Chapter 10 - Conics, Parametric Equations, and Polar Coordinates - Review Exercises - Page 742: 10

$$\frac{{{{\left( {x - 2} \right)}^2}}}{{1/5}} + {y^2} = 1$$

$$\eqalign{ & 5{x^2} + {y^2} - 20x + 19 = 0 \cr & {\text{Group terms}} \cr & 5{x^2} - 20x + {y^2} = - 19 \cr & \left( {5{x^2} - 20x} \right) + {y^2} = - 19 \cr & 5\left( {{x^2} - 4x} \right) + {y^2} = - 19 \cr & {\text{Complete the square and factor}} \cr & 5\left( {{x^2} - 4x + 4} \right) + {y^2} = - 19 + 5\left( 4 \right) \cr & 5{\left( {x - 2} \right)^2} + {y^2} = 1 \cr & \frac{{{{\left( {x - 2} \right)}^2}}}{{1/5}} + {y^2} = 1 \cr & {\text{The equation is in standard form}} \cr & \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} = 1,{\text{ }}a > b \cr & \underbrace {\frac{{{{\left( {x - 2} \right)}^2}}}{{1/5}} + {y^2} = 1}_ \Downarrow \cr & h = 2,k = 0,a = 1,b = \frac{1}{{\sqrt 5 }} \cr & c = \sqrt {1 - \frac{1}{5}} = \frac{{\sqrt 5 }}{5} \cr & {\text{With}} \cr & {\text{Vertex }}\left( {h,k - a} \right){\text{ and }}\left( {h,k + a} \right) \cr & {\text{Vertex }}\left( { - 3,4 - \sqrt {80} } \right){\text{ and }}\left( { - 3,4 + \sqrt {80} } \right) \cr & {\text{Center}}\left( {h,k} \right) \to {\text{Center}}\left( { - 3,4} \right) \cr & {\text{Foci }}\left( {h,k - c} \right){\text{ and }}\left( {h,k + c} \right) \cr & {\text{Foci }}\left( {2, - \frac{{\sqrt 5 }}{5}} \right){\text{ and }}\left( {2,\frac{{\sqrt 5 }}{5}} \right) \cr & {\text{Graph}} \cr} $$