{"id":87625,"date":"2021-08-09T13:26:13","date_gmt":"2021-08-09T18:26:13","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=87625"},"modified":"2026-03-25T11:29:13","modified_gmt":"2026-03-25T16:29:13","slug":"hyperbolas","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/hyperbolas\/","title":{"rendered":"Hyperbolas"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_XZc8cFbGtUw\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_XZc8cFbGtUw\" data-source-videoID=\"XZc8cFbGtUw\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Hyperbolas Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Hyperbolas\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_XZc8cFbGtUw:hover {cursor:pointer;} img#videoThumbnailImage_XZc8cFbGtUw {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/1739-thumb-final-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_XZc8cFbGtUw\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_XZc8cFbGtUw\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Hyperbolas\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_XZc8cFbGtUw\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_XZc8cFbGtUw\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_XZc8cFbGtUw\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction yaP_Function() {\n  var x = document.getElementById(\"yaP\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"yaP_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"yaP\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Horizontal_Hyperbola_Equation\" class=\"smooth-scroll\">Horizontal Hyperbola Equation<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Parts_of_a_Hyperbola\" class=\"smooth-scroll\">Parts of a Hyperbola<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Finding_a_and_b\" class=\"smooth-scroll\">Finding a and b<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Asymptote_Equations\" class=\"smooth-scroll\">Asymptote Equations<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hi, and welcome to this review of hyperbolas!<\/p>\n<p>Not only do hyperbolas have an interesting name, but they\u2019re actually a close relative of the ellipse, even though they look a bit different.<\/p>\n<h2><span id=\"Horizontal_Hyperbola_Equation\" class=\"m-toc-anchor\"><\/span>Horizontal Hyperbola Equation<\/h2>\n<p>\nTo see how they\u2019re related, let\u2019s look at the equation for a horizontal hyperbola: <\/p>\n<div class=\"examplesentence\" style=\"font-size: 125%;\">\\(\\frac{(x-h)^{2}}{a^{2}}-\\frac{(y-k)^{2}}{b^{2}}\\)<span style=\"font-size: 80%;\">\\(=1\\)<\/span><\/div>\n<p>\n&nbsp;<br \/>\nIt looks almost exactly the same as the ellipse equation, except that instead of addition between the two fractions, we see subtraction.<\/p>\n<p>Now, let\u2019s see what it looks like:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-89695\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-1.png\" alt=\"Image of an ellipse with 2 focus points. 2 curved lines drawn to the left and right, with a focus point at the middle of the curve. \" width=\"777\" height=\"437\"style=\"box-shadow: 1.5px 1.5px 3px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-1.png 1920w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-1-300x169.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-1-1024x576.png 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-1-768x432.png 768w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-1-1536x864.png 1536w\" sizes=\"auto, (max-width: 777px) 100vw, 777px\" \/><\/p>\n<p>Okay, it definitely doesn\u2019t look much like an ellipse. But it does have two focus points! It also has one of those math-y definitions we all like so much: All the points on the curves of a <strong>hyperbola<\/strong> lie such that the difference of the distances from any point to both focus points is a constant. <\/p>\n<p>This means that drawing line segments to any point on the hyperbola will result in the longer one always being the same amount longer than the smaller one. The two line segments will both get longer the farther away from the center of the hyperbola we get, but they get longer at the exact same rate so the difference between them is the same.