{"id":1836,"date":"2015-11-12T18:30:44","date_gmt":"2015-11-12T18:30:44","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&#038;p=1836"},"modified":"2015-11-12T18:30:44","modified_gmt":"2015-11-12T18:30:44","slug":"writing-equations-of-ellipses-not-centered-at-the-origin","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/writing-equations-of-ellipses-not-centered-at-the-origin\/","title":{"raw":"Writing Equations of Ellipses Not Centered at the Origin","rendered":"Writing Equations of Ellipses Not Centered at the Origin"},"content":{"raw":"<p>Like the graphs of other equations, the graph of an <strong>ellipse<\/strong> can be translated. If an ellipse is translated [latex]h[\/latex] units horizontally and [latex]k[\/latex] units vertically, the center of the ellipse will be [latex]\\left(h,k\\right)[\/latex]. This <strong>translation<\/strong> results in the standard form of the equation we saw previously, with [latex]x[\/latex] replaced by [latex]\\left(x-h\\right)[\/latex] and <em>y<\/em> replaced by [latex]\\left(y-k\\right)[\/latex].\n<\/p><div class=\"textbox\">\n<h3>A General Note: Standard Forms of the Equation of an Ellipse with Center (<em>h<\/em>, <em>k<\/em>)<\/h3>\nThe standard form of the equation of an ellipse with center [latex]\\left(h,\\text{ }k\\right)[\/latex] and <strong>major axis<\/strong> parallel to the <em>x<\/em>-axis is\n<div style=\"text-align: center;\">[latex]\\frac{{\\left(x-h\\right)}^{2}}{{a}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex]<\/div>\nwhere\n<ul><li>[latex]a&gt;b[\/latex]<\/li>\n\t<li>the length of the major axis is [latex]2a[\/latex]<\/li>\n\t<li>the coordinates of the vertices are [latex]\\left(h\\pm a,k\\right)[\/latex]<\/li>\n\t<li>the length of the minor axis is [latex]2b[\/latex]<\/li>\n\t<li>the coordinates of the co-vertices are [latex]\\left(h,k\\pm b\\right)[\/latex]<\/li>\n\t<li>the coordinates of the foci are [latex]\\left(h\\pm c,k\\right)[\/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[\/latex].<\/li>\n<\/ul>\nThe standard form of the equation of an ellipse with center [latex]\\left(h,k\\right)[\/latex] and major axis parallel to the <em>y<\/em>-axis is\n<div style=\"text-align: center;\">[latex]\\frac{{\\left(x-h\\right)}^{2}}{{b}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}=1[\/latex]<\/div>\nwhere\n<ul><li>[latex]a&gt;b[\/latex]<\/li>\n\t<li>the length of the major axis is [latex]2a[\/latex]<\/li>\n\t<li>the coordinates of the vertices are [latex]\\left(h,k\\pm a\\right)[\/latex]<\/li>\n\t<li>the length of the minor axis is [latex]2b[\/latex]<\/li>\n\t<li>the coordinates of the co-vertices are [latex]\\left(h\\pm b,k\\right)[\/latex]<\/li>\n\t<li>the coordinates of the foci are [latex]\\left(h,k\\pm c\\right)[\/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[\/latex].<\/li>\n<\/ul>\nJust as with ellipses centered at the origin, ellipses that are centered at a point [latex]\\left(h,k\\right)[\/latex] have vertices, co-vertices, and foci that are related by the equation [latex]{c}^{2}={a}^{2}-{b}^{2}[\/latex]. We can use this relationship along with the midpoint and distance formulas to find the equation of the ellipse in standard form when the vertices and foci are given.\n\n[caption id=\"attachment_11286\" align=\"aligncenter\" width=\"663\"]<a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/wp-content\/uploads\/sites\/923\/2015\/09\/Screen-Shot-2015-09-15-at-2.24.16-PM.png\"><img class=\"wp-image-11286 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202204\/Screen-Shot-2015-09-15-at-2.24.16-PM.png\" alt=\"(a) Horizontal ellipse with center (h,k) (b) Vertical ellipse with center (h,k)\" width=\"663\" height=\"316\"\/><\/a> <b>Figure 7.<\/b> (a) Horizontal ellipse with center [latex]\\left(h,k\\right)[\/latex] (b) Vertical ellipse with center [latex]\\left(h,k\\right)[\/latex][\/caption]<\/div>\n<div class=\"textbox\">\n<h3>How To: Given the vertices and foci of an ellipse not centered at the origin, write its equation in standard form.<\/h3>\n<ol><li>Determine whether the major axis is parallel to the <em>x<\/em>- or <em>y<\/em>-axis.\n<ol><li>If the <em>y<\/em>-coordinates of the given vertices and foci are the same, then the major axis is parallel to the <em>x<\/em>-axis. Use the standard form [latex]\\frac{{\\left(x-h\\right)}^{2}}{{a}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex].<\/li>\n\t<li>If the <em>x<\/em>-coordinates of the given vertices and foci are the same, then the major axis is parallel to the <em>y<\/em>-axis. Use the standard form [latex]\\frac{{\\left(x-h\\right)}^{2}}{{b}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}=1[\/latex].<\/li>\n<\/ol><\/li>\n\t<li>Identify the center of the ellipse [latex]\\left(h,k\\right)[\/latex] using the midpoint formula and the given coordinates for the vertices.<\/li>\n\t<li>Find [latex]{a}^{2}[\/latex] by solving for the length of the major axis, [latex]2a[\/latex], which is the distance between the given vertices.<\/li>\n\t<li>Find [latex]{c}^{2}[\/latex] using [latex]h[\/latex] and [latex]k[\/latex], found in Step 2, along with the given coordinates for the foci.<\/li>\n\t<li>Solve for [latex]{b}^{2}[\/latex] using the equation [latex]{c}^{2}={a}^{2}-{b}^{2}[\/latex].<\/li>\n\t<li>Substitute the values for [latex]h,k,{a}^{2}[\/latex], and [latex]{b}^{2}[\/latex] into the standard form of the equation determined in Step 1.<\/li>\n<\/ol><\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Writing the Equation of an Ellipse Centered at a Point Other Than the Origin<\/h3>\nWhat is the standard form equation of the ellipse that has vertices [latex]\\left(-2,-8\\right)[\/latex] and [latex]\\left(-2,\\text{2}\\right)[\/latex]\u00a0and foci [latex]\\left(-2,-7\\right)[\/latex] and [latex]\\left(-2,\\text{1}\\right)?[\/latex]\n\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\nThe <em>x<\/em>-coordinates of the vertices and foci are the same, so the major axis is parallel to the <em>y<\/em>-axis. Thus, the equation of the ellipse will have the form\n<div style=\"text-align: center;\">[latex]\\frac{{\\left(x-h\\right)}^{2}}{{b}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}=1[\/latex]<\/div>\nFirst, we identify the center, [latex]\\left(h,k\\right)[\/latex]. The center is halfway between the vertices, [latex]\\left(-2,-8\\right)[\/latex] and [latex]\\left(-2,\\text{2}\\right)[\/latex]. Applying the midpoint formula, we have:\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(h,k\\right)=\\left(\\frac{-2+\\left(-2\\right)}{2},\\frac{-8+2}{2}\\right)\\hfill \\\\ \\text{ }=\\left(-2,-3\\right)\\hfill \\end{array}[\/latex]<\/div>\nNext, we find [latex]{a}^{2}[\/latex]. The length of the major axis, [latex]2a[\/latex], is bounded by the vertices. We solve for [latex]a[\/latex] by finding the distance between the <em>y<\/em>-coordinates of the vertices.\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}2a=2-\\left(-8\\right)\\\\ 2a=10\\\\ a=5\\end{array}[\/latex]<\/div>\nSo [latex]{a}^{2}=25[\/latex].\n\nNow we find [latex]{c}^{2}[\/latex]. The foci are given by [latex]\\left(h,k\\pm c\\right)[\/latex]. So, [latex]\\left(h,k-c\\right)=\\left(-2,-7\\right)[\/latex] and [latex]\\left(h,k+c\\right)=\\left(-2,\\text{1}\\right)[\/latex]. We substitute [latex]k=-3[\/latex] using either of these points to solve for [latex]c[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}k+c=1\\\\ -3+c=1\\\\ c=4\\end{array}[\/latex]<\/div>\nSo [latex]{c}^{2}=16[\/latex].\n\nNext, we solve for [latex]{b}^{2}[\/latex] using the equation [latex]{c}^{2}={a}^{2}-{b}^{2}[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}{c}^{2}={a}^{2}-{b}^{2}\\\\ 16=25-{b}^{2}\\\\ {b}^{2}=9\\end{array}[\/latex]<\/div>\nFinally, we substitute the values found for [latex]h,k,{a}^{2}[\/latex], and [latex]{b}^{2}[\/latex] into the standard form equation for an ellipse:\n<div style=\"text-align: center;\">[latex]\\frac{{\\left(x+2\\right)}^{2}}{9}+\\frac{{\\left(y+3\\right)}^{2}}{25}=1[\/latex]<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 2<\/h3>\nWhat is the standard form equation of the ellipse that has vertices [latex]\\left(-3,3\\right)[\/latex] and [latex]\\left(5,3\\right)[\/latex] and foci [latex]\\left(1 - 2\\sqrt{3},3\\right)[\/latex] and [latex]\\left(1+2\\sqrt{3},3\\right)?[\/latex]\n\n<a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-23\/\" target=\"_blank\">Solution<\/a>\n\n<\/div>","rendered":"<p>Like the graphs of other equations, the graph of an <strong>ellipse<\/strong> can be translated. If an ellipse is translated [latex]h[\/latex] units horizontally and [latex]k[\/latex] units vertically, the center of the ellipse will be [latex]\\left(h,k\\right)[\/latex]. This <strong>translation<\/strong> results in the standard form of the equation we saw previously, with [latex]x[\/latex] replaced by [latex]\\left(x-h\\right)[\/latex] and <em>y<\/em> replaced by [latex]\\left(y-k\\right)[\/latex].