Chapter 10 - Conics, Parametric Equations, and Polar Coordinates - 10.1 Exercises - Page 692: 40

$${\text{Center}}\left( {2,0} \right)$$ $${\text{Vertices}}\left( { - 3,1/2} \right){\text{ and }}\left( { - 3,3/2} \right){\text{ }}$$ $${\text{Foci}}\left( { - 3,1 - \frac{{\sqrt {13} }}{6}} \right){\text{ and }}\left( { - 3,1 + \frac{{\sqrt {13} }}{6}} \right)$$ $${{\text{Asymptotes}}:{\text{ }}y = \pm \frac{a}{b}\left( {x - h} \right) + k}$$

$$\eqalign{ & 9{x^2} - 4{y^2} + 54x + 8y + 78 = 0 \cr & {\text{Group terms}} \cr & \left( {9{x^2} + 54x} \right) - \left( {4{y^2} - 8y} \right) = - 78 \cr & 9\left( {{x^2} + 6x} \right) - 4\left( {{y^2} - 2y} \right) = - 78 \cr & {\text{Completing the square}} \cr & 9\left( {{x^2} + 6x + 9} \right) - 4\left( {{y^2} - 2y + 1} \right) = - 78 + 9\left( 9 \right) - 4\left( 1 \right) \cr & 9{\left( {x + 3} \right)^2} - 4{\left( {y - 1} \right)^2} = - 1 \cr & 4{\left( {y - 1} \right)^2} - 9{\left( {x + 3} \right)^2} = 1 \cr & \frac{{{{\left( {y - 1} \right)}^2}}}{{1/4}} - \frac{{{{\left( {x + 3} \right)}^2}}}{{1/9}} = 1 \cr & {\text{This equation is written in standard form}} \cr & \frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} = 1 \cr & {\text{Therefore,}} \cr & \frac{{{{\left( {y - 1} \right)}^2}}}{{1/4}} - \frac{{{{\left( {x + 3} \right)}^2}}}{{1/9}} = 1 \to \frac{{{{\left( {y - 1} \right)}^2}}}{{{{\left( {1/2} \right)}^2}}} - \frac{{{{\left( {x + 3} \right)}^2}}}{{{{\left( {1/3} \right)}^2}}} = 1 \cr & \underbrace {\frac{{{{\left( {y - 1} \right)}^2}}}{{{{\left( {1/2} \right)}^2}}} - \frac{{{{\left( {x + 3} \right)}^2}}}{{{{\left( {1/3} \right)}^2}}} = 1}_{\frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} = 1} \cr & h = - 3,{\text{ }}k = 1,{\text{ }}a = \frac{1}{2},{\text{ }}b = \frac{1}{3} \cr & c = \sqrt {{a^2} + {b^2}} = \sqrt {\frac{1}{4} + \frac{1}{9}} \Rightarrow c = \frac{{\sqrt {13} }}{6} \cr & {\text{Characteristics of the hyperbola }} \cr & {\text{Orientation}}:{\text{ Vertical transverse axis}} \cr & {\text{Center}}\left( {h,k} \right) \to {\text{Center}}\left( {2,0} \right) \cr & \underbrace {{\text{Vertices}}\left( {h,k - a} \right){\text{ and }}\left( {h,k + a} \right)}_ \Downarrow \cr & {\text{Vertices}}\left( { - 3,1/2} \right){\text{ and }}\left( { - 3,3/2} \right){\text{ }} \cr & \underbrace {{\text{Foci}}\left( {h,k - c} \right){\text{ and }}\left( {h,k + c} \right)}_ \Downarrow \cr & {\text{Foci}}\left( { - 3,1 - \frac{{\sqrt {13} }}{6}} \right){\text{ and }}\left( { - 3,1 + \frac{{\sqrt {13} }}{6}} \right) \cr & \underbrace {{\text{Asymptotes}}:{\text{ }}y = \pm \frac{a}{b}\left( {x - h} \right) + k}_ \Downarrow \cr & {\text{Asymptotes}}:{\text{ }}y = \pm \frac{3}{2}\left( {x + 3} \right) + 1 \cr & {\text{Graph:}} \cr} $$