Chapter 10 - Analytic Geometry - 10.2 Ellipses - 10.2 Exercises - Page 978: 6

$$\eqalign{ & {\text{domain: }}\left[ { - 4, - 2} \right] \cr & {\text{range: }}\left[ { - 1,5} \right] \cr & {\text{center: }}\left( { - 3,2} \right) \cr & {\text{Vertices: }}\left( { - 3,5} \right){\text{ and }}\left( { - 3, - 1} \right) \cr & {\text{Foci }}\left( { - 3,2 + 2\sqrt 2 } \right){\text{ and }}\left( { - 3,2 - 2\sqrt 2 } \right) \cr} $$

$$\eqalign{ & \frac{{{{\left( {x + 3} \right)}^2}}}{1} + \frac{{{{\left( {y - 2} \right)}^2}}}{9} = 1 \cr & {\text{The equation of the ellipse is in the form }} \cr & \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} = 1\,\,\,\,\left( {a > b} \right) \cr & {\text{Comparing we obtain}} \cr & a = 3,\,\,b = 1 \cr & h = \, - 3,\,\,k = 2 \cr & c = \sqrt {{3^2} - {1^2}} = 2\sqrt 2 \cr & \cr & {\text{The ellipse with center at }}\left( {h,k} \right) = \left( { - 3,2} \right) \cr & {\text{Vertices }}\left( {h,k \pm a} \right) \cr & {\text{Vertices: }}\left( { - 3,5} \right){\text{ and }}\left( { - 3, - 1} \right) \cr & \cr & {\text{Foci }}\left( {h,k \pm c} \right) \cr & {\text{Foci: }}\left( { - 3,2 + 2\sqrt 2 } \right){\text{ and }}\left( { - 3,2 - 2\sqrt 2 } \right) \cr & \cr & {\text{The domain of the ellipse is }}\left[ {h - b,h + b} \right] \cr & {\text{domain }}\left[ { - 4, - 2} \right] \cr & \cr & {\text{The range of the ellipse is }}\left[ {k - a,k + a} \right] \cr & {\text{range }}\left[ { - 1,5} \right] \cr & \cr & {\text{Therefore,}} \cr & {\text{domain: }}\left[ { - 4, - 2} \right] \cr & {\text{range: }}\left[ { - 1,5} \right] \cr & {\text{center: }}\left( { - 3,2} \right) \cr & {\text{Vertices: }}\left( { - 3,5} \right){\text{ and }}\left( { - 3, - 1} \right) \cr & {\text{Foci }}\left( { - 3,2 + 2\sqrt 2 } \right){\text{ and }}\left( { - 3,2 - 2\sqrt 2 } \right) \cr} $$