{"id":1434,"date":"2023-06-05T14:51:42","date_gmt":"2023-06-05T14:51:42","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/chapter\/conic-sections-in-polar-coordinates\/"},"modified":"2023-06-05T14:51:42","modified_gmt":"2023-06-05T14:51:42","slug":"conic-sections-in-polar-coordinates","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/chapter\/conic-sections-in-polar-coordinates\/","title":{"raw":"Conic Sections in Polar Coordinates","rendered":"Conic Sections in Polar Coordinates"},"content":{"raw":"\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n \t<li>Identify a conic in polar form.<\/li>\n \t<li>Graph the polar equations of conics.<\/li>\n \t<li>De\ufb01ne conics in terms of a focus and a directrix.<\/li>\n<\/ul>\n<\/div>\n\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183120\/CNX_Precalc_Figure_10_05_008n2.jpg\" alt=\"The planets and their orbits around the sun. (Pluto is included.)\" width=\"975\" height=\"353\"> <b>Figure 1.<\/b> Planets orbiting the sun follow elliptical paths. (credit: NASA Blueshift, Flickr)[\/caption]\n\nMost of us are familiar with orbital motion, such as the motion of a planet around the sun or an electron around an atomic nucleus. Within the planetary system, orbits of planets, asteroids, and comets around a larger celestial body are often elliptical. Comets, however, may take on a parabolic or hyperbolic orbit instead. And, in reality, the characteristics of the planets\u2019 orbits may vary over time. Each orbit is tied to the location of the celestial body being orbited and the distance and direction of the planet or other object from that body. As a result, we tend to use polar coordinates to represent these orbits.\n\nIn an elliptical orbit, the <strong>periapsis<\/strong> is the point at which the two objects are closest, and the <strong>apoapsis<\/strong> is the point at which they are farthest apart. Generally, the velocity of the orbiting body tends to increase as it approaches the periapsis and decrease as it approaches the apoapsis. Some objects reach an escape velocity, which results in an infinite orbit. These bodies exhibit either a parabolic or a hyperbolic orbit about a body; the orbiting body breaks free of the celestial body\u2019s gravitational pull and fires off into space. Each of these orbits can be modeled by a conic section in the polar coordinate system.\n<h2>Identifying a Conic in Polar Form<\/h2>\nAny conic may be determined by three characteristics: a single <strong>focus<\/strong>, a fixed line called the <strong>directrix<\/strong>, and the ratio of the distances of each to a point on the graph. Consider the <strong>parabola<\/strong> [latex]x=2+{y}^{2}[\/latex] shown in Figure 2.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183123\/CNX_Precalc_Figure_10_05_0012.jpg\" alt=\"\" width=\"487\" height=\"316\"> <b>Figure 2<\/b>[\/caption]\n\nIn <a href=\"https:\/\/courses.lumenlearning.com\/precalctwoxmaster\/chapter\/introduction-to-the-parabola\/\" target=\"_blank\" rel=\"noopener\">The Parabola<\/a>, we learned how a parabola is defined by the focus (a fixed point) and the directrix (a fixed line). In this section, we will learn how to define any conic in the polar coordinate system in terms of a fixed point, the focus [latex]P\\left(r,\\theta \\right)[\/latex] at the pole, and a line, the directrix, which is perpendicular to the polar axis.\n\nIf [latex]F[\/latex] is a fixed point, the focus, and [latex]D[\/latex] is a fixed line, the directrix, then we can let [latex]e[\/latex] be a fixed positive number, called the <strong>eccentricity<\/strong>, which we can define as the ratio of the distances from a point on the graph to the focus and the point on the graph to the directrix. Then the set of all points [latex]P[\/latex] such that [latex]e=\\frac{PF}{PD}[\/latex] is a conic. In other words, we can define a conic as the set of all points [latex]P[\/latex] with the property that the ratio of the distance from [latex]P[\/latex] to [latex]F[\/latex] to the distance from [latex]P[\/latex] to [latex]D[\/latex] is equal to the constant [latex]e[\/latex].