Chapter 10 - Analytic Geometry - Chapter 10 Test Prep - Review Exercises - Page 1001: 41

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

$$\eqalign{ & 5{x^2} + 20x + 2{y^2} - 8y = - 18 \cr & 5\left( {{x^2} + 4x} \right) + 2\left( {{y^2} - 4y} \right) = - 18 \cr & {\text{Complete the square}} \cr & 5\left( {{x^2} + 4x + 4} \right) + 2\left( {{y^2} - 4y + 4} \right) = - 18 + 5\left( 4 \right) + 2\left( 4 \right) \cr & 5{\left( {x + 2} \right)^2} + 2{\left( {y - 2} \right)^2} = 10 \cr & {\text{Divide both sides by 10}} \cr & \frac{{{{\left( {x + 2} \right)}^2}}}{2} + \frac{{{{\left( {y - 2} \right)}^2}}}{5} = 1 \cr & {\text{The equation of the ellipse is in the form }} \cr & \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} = 1\,\,\,\,\left( {a > b} \right) \cr & {\text{Comparing we obtain}} \cr & a = \sqrt 5 ,\,\,b = \sqrt 3 \cr & h = - 2,\,\,k = 2 \cr & \cr & {\text{Vertices }}\left( {h,k \pm a} \right) \cr & {\text{Vertices }}\left( { - 2,2 - \sqrt 5 } \right){\text{ and }}\left( { - 2,2 + \sqrt 5 } \right) \cr & \cr & {\text{The domain of the ellipse is }}\left[ {h - b,h + b} \right] \cr & {\text{domain }}\left[ { - 2 - \sqrt 3 , - 2 + \sqrt 3 } \right] \cr & \cr & {\text{The range of the ellipse is }}\left[ {k - a,k + a} \right] \cr & {\text{range }}\left[ {2 - \sqrt 5 ,2 + \sqrt 5 } \right] \cr & \cr & {\text{Therefore,}} \cr & {\text{domain: }}\left[ { - 2 - \sqrt 3 , - 2 + \sqrt 3 } \right] \cr & {\text{range: }}\left[ {2 - \sqrt 5 ,2 + \sqrt 5 } \right] \cr & {\text{Vertices: }}\left( { - 2,2 - \sqrt 5 } \right){\text{ and }}\left( { - 2,2 + \sqrt 5 } \right) \cr & \cr & {\text{Graph}} \cr} $$