<\/p>\n<h2><span id=\"Parts_of_a_Hyperbola\" class=\"m-toc-anchor\"><\/span>Parts of a Hyperbola<\/h2>\n<p>\nThere\u2019s a lot more to a hyperbola than its definition and foci though. It has vertices, a center, and asymptotes! <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-89701\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-2.png\" alt=\"Image of a hyperbolas with foci, center with perpendicular dashed lines showing the asymptote, and 2 vertexes.\" width=\"777\" height=\"437\"style=\"box-shadow: 1.5px 1.5px 3px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-2-300x168.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-2-1024x574.png 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-2-1536x861.png 1536w\" sizes=\"auto, (max-width: 777px) 100vw, 777px\" \/><\/p>\n<p>There\u2019s a lot going on here! We can see the two asymptote dashed lines intersect at the center, which, just like in an ellipse or circle, gives us the (<span style=\"font-style:normal; font-size:90%\">\\(h\\)<\/span>, <span style=\"font-style:normal; font-size:90%\">\\(k\\)<\/span>) we need in our equation.<\/p>\n<p>The <strong>center<\/strong> is equidistant from the two vertices as well as the two focus points, and all of these points lie on the same horizontal line. That\u2019s why this is a <strong>horizontal hyperbola<\/strong>.<\/p>\n<p>The asymptotes serve as guides for our graph of the hyperbola. As the hyperbola stretches out away from the center, each of its curves will get closer and closer to the nearest asymptote but will never touch. We\u2019ll come back to the asymptotes in a minute.<\/p>\n<p>But first, we need to go back to our equation.<\/p>\n<h2><span id=\"Finding_a_and_b\" class=\"m-toc-anchor\"><\/span>Finding a and b<\/h2>\n<p>\nSo far we\u2019ve identified how to find <span style=\"font-style:normal; font-size:90%\">\\(h\\)<\/span> and <span style=\"font-style:normal; font-size:90%\">\\(k\\)<\/span> in this equation by finding the center of the hyperbola. But what about <span style=\"font-style:normal; font-size:90%\">\\(a\\)<\/span> and <span style=\"font-style:normal; font-size:90%\">\\(b\\)<\/span>? For that, we need to add even more to our drawing: <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-89704\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-3-correct-.png\" alt=\"Image of a hyperbolas with the foci, vertexes, center point with perpendicular dashed lines showing the asymptote, and the major axis with 5 points.  \" width=\"777\" height=\"437\"style=\"box-shadow: 1.5px 1.5px 3px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-3-correct-.png 1917w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-3-correct--300x168.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-3-correct--1024x575.png 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-3-correct--768x431.png 768w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-3-correct--1536x862.png 1536w\" sizes=\"auto, (max-width: 777px) 100vw, 777px\" \/><\/p>\n<p>Now we can see the <strong>major axis<\/strong>, which has five points on it! The two focus points, shown in black; the two vertex points, shown in green; and the red center point. We can also see a yellow box that intersects the two vertex points at its side midpoints, and also has the asymptotes going directly through its corners. This box helps us find the lengths of <span style=\"font-style:normal; font-size:90%\">\\(a\\)<\/span> and <span style=\"font-style:normal; font-size:90%\">\\(b\\)<\/span>! <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-89707\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-4.png\" alt=\"Image of a hyperbolas with 2 foci, center point with perpendicular dashed lines showing the asymptote, vertexes, major axis, and a yellow box with another square inscribed inside with lines B and A.  \" width=\"777\" height=\"437\"style=\"box-shadow: 1.5px 1.5px 3px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-4.png 1920w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-4-300x169.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-4-1024x576.png 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-4-768x432.png 768w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-4-1536x864.png 1536w\" sizes=\"auto, (max-width: 777px) 100vw, 777px\" \/><\/p>\n<p>Now we can see our <span style=\"font-style:normal; font-size:90%\">\\(a\\)<\/span> and <span style=\"font-style:normal; font-size:90%\">\\(b\\)<\/span>! The <span style=\"font-style:normal; font-size:90%\">\\(a\\)<\/span> is always on the major axis, which is called the <strong>transverse axis<\/strong>.