\n\nFor a conic with eccentricity [latex]e[\/latex],\n<ul>\n \t<li>if [latex]0\\le e&lt;1[\/latex], the conic is an ellipse<\/li>\n \t<li>if [latex]e=1[\/latex], the conic is a parabola<\/li>\n \t<li>if [latex]e&gt;1[\/latex], the conic is an hyperbola<\/li>\n<\/ul>\nWith this definition, we may now define a conic in terms of the directrix, [latex]x=\\pm p[\/latex], the eccentricity [latex]e[\/latex], and the angle [latex]\\theta [\/latex]. Thus, each conic may be written as a <strong>polar equation<\/strong>, an equation written in terms of [latex]r[\/latex] and [latex]\\theta [\/latex].\n<div class=\"textbox\">\n<h3>A General Note: The Polar Equation for a Conic<\/h3>\nFor a conic with a focus at the origin, if the directrix is [latex]x=\\pm p[\/latex], where [latex]p[\/latex] is a positive real number, and the <strong>eccentricity<\/strong> is a positive real number [latex]e[\/latex], the conic has a <strong>polar equation<\/strong>\n\n[latex]r=\\frac{ep}{1\\pm e \\cos \\theta }[\/latex]\n\nFor a conic with a focus at the origin, if the directrix is [latex]y=\\pm p[\/latex], where [latex]p[\/latex] is a positive real number, and the eccentricity is a positive real number [latex]e[\/latex], the conic has a polar equation\n\n[latex]r=\\frac{ep}{1\\pm e \\sin \\theta }[\/latex]\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given the polar equation for a conic, identify the type of conic, the directrix, and the eccentricity.<\/h3>\n<ol>\n \t<li>Multiply the numerator and denominator by the reciprocal of the constant in the denominator to rewrite the equation in standard form.<\/li>\n \t<li>Identify the eccentricity [latex]e[\/latex] as the coefficient of the trigonometric function in the denominator.<\/li>\n \t<li>Compare [latex]e[\/latex] with 1 to determine the shape of the conic.<\/li>\n \t<li>Determine the directrix as [latex]x=p[\/latex] if cosine is in the denominator and [latex]y=p[\/latex] if sine is in the denominator. Set [latex]ep[\/latex] equal to the numerator in standard form to solve for [latex]x[\/latex] or [latex]y[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Identifying a Conic Given the Polar Form<\/h3>\nFor each of the following equations, identify the conic with focus at the origin, the <strong>directrix<\/strong>, and the <strong>eccentricity<\/strong>.\n<ol>\n \t<li>[latex]r=\\frac{6}{3+2 \\sin \\theta }[\/latex]<\/li>\n \t<li>[latex]r=\\frac{12}{4+5 \\cos \\theta }[\/latex]<\/li>\n \t<li>[latex]r=\\frac{7}{2 - 2 \\sin \\theta }[\/latex]<\/li>\n<\/ol>\n[reveal-answer q=\"293935\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"293935\"]\n\nFor each of the three conics, we will rewrite the equation in standard form. Standard form has a 1 as the constant in the denominator. Therefore, in all three parts, the first step will be to multiply the numerator and denominator by the reciprocal of the constant of the original equation, [latex]\\frac{1}{c}[\/latex], where [latex]c[\/latex] is that constant.\n<ol>\n \t<li>Multiply the numerator and denominator by [latex]\\frac{1}{3}[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{align}r&amp;=\\frac{6}{3+2\\sin \\theta }\\cdot \\frac{\\left(\\frac{1}{3}\\right)}{\\left(\\frac{1}{3}\\right)} \\\\ &amp;=\\frac{6\\left(\\frac{1}{3}\\right)}{3\\left(\\frac{1}{3}\\right)+2\\left(\\frac{1}{3}\\right)\\sin \\theta } \\\\ r&amp;=\\frac{2}{1+\\frac{2}{3}\\sin \\theta} \\end{align}[\/latex]<\/div>\nBecause [latex]\\sin \\text{ }\\theta [\/latex] is in the denominator, the directrix is [latex]y=p[\/latex]. Comparing to standard form, note that [latex]e=\\frac{2}{3}[\/latex]. Therefore, from the numerator,\n<div style=\"text-align: center;\">[latex]\\begin{align}2&amp;=ep \\\\ 2&amp;=\\frac{2}{3}p \\\\ \\left(\\frac{3}{2}\\right)2&amp;=\\left(\\frac{3}{2}\\right)\\frac{2}{3}p \\\\ 3&amp;=p \\end{align}[\/latex]<\/div>\nSince [latex]e&lt;1[\/latex], the conic is an <strong>ellipse<\/strong>. The eccentricity is [latex]e=\\frac{2}{3}[\/latex] and the directrix is [latex]y=3[\/latex].<\/li>\n \t<li>Multiply the numerator and denominator by [latex]\\frac{1}{4}[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{align} r&amp;=\\frac{12}{4+5 \\cos \\theta }\\cdot \\frac{\\left(\\frac{1}{4}\\right)}{\\left(\\frac{1}{4}\\right)} \\\\ &amp;=\\frac{12\\left(\\frac{1}{4}\\right)}{4\\left(\\frac{1}{4}\\right)+5\\left(\\frac{1}{4}\\right)\\cos \\theta } \\\\ r&amp;=\\frac{3}{1+\\frac{5}{4}\\cos \\theta } \\end{align}[\/latex]<\/div>\nBecause [latex]\\text{ cos}\\theta [\/latex] is in the denominator, the directrix is [latex]x=p[\/latex]. Comparing to standard form, [latex]e=\\frac{5}{4}[\/latex]. Therefore, from the numerator,\n<div style=\"text-align: center;\">[latex]\\begin{align}3&amp;=ep \\\\ 3&amp;=\\frac{5}{4}p \\\\ \\left(\\frac{4}{5}\\right)3&amp;=\\left(\\frac{4}{5}\\right)\\frac{5}{4}p \\\\ \\frac{12}{5}&amp;=p \\end{align}[\/latex]<\/div>\nSince [latex]e&gt;1[\/latex], the conic is a <strong>hyperbola<\/strong>. The eccentricity is [latex]e=\\frac{5}{4}[\/latex] and the directrix is [latex]x=\\frac{12}{5}=2.4[\/latex].