{"id":15776,"date":"2019-09-05T19:46:01","date_gmt":"2019-09-05T19:46:01","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=15776"},"modified":"2025-02-05T05:24:53","modified_gmt":"2025-02-05T05:24:53","slug":"problem-set-69-conic-sections-in-polar-coordinates","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/problem-set-69-conic-sections-in-polar-coordinates\/","title":{"raw":"Problem Set 69: Conic Sections in Polar Coordinates","rendered":"Problem Set 69: Conic Sections in Polar Coordinates"},"content":{"raw":"1. Explain how eccentricity determines which conic section is given.\r\n\r\n2.\u00a0If a conic section is written as a polar equation, what must be true of the denominator?\r\n\r\n3. If a conic section is written as a polar equation, and the denominator involves [latex]\\sin \\text{ }\\theta [\/latex], what conclusion can be drawn about the directrix?\r\n\r\n4.\u00a0If the directrix of a conic section is perpendicular to the polar axis, what do we know about the equation of the graph?\r\n\r\n5. What do we know about the focus\/foci of a conic section if it is written as a polar equation?\r\n\r\nFor the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity.\r\n\r\n6. [latex]r=\\frac{6}{1 - 2\\text{ }\\cos \\text{ }\\theta }[\/latex]\r\n\r\n7. [latex]r=\\frac{3}{4 - 4\\text{ }\\sin \\text{ }\\theta }[\/latex]\r\n\r\n8.\u00a0[latex]r=\\frac{8}{4 - 3\\text{ }\\cos \\text{ }\\theta }[\/latex]\r\n\r\n9. [latex]r=\\frac{5}{1+2\\text{ }\\sin \\text{ }\\theta }[\/latex]\r\n\r\n10.\u00a0[latex]r=\\frac{16}{4+3\\text{ }\\cos \\text{ }\\theta }[\/latex]\r\n\r\n11. [latex]r=\\frac{3}{10+10\\text{ }\\cos \\text{ }\\theta }[\/latex]\r\n\r\n12.\u00a0[latex]r=\\frac{2}{1-\\cos \\text{ }\\theta }[\/latex]\r\n\r\n13. [latex]r=\\frac{4}{7+2\\text{ }\\cos \\text{ }\\theta }[\/latex]\r\n\r\n14.\u00a0[latex]r\\left(1-\\cos \\text{ }\\theta \\right)=3[\/latex]\r\n\r\n15. [latex]r\\left(3+5\\sin \\text{ }\\theta \\right)=11[\/latex]\r\n\r\n16.\u00a0[latex]r\\left(4 - 5\\sin \\text{ }\\theta \\right)=1[\/latex]\r\n\r\n17. [latex]r\\left(7+8\\cos \\text{ }\\theta \\right)=7[\/latex]\r\n\r\nFor the following exercises, convert the polar equation of a conic section to a rectangular equation.\r\n\r\n18. [latex]r=\\frac{4}{1+3\\text{ }\\sin \\text{ }\\theta }[\/latex]\r\n\r\n19. [latex]r=\\frac{2}{5 - 3\\text{ }\\sin \\text{ }\\theta }[\/latex]\r\n\r\n20.\u00a0[latex]r=\\frac{8}{3 - 2\\text{ }\\cos \\text{ }\\theta }[\/latex]\r\n\r\n21. [latex]r=\\frac{3}{2+5\\text{ }\\cos \\text{ }\\theta }[\/latex]\r\n\r\n22.\u00a0[latex]r=\\frac{4}{2+2\\text{ }\\sin \\text{ }\\theta }[\/latex]\r\n\r\n23. [latex]r=\\frac{3}{8 - 8\\text{ }\\cos \\text{ }\\theta }[\/latex]\r\n\r\n24.\u00a0[latex]r=\\frac{2}{6+7\\text{ }\\cos \\text{ }\\theta }[\/latex]\r\n\r\n25. [latex]r=\\frac{5}{5 - 11\\text{ }\\sin \\text{ }\\theta }[\/latex]\r\n\r\n26.\u00a0[latex]r\\left(5+2\\text{ }\\cos \\text{ }\\theta \\right)=6[\/latex]\r\n\r\n27. [latex]r\\left(2-\\cos \\text{ }\\theta \\right)=1[\/latex]\r\n\r\n28.