<\/p>\n<h2><span id=\"Asymptote_Equations\" class=\"m-toc-anchor\"><\/span>Asymptote Equations<\/h2>\n<p>\nThe major or transverse axis is easy to identify since it has the focus points on it as well as the vertex points. So <span style=\"font-style:normal; font-size:90%\">\\(a\\)<\/span> is the distance from the center to either vertex point. And <span style=\"font-style:normal; font-size:90%\">\\(b\\)<\/span> is the distance from the center to the top of the box formed by drawing a rectangle that has the asymptotes cutting through its corners.<\/p>\n<p>That accounts for all the variables in our equation. It also allows us to find the equations of our asymptotes. This is easiest when the center of our hyperbola is at the origin, also known as the point (0,0). Here\u2019s the equation for an asymptote when the center is the origin: <\/p>\n<div class=\"examplesentence\">\\(y=(\\frac{b}{a})x\\)<\/div>\n<p>\n&nbsp;<br \/>\nYou might see it written this way too: <\/p>\n<div class=\"examplesentence\">\\(y=\\frac{bx}{a}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThat gives us one of them. Here\u2019s the other: <\/p>\n<div class=\"examplesentence\">\\(y=-(\\frac{b}{a})x \\text{ or }y=\\frac{-bx}{a}\\)<\/div>\n<p>\n&nbsp;<br \/>\nBut what about when the hyperbola is not centered at the origin? Then the formulas look a bit more complicated: <\/p>\n<div class=\"examplesentence\">\\(y=(\\frac{b}{a})(x-h)+k\\text{ and }\\)\\(y=-(\\frac{b}{a})(x-h)+k\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe don\u2019t actually need to find the asymptotes to find the equation of our hyperbola. But there\u2019s actually one more important variable we need to be able to find, and that\u2019s <span style=\"font-style:normal; font-size:90%\">\\(c\\)<\/span>, which is the distance from the center to the focus points.<\/p>\n<p>Fortunately, we have our old friend the <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/pythagorean-theorem\/\">Pythagorean theorem<\/a> to help us find it from <span style=\"font-style:normal; font-size:90%\">\\(a\\)<\/span> and <span style=\"font-style:normal; font-size:90%\">\\(b\\)<\/span>. <\/p>\n<div class=\"examplesentence\">\\(a^{2}+b^{2}=c^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThat\u2019s awesome! But why does it work? It turns out that the right triangle we can form by drawing a line segment from either ends of <span style=\"font-style:normal; font-size:90%\">\\(b\\)<\/span> and <span style=\"font-style:normal; font-size:90%\">\\(a\\)<\/span> has a hypotenuse that\u2019s the same distance as the distance from the center of the hyperbola to either of its focus points.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-89713\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Hyperbolas-5.png\" alt=\"Image of hyperbolas with a square yellow box that has lines segments c, b, c, and a. These line segments draw out a triangle inscribed in the square with c drawn outward.\" width=\"777\" height=\"437\"style=\"box-shadow: 1.5px 1.5px 3px grey\" \/><\/p>\n<p>In a math problem, we\u2019re more likely to know the focus points than to know the length of <span style=\"font-style:normal; font-size:90%\">\\(b\\)<\/span>, so we have to use the Pythagorean theorem to solve for <span style=\"font-style:normal; font-size:90%\">\\(b\\)<\/span> instead of for <span style=\"font-style:normal; font-size:90%\">\\(c\\)<\/span>. If we rearrange the Pythagorean theorem, we can use it this way:<\/p>\n<div class=\"examplesentence\">\\(b^{2}=c^{2}-a^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNone of this changes too much when our hyperbola is oriented vertically instead of horizontally. Our equation rearranges slightly to:<\/p>\n<div class=\"examplesentence\" style=\"font-size: 125%;\">\\(\\frac{(y-k)^{2}}{a^2}-\\frac{(x-h)^{2}}{b^{2}}\\)<span style=\"font-size: 80%;\">\\(=1\\)<\/span><\/div>\n<p>\n&nbsp;<br \/>\nNotice that the <span style=\"font-style:normal; font-size:90%\">\\(y\\)<\/span>-term comes first and is over the <span style=\"font-style:normal; font-size:90%\">\\(a^{2}\\)<\/span>. That\u2019s because <span style=\"font-style:normal; font-size:90%\">\\(a\\)<\/span> is always on the major (or transverse) axis, and with a vertical hyperbola, our major axis is parallel to the <span style=\"font-style:normal; font-size:90%\">\\(y\\)<\/span>-axis.