<\/li>\n \t<li>Multiply the numerator and denominator by [latex]\\frac{1}{2}[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{align} r&amp;=\\frac{7}{2 - 2 \\sin \\theta }\\cdot \\frac{\\left(\\frac{1}{2}\\right)}{\\left(\\frac{1}{2}\\right)} \\\\ &amp;=\\frac{7\\left(\\frac{1}{2}\\right)}{2\\left(\\frac{1}{2}\\right)-2\\left(\\frac{1}{2}\\right) \\sin \\theta } \\\\ r&amp;=\\frac{\\frac{7}{2}}{1-\\sin \\theta } \\end{align}[\/latex]<\/div>\nBecause sine is in the denominator, the directrix is [latex]y=-p[\/latex]. Comparing to standard form, [latex]e=1[\/latex]. Therefore, from the numerator,\n<div style=\"text-align: center;\">[latex]\\begin{align}\\frac{7}{2}&amp;=ep\\\\ \\frac{7}{2}&amp;=\\left(1\\right)p\\\\ \\frac{7}{2}&amp;=p\\end{align}[\/latex]<\/div>\nBecause [latex]e=1[\/latex], the conic is a <strong>parabola<\/strong>. The eccentricity is [latex]e=1[\/latex] and the directrix is [latex]y=-\\frac{7}{2}=-3.5[\/latex].<\/li>\n<\/ol>\n[\/hidden-answer]\n\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\nIdentify the conic with focus at the origin, the directrix, and the eccentricity for [latex]r=\\frac{2}{3-\\cos \\text{ }\\theta }[\/latex].\n\n[reveal-answer q=\"600388\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"600388\"]\n\nellipse; [latex]e=\\frac{1}{3};x=-2[\/latex]\n\n[\/hidden-answer]\n\n<\/div>\n<h2>Graphing the Polar Equations of Conics<\/h2>\nWhen graphing in Cartesian coordinates, each conic section has a unique equation. This is not the case when graphing in polar coordinates. We must use the eccentricity of a conic section to determine which type of curve to graph, and then determine its specific characteristics. The first step is to rewrite the conic in standard form as we have done in the previous example. In other words, we need to rewrite the equation so that the denominator begins with 1. This enables us to determine [latex]e[\/latex] and, therefore, the shape of the curve. The next step is to substitute values for [latex]\\theta [\/latex] and solve for [latex]r[\/latex] to plot a few key points. Setting [latex]\\theta [\/latex] equal to [latex]0,\\frac{\\pi }{2},\\pi [\/latex], and [latex]\\frac{3\\pi }{2}[\/latex] provides the vertices so we can create a rough sketch of the graph.\n<div class=\"textbox shaded\">\n<h3>Example 2: Graphing a Parabola in Polar Form<\/h3>\nGraph [latex]r=\\frac{5}{3+3\\text{ }\\cos \\text{ }\\theta }[\/latex].\n\n[reveal-answer q=\"944366\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"944366\"]\n\nFirst, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 3, which is [latex]\\frac{1}{3}[\/latex].\n<p style=\"text-align: center;\">[latex]\\begin{align} r&amp;=\\frac{5}{3+3 \\cos \\theta } \\\\ &amp;=\\frac{5\\left(\\frac{1}{3}\\right)}{3\\left(\\frac{1}{3}\\right)+3\\left(\\frac{1}{3}\\right)\\cos \\theta } \\\\ r&amp;=\\frac{\\frac{5}{3}}{1+\\cos \\theta } \\end{align}[\/latex]<\/p>\nBecause [latex]e=1[\/latex], we will graph a <strong>parabola<\/strong> with a focus at the origin. The function has a [latex] \\cos \\text{ }\\theta [\/latex], and there is an addition sign in the denominator, so the directrix is [latex]x=p[\/latex].\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{5}{3}&amp;=ep\\\\ \\frac{5}{3}&amp;=\\left(1\\right)p\\\\ \\frac{5}{3}&amp;=p\\end{align}[\/latex]<\/p>\nThe directrix is [latex]x=\\frac{5}{3}[\/latex].\n\nPlotting a few key points as in the table below&nbsp;will enable us to see the vertices.\n<table id=\"Table_10_05_01\" summary=\"..\">\n<thead>\n<tr>\n<th><\/th>\n<th>A<\/th>\n<th>B<\/th>\n<th>C<\/th>\n<th>D<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]\\theta [\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td>[latex]\\pi [\/latex]<\/td>\n<td>[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]r=\\frac{5}{3+3\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/td>\n<td>[latex]\\frac{5}{6}\\approx 0.83[\/latex]<\/td>\n<td>[latex]\\frac{5}{3}\\approx 1.67[\/latex]<\/td>\n<td>undefined<\/td>\n<td>[latex]\\frac{5}{3}\\approx 1.67[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183125\/CNX_Precalc_Figure_10_05_0022.jpg\" alt=\"\" width=\"487\" height=\"376\"> <b>Figure 3<\/b>[\/caption]\n<h4>Analysis of the Solution<\/h4>\nWe can check our result with a graphing utility.