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr & a = 2,\,\,\,\,b = 4/3 \cr & {\text{This is the equation of a hyperbola ellipse with }}x{\text{ - intercepts at}} \cr & \left( { - a,0} \right){\text{ and }}\left( {a,0} \right) \cr & {\text{The domain is }}\left( { - \infty , - a} \right] \cup \left[ {a,\infty } \right) \cr & {\text{The range is }}\left( { - \infty ,\infty } \right) \cr & {\text{In the original equation}},{\text{ the radical expression }} - \sqrt {1 + 4{y^2}} {\text{ represents}} \cr & {\text{a negative number}},{\text{ so the only possible values of }}x{\text{ are negative}} \cr & ,{\text{then the domain is }}\left( { - \infty , - a} \right]. \cr & \cr & {\text{Domain: }}\left( { - \infty , - 2} \right] \cr & {\text{Range: }}\left( { - \infty ,\infty } \right) \cr & \cr & {\text{Graph}} \cr} $$