Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 10 - Parametric And Polar Curves; Conic Sections - 10.4 Conic Sections - Exercises Set 10.4 - Page 745: 9

Answer

$$\eqalign{ & \left( {\bf{a}} \right) \cr & {\text{foci : }}\left( { - 3 - 2\sqrt 3 ,5} \right){\text{ and }}\left( { - 3 + 2\sqrt 3 ,5} \right) \cr & {\text{Vertices: }}\left( { - 7,5} \right){\text{ and }}\left( {1,5} \right) \cr & {\text{Ends of the minor axis: }}\left( { - 3,3} \right){\text{ and }}\left( { - 3,7} \right) \cr & \left( {\bf{b}} \right) \cr & {\text{foci : }}\left( {0, - 2 - \sqrt 5 } \right){\text{ and }}\left( {0, - 2 + \sqrt 5 } \right) \cr & {\text{Vertices: }}\left( {0, - 5} \right){\text{ and }}\left( {0,1} \right) \cr & {\text{Ends of the minor axis: }}\left( { - 2, - 2} \right){\text{ and }}\left( {2, - 2} \right) \cr} $$

Work Step by Step

$$\eqalign{ & \left( {\bf{a}} \right){\left( {x + 3} \right)^2} + 4{\left( {y - 5} \right)^2} = 16 \cr & \frac{{{{\left( {x + 3} \right)}^2}}}{{16}} + \frac{{4{{\left( {y - 5} \right)}^2}}}{{16}} = \frac{{16}}{{16}} \cr & \frac{{{{\left( {x + 3} \right)}^2}}}{{16}} + \frac{{{{\left( {y - 5} \right)}^2}}}{4} = 1 \cr & {\text{The standard form of the ellipse is }}\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 \cr & \frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1\,\,\,:\,\,\,\,\frac{{{{\left( {x + 3} \right)}^2}}}{{16}} + \frac{{{{\left( {y - 5} \right)}^2}}}{4} = 1 \cr & \,\, \to \,\,a = 4,\,\,\,\,b = 2,\,\,\,\,\,h = - 3,\,\,\,\,\,k = 5 \cr & {c^2} = {a^2} - {b^2} \cr & {c^2} = 16 - 4 \cr & {c^2} = 12 \cr & c = 2\sqrt 3 \cr & \cr & {\text{With:}} \cr & {\text{foci: }}\left( {h - c,k} \right){\text{ and }}\left( {h + c,k} \right) \cr & {\text{foci : }}\left( { - 3 - 2\sqrt 3 ,5} \right){\text{ and }}\left( { - 3 + 2\sqrt 3 ,5} \right) \cr & {\text{Vertices: }}\left( {h - a,k} \right){\text{ and }}\left( {h + a,k} \right) \cr & {\text{Vertices: }}\left( { - 7,5} \right){\text{ and }}\left( {1,5} \right) \cr & {\text{Ends of the minor axis: }}\left( {h,k - b} \right){\text{ and }}\left( {h,k + b} \right) \cr & {\text{Ends of the minor axis: }}\left( { - 3,3} \right){\text{ and }}\left( { - 3,7} \right) \cr & \cr & \cr & \left( {\bf{b}} \right)\frac{1}{4}{x^2} + \frac{1}{9}{\left( {y + 2} \right)^2} - 1 = 0 \cr & \frac{{{x^2}}}{4} + \frac{{{{\left( {y + 2} \right)}^2}}}{9} = 1 \cr & {\text{The standard form of the ellipse is }}\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 \cr & \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} = 1\,\,\,:\,\,\,\,\frac{{{x^2}}}{4} + \frac{{{{\left( {y + 2} \right)}^2}}}{9} = 1 \cr & \,\, \to \,\,a = 3,\,\,\,\,b = 2,\,\,\,\,\,h = 0,\,\,\,\,\,k = - 2 \cr & {c^2} = {a^2} - {b^2} \cr & {c^2} = 9 - 4 \cr & c = \sqrt 5 \cr & \cr & {\text{With:}} \cr & {\text{foci: }}\left( {h,k - c} \right){\text{ and }}\left( {h,k + c} \right) \cr & {\text{foci : }}\left( {0, - 2 - \sqrt 5 } \right){\text{ and }}\left( {0, - 2 + \sqrt 5 } \right) \cr & {\text{Vertices: }}\left( {h,k - a} \right){\text{ and }}\left( {h,k + a} \right) \cr & {\text{Vertices: }}\left( {0, - 5} \right){\text{ and }}\left( {0,1} \right) \cr & {\text{Ends of the minor axis: }}\left( {h - b,k} \right){\text{ and }}\left( {h + b,k} \right) \cr & {\text{Ends of the minor axis: }}\left( { - 2, - 2} \right){\text{ and }}\left( {2, - 2} \right) \cr} $$
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