Chapter 10 - Conics, Parametric Equations, and Polar Coordinates - 10.1 Exercises - Page 692: 28

$$\eqalign{ & {\text{Center}}\left( {2, - 3} \right) \cr & {\text{Vertex }}\left( {2 - \sqrt {\frac{5}{8}} , - 3} \right){\text{ and }}\left( {2 + \sqrt {\frac{5}{8}} , - 3} \right) \cr & {\text{Foci }}\left( {2 - \frac{3}{{\sqrt {40} }}, - 3} \right){\text{ and }}\left( {2 + \frac{3}{{\sqrt {40} }}, - 3} \right) \cr} $$

$$\eqalign{ & 16{x^2} + 25{y^2} - 64x + 150y + 279 = 0 \cr & {\text{Group terms}} \cr & 16{x^2} + 25{y^2} - 64x + 150y = - 279 \cr & \left( {16{x^2} - 64x} \right) + \left( {25{y^2} + 150y} \right) = - 279 \cr & 16\left( {{x^2} - 4x} \right) + 25\left( {{y^2} + 6y} \right) = - 279 \cr & {\text{Completing the square}} \cr & 16\left( {{x^2} - 4x + 4} \right) + 25\left( {{y^2} + 6y + 9} \right) = - 279 + 16\left( 4 \right) + 25\left( 9 \right) \cr & 16{\left( {x - 2} \right)^2} + 25{\left( {y + 3} \right)^2} = 10 \cr & {\text{Divide both sides of the equation by 10}} \cr & \frac{{16{{\left( {x - 2} \right)}^2}}}{{10}} + \frac{{25{{\left( {y + 3} \right)}^2}}}{{10}} = 1 \cr & \frac{{8{{\left( {x - 2} \right)}^2}}}{5} + \frac{{5{{\left( {y + 3} \right)}^2}}}{2} = 1 \cr & \frac{{{{\left( {x - 2} \right)}^2}}}{{5/8}} + \frac{{{{\left( {y + 3} \right)}^2}}}{{2/5}} = 1 \cr & \frac{{{{\left( {x - 2} \right)}^2}}}{{{{\left( {\sqrt {5/8} } \right)}^2}}} + \frac{{{{\left( {y + 3} \right)}^2}}}{{{{\left( {\sqrt {2/5} } \right)}^2}}} = 1 \cr & {\text{The equation has the standard form}} \cr & \frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1,{\text{ }}a > b \cr & {\text{Comparing}} \cr & \frac{{{{\left( {x - 2} \right)}^2}}}{{{{\left( {\sqrt {5/8} } \right)}^2}}} + \frac{{{{\left( {y + 3} \right)}^2}}}{{{{\left( {\sqrt {2/5} } \right)}^2}}} = 1 \cr & \to h = 2,{\text{ }}k = - 3,{\text{ }}a = \sqrt {5/8} ,{\text{ }}b = \sqrt {2/5} ,{\text{ }}c = \frac{3}{{\sqrt {40} }} \cr & {\text{With}} \cr & {\text{Orientation of Major Axis Vertical along the }}y{\text{ - axis}} \cr & \underbrace {{\text{Vertex }}\left( {h - a,k} \right){\text{ and }}\left( {h + a,k} \right)}_ \Downarrow \cr & {\text{Vertex }}\left( {2 - \sqrt {\frac{5}{8}} , - 3} \right){\text{ and }}\left( {2 + \sqrt {\frac{5}{8}} , - 3} \right) \cr & \underbrace {{\text{Foci }}\left( {h - c,k} \right){\text{ and }}\left( {h + c,k} \right)}_ \Downarrow \cr & {\text{Foci }}\left( {2 - \frac{3}{{\sqrt {40} }}, - 3} \right){\text{ and }}\left( {2 + \frac{3}{{\sqrt {40} }}, - 3} \right) \cr & {\text{Excentricity }}e = \frac{c}{a} = \frac{{3/\sqrt {40} }}{{\sqrt {5/8} }} = \frac{3}{5} \cr & {\text{Center}}\left( {h,k} \right) \to {\text{Center}}\left( {2, - 3} \right) \cr & {\text{Graph:}} \cr} $$