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183127\/CNX_Precalc_Figure_10_05_0032.jpg\" alt=\"\" width=\"487\" height=\"376\"> <b>Figure 4<\/b>[\/caption]\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Graphing a Hyperbola in Polar Form<\/h3>\nGraph [latex]r=\\frac{8}{2 - 3 \\sin \\theta }[\/latex].\n\n[reveal-answer q=\"930942\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"930942\"]\n\nFirst, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 2, which is [latex]\\frac{1}{2}[\/latex].\n<p style=\"text-align: center;\">[latex]\\begin{align} r&amp;=\\frac{8}{2 - 3\\sin \\theta } \\\\ &amp;=\\frac{8\\left(\\frac{1}{2}\\right)}{2\\left(\\frac{1}{2}\\right)-3\\left(\\frac{1}{2}\\right)\\sin \\theta } \\\\ r&amp;=\\frac{4}{1-\\frac{3}{2} \\sin \\theta } \\end{align}[\/latex]<\/p>\nBecause [latex]e=\\frac{3}{2},e&gt;1[\/latex], so we will graph a <strong>hyperbola<\/strong> with a focus at the origin. The function has a [latex]\\sin \\text{ }\\theta [\/latex] term and there is a subtraction sign in the denominator, so the directrix is [latex]y=-p[\/latex].\n<p style=\"text-align: center;\">[latex]\\begin{align}4&amp;=ep \\\\ 4&amp;=\\left(\\frac{3}{2}\\right)p \\\\ 4\\left(\\frac{2}{3}\\right)&amp;=p \\\\ \\frac{8}{3}&amp;=p \\end{align}[\/latex]<\/p>\nThe directrix is [latex]y=-\\frac{8}{3}[\/latex].\n\nPlotting a few key points as in the table below&nbsp;will enable us to see the vertices.\n<table id=\"Table_10_05_02\" summary=\"..\">\n<thead>\n<tr>\n<th><\/th>\n<th>A<\/th>\n<th>B<\/th>\n<th>C<\/th>\n<th>D<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]\\theta [\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td>[latex]\\pi [\/latex]<\/td>\n<td>[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]r=\\frac{8}{2 - 3\\sin \\theta }[\/latex]<\/td>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]-8[\/latex]<\/td>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]\\frac{8}{5}=1.6[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183130\/CNX_Precalc_Figure_10_05_0042.jpg\" alt=\"\" width=\"975\" height=\"810\"> <b>Figure 5<\/b>[\/caption]\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 4: Graphing an Ellipse in Polar Form<\/h3>\nGraph [latex]r=\\frac{10}{5 - 4 \\cos \\theta }[\/latex].\n\n[reveal-answer q=\"252925\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"252925\"]\n\nFirst, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 5, which is [latex]\\frac{1}{5}[\/latex].\n<p style=\"text-align: center;\">[latex]\\begin{align}r&amp;=\\frac{10}{5 - 4\\cos \\theta } \\\\ &amp;=\\frac{10\\left(\\frac{1}{5}\\right)}{5\\left(\\frac{1}{5}\\right)-4\\left(\\frac{1}{5}\\right)\\cos \\theta } \\\\ r&amp;=\\frac{2}{1-\\frac{4}{5}\\cos \\theta } \\end{align}[\/latex]<\/p>\nBecause [latex]e=\\frac{4}{5},e&lt;1[\/latex], so we will graph an <strong>ellipse<\/strong> with a <strong>focus<\/strong> at the origin. The function has a [latex]\\text{cos}\\theta [\/latex], and there is a subtraction sign in the denominator, so the <strong>directrix<\/strong> is [latex]x=-p[\/latex].\n<p style=\"text-align: center;\">[latex]\\begin{align}2&amp;=ep \\\\ 2&amp;=\\left(\\frac{4}{5}\\right)p \\\\ 2\\left(\\frac{5}{4}\\right)&amp;=p \\\\ \\frac{5}{2}&amp;=p \\end{align}[\/latex]<\/p>\nThe directrix is [latex]x=-\\frac{5}{2}[\/latex].\n\nPlotting a few key points as in the table below&nbsp;will enable us to see the vertices.\n<table id=\"Table_10_05_03\" summary=\"..\">\n<thead>\n<tr>\n<th><\/th>\n<th>A<\/th>\n<th>B<\/th>\n<th>C<\/th>\n<th>D<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]\\theta [\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td>[latex]\\pi [\/latex]<\/td>\n<td>[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]r=\\frac{10}{5 - 4\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/td>\n<td>[latex]10[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]\\frac{10}{9}\\approx 1.1[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183132\/CNX_Precalc_Figure_10_05_0062.jpg\" alt=\"\" width=\"487\" height=\"431\"> <b>Figure 6<\/b>[\/caption]\n<h4>Analysis of the Solution<\/h4>\nWe can check our result using a graphing utility.