<\/p>\n<p>Let\u2019s try a problem that will require us to find the <span style=\"font-style:normal; font-size:90%\">\\(h\\)<\/span>, <span style=\"font-style:normal; font-size:90%\">\\(k\\)<\/span>, <span style=\"font-style:normal; font-size:90%\">\\(a\\)<\/span>, and <span style=\"font-style:normal; font-size:90%\">\\(b\\)<\/span> when given focus point and the two vertex points.<\/p>\n<p>What is the equation of a hyperbola that has vertices at (2,5) and (2,-3) and focus points at (2,6) and (2,-4)?<\/p>\n<p><em>Hint<\/em>: Sketching the points and then finding the center will help us in finding <span style=\"font-style:normal; font-size:90%\">\\(a\\)<\/span> and <span style=\"font-style:normal; font-size:90%\">\\(c\\)<\/span>, as well as giving us <span style=\"font-style:normal; font-size:90%\">\\(h\\)<\/span> and <span style=\"font-style:normal; font-size:90%\">\\(k\\)<\/span>.<\/p>\n<div class=\"examplesentence\" style=\"font-size: 125%;\">\\(\\frac{(y-1)^{2}}{16}-\\frac{(x-2)^{2}}{9}\\)<span style=\"font-size: 80%;\">\\(=1\\)<\/span><\/div>\n<p>\n&nbsp;<br \/>\nWe start with the general equation of a vertical hyperbola that is taller than it is wide. We know to use this one because all the points given are on the same vertical line. Next, we find the center of the hyperbola by finding the midpoint of the two vertex points.<\/p>\n<p>Once we know the center is at the point (2,1), we substitute the <span style=\"font-style:normal; font-size:90%\">\\(x\\)<\/span>&#8211; and <span style=\"font-style:normal; font-size:90%\">\\(y\\)<\/span>-values of the center for <span style=\"font-style:normal; font-size:90%\">\\(h\\)<\/span> and <span style=\"font-style:normal; font-size:90%\">\\(k\\)<\/span> in our general equation.<\/p>\n<p>To find the value for <span style=\"font-style:normal; font-size:90%\">\\(a\\)<\/span>, we find the distance between our center that we just found and either of the vertex points, which is a distance of 4 units. To find <span style=\"font-style:normal; font-size:90%\">\\(b\\)<\/span>, we need to use our focus points to find <span style=\"font-style:normal; font-size:90%\">\\(c\\)<\/span> first. Taking the distance from the center to either focus point, we find a distance of 5.<\/p>\n<p>From there, we find that <span style=\"font-style:normal; font-size:90%\">\\(b=3\\)<\/span> using the Pythagorean theorem. Now we plug <span style=\"font-style:normal; font-size:90%\">\\(a\\)<\/span> into our equation and square it to get our first denominator, which is 16. We do the same thing for <span style=\"font-style:normal; font-size:90%\">\\(b\\)<\/span> to find our second denominator is 9. Done!<\/p>\n<p>That\u2019s all there is to it! I hope this review of hyperbolas was helpful! Thanks for watching, and happy studying!<\/p>\n<ul class=\"citelist\">\n<li><a href=\"https:\/\/www.mathsisfun.com\/geometry\/hyperbola.html\"target=\"_blank\">\u201cHyperbola.\u201d n.d.<\/a><\/li>\n<li><a href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/chapter\/equations-of-hyperbolas\/\"target=\"_blank\">\u201cEquations of Hyperbolas | College Algebra.\u201d n.d.<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/geometry\/\">Return to Geometry Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Geometry Videos<\/p>\n","protected":false},"author":1,"featured_media":100750,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-87625","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-graph-videos","7":"page_type-video","8":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/87625","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=87625"}],"version-history":[{"count":6,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/87625\/revisions"}],"predecessor-version":[{"id":262024,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/87625\/revisions\/262024"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/100750"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=87625"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}