Chapter 10 - Analytic Geometry - Chapter 10 Test Prep - Review Exercises - Page 1001: 40

$$\eqalign{ & {\text{domain: }}\left[ { - 6,4} \right] \cr & {\text{range: }}\left[ {3,7} \right] \cr & {\text{Vertices: }}\left( { - 6, - 5} \right){\text{ and }}\left( {4,5} \right) \cr} $$

$$\eqalign{ & 4{x^2} + 8x + 25{y^2} - 250y = - 529 \cr & \left( {4{x^2} + 8x} \right) + \left( {25{y^2} - 250y} \right) = - 529 \cr & 4\left( {{x^2} + 2x} \right) + 25\left( {{y^2} - 10y} \right) = - 529 \cr & {\text{Complete the square}} \cr & 4\left( {{x^2} + 2x + 1} \right) + 25\left( {{y^2} - 10y + 25} \right) = - 529 + 4\left( 1 \right) + 25\left( {25} \right) \cr & 4{\left( {x + 1} \right)^2} + 25{\left( {y - 5} \right)^2} = 100 \cr & {\text{Divide both sides by 100}} \cr & \frac{{{{\left( {x + 1} \right)}^2}}}{{25}} + \frac{{{{\left( {y - 5} \right)}^2}}}{4} = 1 \cr & {\text{The equation of the ellipse is in the form }} \cr & \frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1\,\,\,\,\left( {a > b} \right) \cr & a = 5,\,\,b = 2 \cr & h = \, - 1,\,\,k = 5 \cr & \cr & {\text{Vertices: }}\left( {h \pm a,k} \right) \cr & {\text{Vertices: }}\left( { - 1 - 5,5} \right){\text{ and }}\left( { - 1 + 5,5} \right) \cr & {\text{Vertices: }}\left( { - 6, - 5} \right){\text{ and }}\left( {4,5} \right) \cr & \cr & {\text{The domain of the ellipse is }}\left[ {h - a,h + a} \right] \cr & {\text{domain }}\left[ { - 6,4} \right] \cr & \cr & {\text{The range of the ellipse is }}\left[ {k - b,k + b} \right] \cr & {\text{range }}\left[ {3,7} \right] \cr & \cr & {\text{Therefore,}} \cr & {\text{domain: }}\left[ { - 6,4} \right] \cr & {\text{range: }}\left[ {3,7} \right] \cr & {\text{Vertices: }}\left( { - 6, - 5} \right){\text{ and }}\left( {4,5} \right) \cr & \cr & {\text{Graph}} \cr} $$