\u00a0[latex]r\\left(2.5 - 2.5\\text{ }\\sin \\text{ }\\theta \\right)=5[\/latex]\r\n\r\n29. [latex]r=\\frac{6\\sec \\text{ }\\theta }{-2+3\\text{ }\\sec \\text{ }\\theta }[\/latex]\r\n\r\n30.\u00a0[latex]r=\\frac{6\\csc \\text{ }\\theta }{3+2\\text{ }\\csc \\text{ }\\theta }[\/latex]\r\n\r\nFor the following exercises, graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci.\r\n\r\n31. [latex]r=\\frac{5}{2+\\cos \\text{ }\\theta }[\/latex]\r\n\r\n32. [latex]r=\\frac{2}{3+3\\text{ }\\sin \\text{ }\\theta }[\/latex]\r\n\r\n33. [latex]r=\\frac{10}{5 - 4\\text{ }\\sin \\text{ }\\theta }[\/latex]\r\n\r\n34. [latex]r=\\frac{3}{1+2\\text{ }\\cos \\text{ }\\theta }[\/latex]\r\n\r\n35. [latex]r=\\frac{8}{4 - 5\\text{ }\\cos \\text{ }\\theta }[\/latex]\r\n\r\n36. [latex]r=\\frac{3}{4 - 4\\text{ }\\cos \\text{ }\\theta }[\/latex]\r\n\r\n37. [latex]r=\\frac{2}{1-\\sin \\text{ }\\theta }[\/latex]\r\n\r\n38. [latex]r=\\frac{6}{3+2\\text{ }\\sin \\text{ }\\theta }[\/latex]\r\n\r\n39. [latex]r\\left(1+\\cos \\text{ }\\theta \\right)=5[\/latex]\r\n\r\n40. [latex]r\\left(3 - 4\\sin \\text{ }\\theta \\right)=9[\/latex]\r\n\r\n41. [latex]r\\left(3 - 2\\sin \\text{ }\\theta \\right)=6[\/latex]\r\n\r\n42. [latex]r\\left(6 - 4\\cos \\text{ }\\theta \\right)=5[\/latex]\r\n\r\nFor the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix.\r\n\r\n43. Directrix: [latex]x=4;e=\\frac{1}{5}[\/latex]\r\n\r\n44.\u00a0Directrix: [latex]x=-4;e=5[\/latex]\r\n\r\n45. Directrix: [latex]y=2;e=2[\/latex]\r\n\r\n46.\u00a0Directrix: [latex]y=-2;e=\\frac{1}{2}[\/latex]\r\n\r\n47. Directrix: [latex]x=1;e=1[\/latex]\r\n\r\n48.\u00a0Directrix: [latex]x=-1;e=1[\/latex]\r\n\r\n49. Directrix: [latex]x=-\\frac{1}{4};e=\\frac{7}{2}[\/latex]\r\n\r\n50.\u00a0Directrix: [latex]y=\\frac{2}{5};e=\\frac{7}{2}[\/latex]\r\n\r\n51. Directrix: [latex]y=4;e=\\frac{3}{2}[\/latex]\r\n\r\n52.\u00a0Directrix: [latex]x=-2;e=\\frac{8}{3}[\/latex]\r\n\r\n53. Directrix: [latex]x=-5;e=\\frac{3}{4}[\/latex]\r\n\r\n54.\u00a0Directrix: [latex]y=2;e=2.5[\/latex]\r\n\r\n55. Directrix: [latex]x=-3;e=\\frac{1}{3}[\/latex]\r\n\r\nEquations of conics with an [latex]xy[\/latex] term have rotated graphs. For the following exercises, express each equation in polar form with [latex]r[\/latex] as a function of [latex]\\theta [\/latex].\r\n\r\n56. [latex]xy=2[\/latex]\r\n\r\n57. [latex]{x}^{2}+xy+{y}^{2}=4[\/latex]\r\n\r\n58.\u00a0[latex]2{x}^{2}+4xy+2{y}^{2}=9[\/latex]\r\n\r\n59. [latex]16{x}^{2}+24xy+9{y}^{2}=4[\/latex]\r\n\r\n60.\u00a0[latex]2xy+y=1[\/latex]","rendered":"<p>1. Explain how eccentricity determines which conic section is given.<\/p>\n<p>2.\u00a0If a conic section is written as a polar equation, what must be true of the denominator?<\/p>\n<p>3. If a conic section is written as a polar equation, and the denominator involves [latex]\\sin \\text{ }\\theta[\/latex], what conclusion can be drawn about the directrix?<\/p>\n<p>4.