Chapter 10 - Conics, Parametric Equations, and Polar Coordinates - 10.1 Exercises - Page 692: 31

$$\frac{{{{\left( {x - 3} \right)}^2}}}{9} + \frac{{{{\left( {y - 5} \right)}^2}}}{{16}} = 1$$

$$\eqalign{ & {\text{Vertices }}\left( {3,1} \right),\left( {3,9} \right) \cr & {\text{Minor axis length: 6}} \cr & {\text{Vertices }}\left( {\underbrace 3_x,\underbrace 1_y} \right){\text{ and }}\left( {\underbrace 3_x,\underbrace 9_y} \right) \cr & {\text{Form of the equation}}:\frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} = 1 \cr & {\text{Vertices }}\left( {h,k - a} \right){\text{ and }}\left( {h,k + a} \right) \cr & b = \sqrt {{a^2} - {c^2}} \cr & \cr & \underbrace {{\text{Vertices }}\left( {\underbrace 3_h,\underbrace 1_{k - a}} \right){\text{ and }}\left( {\underbrace 3_h,\underbrace 9_{k + a}} \right)}_ \Downarrow \cr & k - a = 1;{\text{ }}k + a = 9;{\text{ }}h = 3 \cr & {\text{Find }}a{\text{ and }}k \cr & k - a = 1 \cr & \underline {k + a = 9} \cr & 2k = 10 \cr & k = 5 \cr & a = 9 - k \Rightarrow a = 4 \cr & {\text{Minor axis length }}2b = 6 \cr & b = 3 \cr & \cr & {\text{The Standard form of the equation is}} \cr & \underbrace {\frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} = 1}_ \Downarrow \cr & \frac{{{{\left( {x - 3} \right)}^2}}}{{{{\left( 3 \right)}^2}}} + \frac{{{{\left( {y - 5} \right)}^2}}}{{{{\left( 4 \right)}^2}}} = 1 \cr & {\text{Simplify}} \cr & \frac{{{{\left( {x - 3} \right)}^2}}}{9} + \frac{{{{\left( {y - 5} \right)}^2}}}{{16}} = 1 \cr} $$