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183134\/CNX_Precalc_Figure_10_05_0072.jpg\" alt=\"\" width=\"487\" height=\"431\"> <b>Figure 7.<\/b> [latex]r=\\frac{10}{5 - 4 \\cos \\theta }[\/latex] graphed on a viewing window of [latex]\\left[-3,12,1\\right][\/latex] by [latex]\\left[-4,4,1\\right],\\theta \\text{min =}0[\/latex] and [latex]\\theta \\text{max =}2\\pi [\/latex].[\/caption][\/hidden-answer]<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\nGraph [latex]r=\\frac{2}{4-\\cos \\theta }[\/latex].\n\n[reveal-answer q=\"275168\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"275168\"]\n\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183136\/CNX_Precalc_Figure_10_05_0092.jpg\" alt=\"\">\n\n[\/hidden-answer]\n\n<\/div>\n<h2>Defining Conics in Terms of a Focus and a Directrix<\/h2>\nSo far we have been using polar equations of conics to describe and graph the curve. Now we will work in reverse; we will use information about the origin, eccentricity, and directrix to determine the polar equation.\n<div class=\"textbox\">\n<h3>How To: Given the focus, eccentricity, and directrix of a conic, determine the polar equation.<strong>\n<\/strong><\/h3>\n<ol>\n \t<li>Determine whether the directrix is horizontal or vertical. If the directrix is given in terms of [latex]y[\/latex], we use the general polar form in terms of sine. If the directrix is given in terms of [latex]x[\/latex], we use the general polar form in terms of cosine.<\/li>\n \t<li>Determine the sign in the denominator. If [latex]p&lt;0[\/latex], use subtraction. If [latex]p&gt;0[\/latex], use addition.<\/li>\n \t<li>Write the coefficient of the trigonometric function as the given eccentricity.<\/li>\n \t<li>Write the absolute value of [latex]p[\/latex] in the numerator, and simplify the equation.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Finding the Polar Form of a Vertical Conic Given a Focus at the Origin and the Eccentricity and Directrix<\/h3>\nFind the polar form of the <strong>conic<\/strong> given a <strong>focus<\/strong> at the origin, [latex]e=3[\/latex] and <strong>directrix<\/strong> [latex]y=-2[\/latex].\n\n[reveal-answer q=\"155638\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"155638\"]\n\nThe directrix is [latex]y=-p[\/latex], so we know the trigonometric function in the denominator is sine.\n\nBecause [latex]y=-2,-2&lt;0[\/latex], so we know there is a subtraction sign in the denominator. We use the standard form of\n<p style=\"text-align: center;\">[latex]r=\\frac{ep}{1-e \\sin \\theta }[\/latex]<\/p>\nand [latex]e=3[\/latex] and [latex]|-2|=2=p[\/latex].\n\nTherefore,\n<p style=\"text-align: center;\">[latex]\\begin{align}r&amp;=\\frac{\\left(3\\right)\\left(2\\right)}{1 - 3 \\sin \\theta } \\\\ r&amp;=\\frac{6}{1 - 3 \\sin \\theta } \\end{align}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 6: Finding the Polar Form of a Horizontal Conic Given a Focus at the Origin and the Eccentricity and Directrix<\/h3>\nFind the <strong>polar form of a conic<\/strong> given a <strong>focus<\/strong> at the origin, [latex]e=\\frac{3}{5}[\/latex], and <strong>directrix<\/strong> [latex]x=4[\/latex].\n\n[reveal-answer q=\"323889\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"323889\"]\n\nBecause the directrix is [latex]x=p[\/latex], we know the function in the denominator is cosine. Because [latex]x=4,4&gt;0[\/latex], so we know there is an addition sign in the denominator. We use the standard form of\n<p style=\"text-align: center;\">[latex]r=\\frac{ep}{1+e \\cos \\theta }[\/latex]<\/p>\nand [latex]e=\\frac{3}{5}[\/latex] and [latex]|4|=4=p[\/latex].\n\nTherefore,\n<p style=\"text-align: center;\">[latex]\\begin{align} r&amp;=\\frac{\\left(\\frac{3}{5}\\right)\\left(4\\right)}{1+\\frac{3}{5}\\cos \\theta } \\\\ &amp;=\\frac{\\frac{12}{5}}{1+\\frac{3}{5}\\cos \\theta } \\\\ &amp;=\\frac{\\frac{12}{5}}{1\\left(\\frac{5}{5}\\right)+\\frac{3}{5}\\cos \\theta } \\\\ &amp;=\\frac{\\frac{12}{5}}{\\frac{5}{5}+\\frac{3}{5}\\cos \\theta } \\\\ &amp;=\\frac{12}{5}\\cdot \\frac{5}{5+3\\cos \\theta } \\\\ r&amp;=\\frac{12}{5+3\\cos \\theta } \\end{align}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\nFind the polar form of the conic given a focus at the origin, [latex]e=1[\/latex], and directrix [latex]x=-1[\/latex].