\n<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Standard Forms of the Equation of an Ellipse with Center (<em>h<\/em>, <em>k<\/em>)<\/h3>\n<p>The standard form of the equation of an ellipse with center [latex]\\left(h,\\text{ }k\\right)[\/latex] and <strong>major axis<\/strong> parallel to the <em>x<\/em>-axis is<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{{\\left(x-h\\right)}^{2}}{{a}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex]<\/div>\n<p>where<\/p>\n<ul>\n<li>[latex]a>b[\/latex]<\/li>\n<li>the length of the major axis is [latex]2a[\/latex]<\/li>\n<li>the coordinates of the vertices are [latex]\\left(h\\pm a,k\\right)[\/latex]<\/li>\n<li>the length of the minor axis is [latex]2b[\/latex]<\/li>\n<li>the coordinates of the co-vertices are [latex]\\left(h,k\\pm b\\right)[\/latex]<\/li>\n<li>the coordinates of the foci are [latex]\\left(h\\pm c,k\\right)[\/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[\/latex].<\/li>\n<\/ul>\n<p>The standard form of the equation of an ellipse with center [latex]\\left(h,k\\right)[\/latex] and major axis parallel to the <em>y<\/em>-axis is<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{{\\left(x-h\\right)}^{2}}{{b}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}=1[\/latex]<\/div>\n<p>where<\/p>\n<ul>\n<li>[latex]a>b[\/latex]<\/li>\n<li>the length of the major axis is [latex]2a[\/latex]<\/li>\n<li>the coordinates of the vertices are [latex]\\left(h,k\\pm a\\right)[\/latex]<\/li>\n<li>the length of the minor axis is [latex]2b[\/latex]<\/li>\n<li>the coordinates of the co-vertices are [latex]\\left(h\\pm b,k\\right)[\/latex]<\/li>\n<li>the coordinates of the foci are [latex]\\left(h,k\\pm c\\right)[\/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[\/latex].<\/li>\n<\/ul>\n<p>Just as with ellipses centered at the origin, ellipses that are centered at a point [latex]\\left(h,k\\right)[\/latex] have vertices, co-vertices, and foci that are related by the equation [latex]{c}^{2}={a}^{2}-{b}^{2}[\/latex]. We can use this relationship along with the midpoint and distance formulas to find the equation of the ellipse in standard form when the vertices and foci are given.<\/p>\n<div id=\"attachment_11286\" style=\"width: 673px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/wp-content\/uploads\/sites\/923\/2015\/09\/Screen-Shot-2015-09-15-at-2.24.16-PM.png\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-11286\" class=\"wp-image-11286 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202204\/Screen-Shot-2015-09-15-at-2.24.16-PM.png\" alt=\"(a) Horizontal ellipse with center (h,k) (b) Vertical ellipse with center (h,k)\" width=\"663\" height=\"316\" \/><\/a><\/p>\n<p id=\"caption-attachment-11286\" class=\"wp-caption-text\"><b>Figure 7.<\/b> (a) Horizontal ellipse with center [latex]\\left(h,k\\right)[\/latex] (b) Vertical ellipse with center [latex]\\left(h,k\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given the vertices and foci of an ellipse not centered at the origin, write its equation in standard form.<\/h3>\n<ol>\n<li>Determine whether the major axis is parallel to the <em>x<\/em>&#8211; or <em>y<\/em>-axis.\n<ol>\n<li>If the <em>y<\/em>-coordinates of the given vertices and foci are the same, then the major axis is parallel to the <em>x<\/em>-axis. Use the standard form [latex]\\frac{{\\left(x-h\\right)}^{2}}{{a}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex].<\/li>\n<li>If the <em>x<\/em>-coordinates of the given vertices and foci are the same, then the major axis is parallel to the <em>y<\/em>-axis. Use the standard form [latex]\\frac{{\\left(x-h\\right)}^{2}}{{b}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}=1[\/latex].<\/li>\n<\/ol>\n<\/li>\n<li>Identify the center of the ellipse [latex]\\left(h,k\\right)[\/latex] using the midpoint formula and the given coordinates for the vertices.<\/li>\n<li>Find [latex]{a}^{2}[\/latex] by solving for the length of the major axis, [latex]2a[\/latex], which is the distance between the given vertices.<\/li>\n<li>Find [latex]{c}^{2}[\/latex] using [latex]h[\/latex] and [latex]k[\/latex], found in Step 2, along with the given coordinates for the foci.