\u00a0If the directrix of a conic section is perpendicular to the polar axis, what do we know about the equation of the graph?<\/p>\n<p>5. What do we know about the focus\/foci of a conic section if it is written as a polar equation?<\/p>\n<p>For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity.<\/p>\n<p>6. [latex]r=\\frac{6}{1 - 2\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/p>\n<p>7. [latex]r=\\frac{3}{4 - 4\\text{ }\\sin \\text{ }\\theta }[\/latex]<\/p>\n<p>8.\u00a0[latex]r=\\frac{8}{4 - 3\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/p>\n<p>9. [latex]r=\\frac{5}{1+2\\text{ }\\sin \\text{ }\\theta }[\/latex]<\/p>\n<p>10.\u00a0[latex]r=\\frac{16}{4+3\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/p>\n<p>11. [latex]r=\\frac{3}{10+10\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/p>\n<p>12.\u00a0[latex]r=\\frac{2}{1-\\cos \\text{ }\\theta }[\/latex]<\/p>\n<p>13. [latex]r=\\frac{4}{7+2\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/p>\n<p>14.\u00a0[latex]r\\left(1-\\cos \\text{ }\\theta \\right)=3[\/latex]<\/p>\n<p>15. [latex]r\\left(3+5\\sin \\text{ }\\theta \\right)=11[\/latex]<\/p>\n<p>16.\u00a0[latex]r\\left(4 - 5\\sin \\text{ }\\theta \\right)=1[\/latex]<\/p>\n<p>17. [latex]r\\left(7+8\\cos \\text{ }\\theta \\right)=7[\/latex]<\/p>\n<p>For the following exercises, convert the polar equation of a conic section to a rectangular equation.<\/p>\n<p>18. [latex]r=\\frac{4}{1+3\\text{ }\\sin \\text{ }\\theta }[\/latex]<\/p>\n<p>19. [latex]r=\\frac{2}{5 - 3\\text{ }\\sin \\text{ }\\theta }[\/latex]<\/p>\n<p>20.\u00a0[latex]r=\\frac{8}{3 - 2\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/p>\n<p>21. [latex]r=\\frac{3}{2+5\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/p>\n<p>22.\u00a0[latex]r=\\frac{4}{2+2\\text{ }\\sin \\text{ }\\theta }[\/latex]<\/p>\n<p>23. [latex]r=\\frac{3}{8 - 8\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/p>\n<p>24.\u00a0[latex]r=\\frac{2}{6+7\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/p>\n<p>25. [latex]r=\\frac{5}{5 - 11\\text{ }\\sin \\text{ }\\theta }[\/latex]<\/p>\n<p>26.\u00a0[latex]r\\left(5+2\\text{ }\\cos \\text{ }\\theta \\right)=6[\/latex]<\/p>\n<p>27. [latex]r\\left(2-\\cos \\text{ }\\theta \\right)=1[\/latex]<\/p>\n<p>28.\u00a0[latex]r\\left(2.5 - 2.5\\text{ }\\sin \\text{ }\\theta \\right)=5[\/latex]<\/p>\n<p>29. [latex]r=\\frac{6\\sec \\text{ }\\theta }{-2+3\\text{ }\\sec \\text{ }\\theta }[\/latex]<\/p>\n<p>30.\u00a0[latex]r=\\frac{6\\csc \\text{ }\\theta }{3+2\\text{ }\\csc \\text{ }\\theta }[\/latex]<\/p>\n<p>For the following exercises, graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci.<\/p>\n<p>31. [latex]r=\\frac{5}{2+\\cos \\text{ }\\theta }[\/latex]<\/p>\n<p>32. [latex]r=\\frac{2}{3+3\\text{ }\\sin \\text{ }\\theta }[\/latex]<\/p>\n<p>33. [latex]r=\\frac{10}{5 - 4\\text{ }\\sin \\text{ }\\theta }[\/latex]<\/p>\n<p>34. [latex]r=\\frac{3}{1+2\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/p>\n<p>35. [latex]r=\\frac{8}{4 - 5\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/p>\n<p>36. [latex]r=\\frac{3}{4 - 4\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/p>\n<p>37. [latex]r=\\frac{2}{1-\\sin \\text{ }\\theta }[\/latex]<\/p>\n<p>38. [latex]r=\\frac{6}{3+2\\text{ }\\sin \\text{ }\\theta }[\/latex]<\/p>\n<p>39. [latex]r\\left(1+\\cos \\text{ }\\theta \\right)=5[\/latex]<\/p>\n<p>40. [latex]r\\left(3 - 4\\sin \\text{ }\\theta \\right)=9[\/latex]<\/p>\n<p>41. [latex]r\\left(3 - 2\\sin \\text{ }\\theta \\right)=6[\/latex]<\/p>\n<p>42. [latex]r\\left(6 - 4\\cos \\text{ }\\theta \\right)=5[\/latex]<\/p>\n<p>For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix.<\/p>\n<p>43. Directrix: [latex]x=4;e=\\frac{1}{5}[\/latex]<\/p>\n<p>44.\u00a0Directrix: [latex]x=-4;e=5[\/latex]<\/p>\n<p>45. Directrix: [latex]y=2;e=2[\/latex]<\/p>\n<p>46.\u00a0Directrix: [latex]y=-2;e=\\frac{1}{2}[\/latex]<\/p>\n<p>47. Directrix: [latex]x=1;e=1[\/latex]<\/p>\n<p>48.\u00a0Directrix: [latex]x=-1;e=1[\/latex]<\/p>\n<p>49. Directrix: [latex]x=-\\frac{1}{4};e=\\frac{7}{2}[\/latex]<\/p>\n<p>50.\u00a0Directrix: [latex]y=\\frac{2}{5};e=\\frac{7}{2}[\/latex]<\/p>\n<p>51. Directrix: [latex]y=4;e=\\frac{3}{2}[\/latex]<\/p>\n<p>52.\u00a0Directrix: [latex]x=-2;e=\\frac{8}{3}[\/latex]<\/p>\n<p>53. Directrix: [latex]x=-5;e=\\frac{3}{4}[\/latex]<\/p>\n<p>54.\u00a0Directrix: [latex]y=2;e=2.5[\/latex]<\/p>\n<p>55. Directrix: [latex]x=-3;e=\\frac{1}{3}[\/latex]<\/p>\n<p>Equations of conics with an [latex]xy[\/latex] term have rotated graphs. For the following exercises, express each equation in polar form with [latex]r[\/latex] as a function of [latex]\\theta[\/latex].<\/p>\n<p>56. [latex]xy=2[\/latex]<\/p>\n<p>57. [latex]{x}^{2}+xy+{y}^{2}=4[\/latex]<\/p>\n<p>58.\u00a0[latex]2{x}^{2}+4xy+2{y}^{2}=9[\/latex]<\/p>\n<p>59. [latex]16{x}^{2}+24xy+9{y}^{2}=4[\/latex]<\/p>\n<p>60.\u00a0[latex]2xy+y=1[\/latex]<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-15776\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at: http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":169554,"menu_order":14,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at: http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-15776","chapter","type-chapter","status-publish","hentry"],"part":14618,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/15776","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/169554"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/15776\/revisions"}],"predecessor-version":[{"id":15779,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/15776\/revisions\/15779"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/14618"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/15776\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=15776"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=15776"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=15776"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=15776"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}