{"id":2525,"date":"2025-08-13T17:48:18","date_gmt":"2025-08-13T17:48:18","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=2525"},"modified":"2025-08-13T17:49:11","modified_gmt":"2025-08-13T17:49:11","slug":"ellipses-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/ellipses-learn-it-2\/","title":{"raw":"Ellipses: Learn It 2","rendered":"Ellipses: Learn It 2"},"content":{"raw":"<h2>Writing Equations of Ellipses Not Centered at the Origin<\/h2>\r\nLike 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].\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>ellipse with a center (h,k)<\/h3>\r\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\r\n<p style=\"text-align: center;\">[latex]\\frac{{\\left(x-h\\right)}^{2}}{{a}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex]<\/p>\r\nwhere\r\n<ul>\r\n \t<li>[latex]a&gt;b[\/latex]<\/li>\r\n \t<li>the length of the major axis is [latex]2a[\/latex]<\/li>\r\n \t<li>the coordinates of the vertices are [latex]\\left(h\\pm a,k\\right)[\/latex]<\/li>\r\n \t<li>the length of the minor axis is [latex]2b[\/latex]<\/li>\r\n \t<li>the coordinates of the co-vertices are [latex]\\left(h,k\\pm b\\right)[\/latex]<\/li>\r\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>\r\n<\/ul>\r\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\r\n<p style=\"text-align: center;\">[latex]\\frac{{\\left(x-h\\right)}^{2}}{{b}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}=1[\/latex]<\/p>\r\nwhere\r\n<ul>\r\n \t<li>[latex]a&gt;b[\/latex]<\/li>\r\n \t<li>the length of the major axis is [latex]2a[\/latex]<\/li>\r\n \t<li>the coordinates of the vertices are [latex]\\left(h,k\\pm a\\right)[\/latex]<\/li>\r\n \t<li>the length of the minor axis is [latex]2b[\/latex]<\/li>\r\n \t<li>the coordinates of the co-vertices are [latex]\\left(h\\pm b,k\\right)[\/latex]<\/li>\r\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>\r\n<\/ul>\r\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.\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/precalctwoxmaster\/wp-content\/uploads\/sites\/145\/2015\/09\/Screen-Shot-2015-09-15-at-2.24.16-PM.png\"><img class=\"wp-image-11286 size-full aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182716\/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>\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given the vertices and foci of an ellipse not centered at the origin, write its equation in standard form.<\/strong>\r\n<ol>\r\n \t<li>Determine whether the major axis is parallel to the <em>x<\/em>- or <em>y<\/em>-axis.\r\n<ol>\r\n \t<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>\r\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>\r\n<\/ol>\r\n<\/li>\r\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>\r\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>\r\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>\r\n \t<li>Solve for [latex]{b}^{2}[\/latex] using the equation [latex]{c}^{2}={a}^{2}-{b}^{2}[\/latex].<\/li>\r\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>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">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][reveal-answer q=\"738186\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"738186\"]\r\n\r\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\r\n<p style=\"text-align: center;\">[latex]\\frac{{\\left(x-h\\right)}^{2}}{{b}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}=1[\/latex]<\/p>\r\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:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\left(h,k\\right)&amp;=\\left(\\frac{-2+\\left(-2\\right)}{2},\\frac{-8+2}{2}\\right) \\\\ &amp;=\\left(-2,-3\\right) \\end{align}[\/latex]<\/p>\r\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.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2a=2-\\left(-8\\right)\\\\ 2a=10\\\\ a=5\\end{gathered}[\/latex]<\/p>\r\nSo [latex]{a}^{2}=25[\/latex].\r\n\r\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].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}k+c=1\\\\ -3+c=1\\\\ c=4\\end{gathered}[\/latex]<\/p>\r\nSo [latex]{c}^{2}=16[\/latex].\r\n\r\nNext, we solve for [latex]{b}^{2}[\/latex] using the equation [latex]{c}^{2}={a}^{2}-{b}^{2}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{c}^{2}={a}^{2}-{b}^{2}\\\\ 16=25-{b}^{2}\\\\ {b}^{2}=9\\end{gathered}[\/latex]<\/p>\r\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:\r\n<p style=\"text-align: center;\">[latex]\\frac{{\\left(x+2\\right)}^{2}}{9}+\\frac{{\\left(y+3\\right)}^{2}}{25}=1[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">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][reveal-answer q=\"653759\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"653759\"][latex]\\frac{{\\left(x - 1\\right)}^{2}}{16}+\\frac{{\\left(y - 3\\right)}^{2}}{4}=1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]173951[\/ohm_question]<\/section>","rendered":"<h2>Writing Equations of Ellipses Not Centered at the Origin<\/h2>\n<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].<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>ellipse with a center (h,k)<\/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<p style=\"text-align: center;\">[latex]\\frac{{\\left(x-h\\right)}^{2}}{{a}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex]<\/p>\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<p style=\"text-align: center;\">[latex]\\frac{{\\left(x-h\\right)}^{2}}{{b}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}=1[\/latex]<\/p>\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<p><a href=\"https:\/\/courses.lumenlearning.com\/precalctwoxmaster\/wp-content\/uploads\/sites\/145\/2015\/09\/Screen-Shot-2015-09-15-at-2.24.16-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-11286 size-full aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182716\/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<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given the vertices and foci of an ellipse not centered at the origin, write its equation in standard form.<\/strong><\/p>\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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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 class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q738186\">Show Solution<\/button><\/p>\n<div id=\"q738186\" class=\"hidden-answer\" style=\"display: none\">\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<p style=\"text-align: center;\">[latex]\\frac{{\\left(x-h\\right)}^{2}}{{b}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}=1[\/latex]<\/p>\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<p style=\"text-align: center;\">[latex]\\begin{align}\\left(h,k\\right)&=\\left(\\frac{-2+\\left(-2\\right)}{2},\\frac{-8+2}{2}\\right) \\\\ &=\\left(-2,-3\\right) \\end{align}[\/latex]<\/p>\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<p style=\"text-align: center;\">[latex]\\begin{gathered}2a=2-\\left(-8\\right)\\\\ 2a=10\\\\ a=5\\end{gathered}[\/latex]<\/p>\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<p style=\"text-align: center;\">[latex]\\begin{gathered}k+c=1\\\\ -3+c=1\\\\ c=4\\end{gathered}[\/latex]<\/p>\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<p style=\"text-align: center;\">[latex]\\begin{gathered}{c}^{2}={a}^{2}-{b}^{2}\\\\ 16=25-{b}^{2}\\\\ {b}^{2}=9\\end{gathered}[\/latex]<\/p>\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<p style=\"text-align: center;\">[latex]\\frac{{\\left(x+2\\right)}^{2}}{9}+\\frac{{\\left(y+3\\right)}^{2}}{25}=1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">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<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q653759\">Show Solution<\/button><\/p>\n<div id=\"q653759\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac{{\\left(x - 1\\right)}^{2}}{16}+\\frac{{\\left(y - 3\\right)}^{2}}{4}=1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm173951\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173951&theme=lumen&iframe_resize_id=ohm173951&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":13,"menu_order":6,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":522,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2525"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/13"}],"version-history":[{"count":2,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2525\/revisions"}],"predecessor-version":[{"id":2536,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2525\/revisions\/2536"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/522"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2525\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=2525"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=2525"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=2525"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=2525"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}