\n\n[reveal-answer q=\"733852\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"733852\"]\n\n[latex]r=\\frac{1}{1-\\cos \\theta }[\/latex]\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 7: Converting a Conic in Polar Form to Rectangular Form<\/h3>\nConvert the conic [latex]r=\\frac{1}{5 - 5\\sin \\theta }[\/latex] to rectangular form.\n\n[reveal-answer q=\"462269\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"462269\"]\n\nWe will rearrange the formula to use the identities [latex] r=\\sqrt{{x}^{2}+{y}^{2}},x=r\\cos \\theta ,\\text{and }y=r\\sin \\theta [\/latex].\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;r=\\frac{1}{5 - 5\\sin \\theta } \\\\ &amp;r\\cdot \\left(5 - 5\\sin \\theta \\right)=\\frac{1}{5 - 5\\sin \\theta }\\cdot \\left(5 - 5\\sin \\theta \\right)&amp;&amp; \\text{Eliminate the fraction}. \\\\ &amp;5r - 5r\\sin \\theta =1&amp;&amp; \\text{Distribute}. \\\\ &amp;5r=1+5r\\sin \\theta &amp;&amp; \\text{Isolate }5r. \\\\ &amp;25{r}^{2}={\\left(1+5r\\sin \\theta \\right)}^{2}&amp;&amp; \\text{Square both sides}. \\\\ &amp;25\\left({x}^{2}+{y}^{2}\\right)={\\left(1+5y\\right)}^{2}&amp;&amp; \\text{Substitute }r=\\sqrt{{x}^{2}+{y}^{2}}\\text{ and }y=r\\sin \\theta . \\\\ &amp;25{x}^{2}+25{y}^{2}=1+10y+25{y}^{2}&amp;&amp; \\text{Distribute and use FOIL}. \\\\ &amp;25{x}^{2}-10y=1&amp;&amp; \\text{Rearrange terms and set equal to 1}. \\end{align}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\nConvert the conic [latex]r=\\frac{2}{1+2\\text{ }\\cos \\text{ }\\theta }[\/latex] to rectangular form.\n\n[reveal-answer q=\"554425\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"554425\"]\n\n[latex]4 - 8x+3{x}^{2}-{y}^{2}=0[\/latex]\n\n[\/hidden-answer]\n\n<\/div>\n<h2>Key Concepts<\/h2>\n<ul>\n \t<li>Any conic may be determined by a single focus, the corresponding eccentricity, and the directrix. We can also define a conic in terms of a fixed point, the focus [latex]P\\left(r,\\theta \\right)[\/latex] at the pole, and a line, the directrix, which is perpendicular to the polar axis.<\/li>\n \t<li>A conic is the set of all points [latex]e=\\frac{PF}{PD}[\/latex], where eccentricity [latex]e[\/latex] is a positive real number. Each conic may be written in terms of its polar equation.<\/li>\n \t<li>The polar equations of conics can be graphed.<\/li>\n \t<li>Conics can be defined in terms of a focus, a directrix, and eccentricity.<\/li>\n \t<li>We can use the identities [latex]r=\\sqrt{{x}^{2}+{y}^{2}},x=r\\text{ }\\cos \\text{ }\\theta [\/latex], and [latex]y=r\\text{ }\\sin \\text{ }\\theta [\/latex] to convert the equation for a conic from polar to rectangular form.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id2172786\" class=\"definition\">\n \t<dt>eccentricity<\/dt>\n \t<dd id=\"fs-id2172791\">the ratio of the distances from a point [latex]P[\/latex] on the graph to the focus [latex]F[\/latex] and to the directrix [latex]D[\/latex] represented by [latex]e=\\frac{PF}{PD}[\/latex], where [latex]e[\/latex] is a positive real number<\/dd>\n<\/dl>\n<dl id=\"fs-id2172885\" class=\"definition\">\n \t<dt>polar equation<\/dt>\n \t<dd id=\"fs-id1271646\">an equation of a curve in polar coordinates [latex]r[\/latex] and [latex]\\theta [\/latex]<\/dd>\n<\/dl>\n","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Identify a conic in polar form.<\/li>\n<li>Graph the polar equations of conics.<\/li>\n<li>De\ufb01ne conics in terms of a focus and a directrix.<\/li>\n<\/ul>\n<\/div>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183120\/CNX_Precalc_Figure_10_05_008n2.jpg\" alt=\"The planets and their orbits around the sun. (Pluto is included.)\" width=\"975\" height=\"353\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1.<\/b> Planets orbiting the sun follow elliptical paths. (credit: NASA Blueshift, Flickr)<\/p>\n<\/div>\n<p>Most of us are familiar with orbital motion, such as the motion of a planet around the sun or an electron around an atomic nucleus. Within the planetary system, orbits of planets, asteroids, and comets around a larger celestial body are often elliptical. Comets, however, may take on a parabolic or hyperbolic orbit instead. And, in reality, the characteristics of the planets\u2019 orbits may vary over time. Each orbit is tied to the location of the celestial body being orbited and the distance and direction of the planet or other object from that body. As a result, we tend to use polar coordinates to represent these orbits.