Chapter 10 - Analytic Geometry - 10.2 Ellipses - 10.2 Exercises - Page 979: 27

$$\frac{{{x^2}}}{{72}} + \frac{{{y^2}}}{{81}} = 1$$

$$\eqalign{ & {\text{foci at }}\left( {0, - 3} \right){\text{ and }}\left( {0,3} \right) \cr & {\text{The }}x{\text{ coordinate is the same, then the ellipse has the equation}} \cr & \frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1 \cr & {\text{With foci }}\left( {0, \pm c} \right),\,\,\left( {0, \pm 3} \right),\,\,\,c = 3 \cr & {c^2} = {a^2} - {b^2} \cr & {a^2} - {b^2} = 9\,\,\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{We know the point }}\left( {8,3} \right){\text{ lies on the ellipse, then}} \cr & \frac{{{{\left( 8 \right)}^2}}}{{{b^2}}} + \frac{{{{\left( 3 \right)}^2}}}{{{a^2}}} = 1 \cr & \frac{{64}}{{{b^2}}} + \frac{9}{{{a^2}}} = 1\,\,\,\, \cr & 64{a^2} + 9{b^2} = {a^2}{b^2}\,\,\,\,\,\left( {\bf{2}} \right) \cr & \cr & {\text{Solve the equation }}\left( {\bf{1}} \right){\text{ for }}{b^2} \cr & {b^2} = {a^2} - 9 \cr & \cr & {\text{Substitute }}{b^2}{\text{ into the equation }}\left( {\bf{2}} \right) \cr & 64{a^2} + 9\left( {{a^2} - 9} \right) = {a^2}{b^2}\, \cr & 64{a^2} + 9{a^2} - 81 = {a^2}\left( {{a^2} - 9} \right) \cr & 64{a^2} + 9{a^2} - 81 = {a^4} - 9{a^2} \cr & 73{a^2} - 81 = {a^4} - 9{a^2} \cr & {a^4} - 82{a^2} + 81 = 0 \cr & \left( {{a^2} - 81} \right)\left( {{a^2} - 1} \right) = 0 \cr & {a^2} = 1,\,\,81 \cr & {\text{Where }}{a^2} > {c^2},{\text{ }}\,{\text{Then }}{a^2} = 81 \cr & {\text{Calculating }}{b^2} \cr & {b^2} = {\left( 9 \right)^2} - \left( 9 \right) \cr & {b^2} = 72 \cr & \cr & {\text{The equation of the ellipse is}} \cr & \frac{{{x^2}}}{{72}} + \frac{{{y^2}}}{{81}} = 1 \cr} $$