<\/li>\n<li>Solve for [latex]{b}^{2}[\/latex] using the equation [latex]{c}^{2}={a}^{2}-{b}^{2}[\/latex].<\/li>\n<li>Substitute the values for [latex]h,k,{a}^{2}[\/latex], and [latex]{b}^{2}[\/latex] into the standard form of the equation determined in Step 1.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Writing the Equation of an Ellipse Centered at a Point Other Than the Origin<\/h3>\n<p>What is the standard form equation of the ellipse that has vertices [latex]\\left(-2,-8\\right)[\/latex] and [latex]\\left(-2,\\text{2}\\right)[\/latex]\u00a0and foci [latex]\\left(-2,-7\\right)[\/latex] and [latex]\\left(-2,\\text{1}\\right)?[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\n<p>The <em>x<\/em>-coordinates of the vertices and foci are the same, so the major axis is parallel to the <em>y<\/em>-axis. Thus, the equation of the ellipse will have the form<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{{\\left(x-h\\right)}^{2}}{{b}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}=1[\/latex]<\/div>\n<p>First, we identify the center, [latex]\\left(h,k\\right)[\/latex]. The center is halfway between the vertices, [latex]\\left(-2,-8\\right)[\/latex] and [latex]\\left(-2,\\text{2}\\right)[\/latex]. Applying the midpoint formula, we have:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(h,k\\right)=\\left(\\frac{-2+\\left(-2\\right)}{2},\\frac{-8+2}{2}\\right)\\hfill \\\\ \\text{ }=\\left(-2,-3\\right)\\hfill \\end{array}[\/latex]<\/div>\n<p>Next, we find [latex]{a}^{2}[\/latex]. The length of the major axis, [latex]2a[\/latex], is bounded by the vertices. We solve for [latex]a[\/latex] by finding the distance between the <em>y<\/em>-coordinates of the vertices.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}2a=2-\\left(-8\\right)\\\\ 2a=10\\\\ a=5\\end{array}[\/latex]<\/div>\n<p>So [latex]{a}^{2}=25[\/latex].<\/p>\n<p>Now we find [latex]{c}^{2}[\/latex]. The foci are given by [latex]\\left(h,k\\pm c\\right)[\/latex]. So, [latex]\\left(h,k-c\\right)=\\left(-2,-7\\right)[\/latex] and [latex]\\left(h,k+c\\right)=\\left(-2,\\text{1}\\right)[\/latex]. We substitute [latex]k=-3[\/latex] using either of these points to solve for [latex]c[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}k+c=1\\\\ -3+c=1\\\\ c=4\\end{array}[\/latex]<\/div>\n<p>So [latex]{c}^{2}=16[\/latex].<\/p>\n<p>Next, we solve for [latex]{b}^{2}[\/latex] using the equation [latex]{c}^{2}={a}^{2}-{b}^{2}[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}{c}^{2}={a}^{2}-{b}^{2}\\\\ 16=25-{b}^{2}\\\\ {b}^{2}=9\\end{array}[\/latex]<\/div>\n<p>Finally, we substitute the values found for [latex]h,k,{a}^{2}[\/latex], and [latex]{b}^{2}[\/latex] into the standard form equation for an ellipse:<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{{\\left(x+2\\right)}^{2}}{9}+\\frac{{\\left(y+3\\right)}^{2}}{25}=1[\/latex]<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 2<\/h3>\n<p>What is the standard form equation of the ellipse that has vertices [latex]\\left(-3,3\\right)[\/latex] and [latex]\\left(5,3\\right)[\/latex] and foci [latex]\\left(1 - 2\\sqrt{3},3\\right)[\/latex] and [latex]\\left(1+2\\sqrt{3},3\\right)?[\/latex]<\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-23\/\" target=\"_blank\">Solution<\/a><\/p>\n<\/div>\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-1836\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <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><\/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":276,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1836","chapter","type-chapter","status-publish","hentry"],"part":1825,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1836","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/users\/276"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1836\/revisions"}],"predecessor-version":[{"id":2224,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1836\/revisions\/2224"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/1825"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1836\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/media?parent=1836"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1836"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/contributor?post=1836"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/license?post=1836"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}