<\/p>\n<p>In an elliptical orbit, the <strong>periapsis<\/strong> is the point at which the two objects are closest, and the <strong>apoapsis<\/strong> is the point at which they are farthest apart. Generally, the velocity of the orbiting body tends to increase as it approaches the periapsis and decrease as it approaches the apoapsis. Some objects reach an escape velocity, which results in an infinite orbit. These bodies exhibit either a parabolic or a hyperbolic orbit about a body; the orbiting body breaks free of the celestial body\u2019s gravitational pull and fires off into space. Each of these orbits can be modeled by a conic section in the polar coordinate system.<\/p>\n<h2>Identifying a Conic in Polar Form<\/h2>\n<p>Any conic may be determined by three characteristics: a single <strong>focus<\/strong>, a fixed line called the <strong>directrix<\/strong>, and the ratio of the distances of each to a point on the graph. Consider the <strong>parabola<\/strong> [latex]x=2+{y}^{2}[\/latex] shown in Figure 2.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183123\/CNX_Precalc_Figure_10_05_0012.jpg\" alt=\"\" width=\"487\" height=\"316\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<p>In <a href=\"https:\/\/courses.lumenlearning.com\/precalctwoxmaster\/chapter\/introduction-to-the-parabola\/\" target=\"_blank\" rel=\"noopener\">The Parabola<\/a>, we learned how a parabola is defined by the focus (a fixed point) and the directrix (a fixed line). In this section, we will learn how to define any conic in the polar coordinate system in terms of a fixed point, the focus [latex]P\\left(r,\\theta \\right)[\/latex] at the pole, and a line, the directrix, which is perpendicular to the polar axis.<\/p>\n<p>If [latex]F[\/latex] is a fixed point, the focus, and [latex]D[\/latex] is a fixed line, the directrix, then we can let [latex]e[\/latex] be a fixed positive number, called the <strong>eccentricity<\/strong>, which we can define as the ratio of the distances from a point on the graph to the focus and the point on the graph to the directrix. Then the set of all points [latex]P[\/latex] such that [latex]e=\\frac{PF}{PD}[\/latex] is a conic. In other words, we can define a conic as the set of all points [latex]P[\/latex] with the property that the ratio of the distance from [latex]P[\/latex] to [latex]F[\/latex] to the distance from [latex]P[\/latex] to [latex]D[\/latex] is equal to the constant [latex]e[\/latex].<\/p>\n<p>For a conic with eccentricity [latex]e[\/latex],<\/p>\n<ul>\n<li>if [latex]0\\le e<1[\/latex], the conic is an ellipse<\/li>\n<li>if [latex]e=1[\/latex], the conic is a parabola<\/li>\n<li>if [latex]e>1[\/latex], the conic is an hyperbola<\/li>\n<\/ul>\n<p>With this definition, we may now define a conic in terms of the directrix, [latex]x=\\pm p[\/latex], the eccentricity [latex]e[\/latex], and the angle [latex]\\theta[\/latex]. Thus, each conic may be written as a <strong>polar equation<\/strong>, an equation written in terms of [latex]r[\/latex] and [latex]\\theta[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Polar Equation for a Conic<\/h3>\n<p>For a conic with a focus at the origin, if the directrix is [latex]x=\\pm p[\/latex], where [latex]p[\/latex] is a positive real number, and the <strong>eccentricity<\/strong> is a positive real number [latex]e[\/latex], the conic has a <strong>polar equation<\/strong><\/p>\n<p>[latex]r=\\frac{ep}{1\\pm e \\cos \\theta }[\/latex]<\/p>\n<p>For a conic with a focus at the origin, if the directrix is [latex]y=\\pm p[\/latex], where [latex]p[\/latex] is a positive real number, and the eccentricity is a positive real number [latex]e[\/latex], the conic has a polar equation<\/p>\n<p>[latex]r=\\frac{ep}{1\\pm e \\sin \\theta }[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given the polar equation for a conic, identify the type of conic, the directrix, and the eccentricity.<\/h3>\n<ol>\n<li>Multiply the numerator and denominator by the reciprocal of the constant in the denominator to rewrite the equation in standard form.<\/li>\n<li>Identify the eccentricity [latex]e[\/latex] as the coefficient of the trigonometric function in the denominator.<\/li>\n<li>Compare [latex]e[\/latex] with 1 to determine the shape of the conic.<\/li>\n<li>Determine the directrix as [latex]x=p[\/latex] if cosine is in the denominator and [latex]y=p[\/latex] if sine is in the denominator. Set [latex]ep[\/latex] equal to the numerator in standard form to solve for [latex]x[\/latex] or [latex]y[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Identifying a Conic Given the Polar Form<\/h3>\n<p>For each of the following equations, identify the conic with focus at the origin, the <strong>directrix<\/strong>, and the <strong>eccentricity<\/strong>.<\/p>\n<ol>\n<li>[latex]r=\\frac{6}{3+2 \\sin \\theta }[\/latex]<\/li>\n<li>[latex]r=\\frac{12}{4+5 \\cos \\theta }[\/latex]<\/li>\n<li>[latex]r=\\frac{7}{2 - 2 \\sin \\theta }[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q293935\">Show Solution<\/span><\/p>\n<div id=\"q293935\" class=\"hidden-answer\" style=\"display: none\">\n<p>For each of the three conics, we will rewrite the equation in standard form. Standard form has a 1 as the constant in the denominator. Therefore, in all three parts, the first step will be to multiply the numerator and denominator by the reciprocal of the constant of the original equation, [latex]\\frac{1}{c}[\/latex], where [latex]c[\/latex] is that constant.<\/p>\n<ol>\n<li>Multiply the numerator and denominator by [latex]\\frac{1}{3}[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{align}r&=\\frac{6}{3+2\\sin \\theta }\\cdot \\frac{\\left(\\frac{1}{3}\\right)}{\\left(\\frac{1}{3}\\right)} \\\\ &=\\frac{6\\left(\\frac{1}{3}\\right)}{3\\left(\\frac{1}{3}\\right)+2\\left(\\frac{1}{3}\\right)\\sin \\theta } \\\\ r&=\\frac{2}{1+\\frac{2}{3}\\sin \\theta} \\end{align}[\/latex]<\/div>\n<p>Because [latex]\\sin \\text{ }\\theta[\/latex] is in the denominator, the directrix is [latex]y=p[\/latex]. Comparing to standard form, note that [latex]e=\\frac{2}{3}[\/latex]. Therefore, from the numerator,<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}2&=ep \\\\ 2&=\\frac{2}{3}p \\\\ \\left(\\frac{3}{2}\\right)2&=\\left(\\frac{3}{2}\\right)\\frac{2}{3}p \\\\ 3&=p \\end{align}[\/latex]<\/div>\n<p>Since [latex]e<1[\/latex], the conic is an <strong>ellipse<\/strong>. The eccentricity is [latex]e=\\frac{2}{3}[\/latex] and the directrix is [latex]y=3[\/latex].<\/li>\n<li>Multiply the numerator and denominator by [latex]\\frac{1}{4}[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{align} r&=\\frac{12}{4+5 \\cos \\theta }\\cdot \\frac{\\left(\\frac{1}{4}\\right)}{\\left(\\frac{1}{4}\\right)} \\\\ &=\\frac{12\\left(\\frac{1}{4}\\right)}{4\\left(\\frac{1}{4}\\right)+5\\left(\\frac{1}{4}\\right)\\cos \\theta } \\\\ r&=\\frac{3}{1+\\frac{5}{4}\\cos \\theta } \\end{align}[\/latex]<\/div>\n<p>Because [latex]\\text{ cos}\\theta[\/latex] is in the denominator, the directrix is [latex]x=p[\/latex]. Comparing to standard form, [latex]e=\\frac{5}{4}[\/latex]. Therefore, from the numerator,<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}3&=ep \\\\ 3&=\\frac{5}{4}p \\\\ \\left(\\frac{4}{5}\\right)3&=\\left(\\frac{4}{5}\\right)\\frac{5}{4}p \\\\ \\frac{12}{5}&=p \\end{align}[\/latex]<\/div>\n<p>Since [latex]e>1[\/latex], the conic is a <strong>hyperbola<\/strong>. The eccentricity is [latex]e=\\frac{5}{4}[\/latex] and the directrix is [latex]x=\\frac{12}{5}=2.4[\/latex].<\/li>\n<li>Multiply the numerator and denominator by [latex]\\frac{1}{2}[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{align} r&=\\frac{7}{2 - 2 \\sin \\theta }\\cdot \\frac{\\left(\\frac{1}{2}\\right)}{\\left(\\frac{1}{2}\\right)} \\\\ &=\\frac{7\\left(\\frac{1}{2}\\right)}{2\\left(\\frac{1}{2}\\right)-2\\left(\\frac{1}{2}\\right) \\sin \\theta } \\\\ r&=\\frac{\\frac{7}{2}}{1-\\sin \\theta } \\end{align}[\/latex]<\/div>\n<p>Because sine is in the denominator, the directrix is [latex]y=-p[\/latex]. Comparing to standard form, [latex]e=1[\/latex]. Therefore, from the numerator,<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\frac{7}{2}&=ep\\\\ \\frac{7}{2}&=\\left(1\\right)p\\\\ \\frac{7}{2}&=p\\end{align}[\/latex]<\/div>\n<p>Because [latex]e=1[\/latex], the conic is a <strong>parabola<\/strong>. The eccentricity is [latex]e=1[\/latex] and the directrix is [latex]y=-\\frac{7}{2}=-3.5[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Identify the conic with focus at the origin, the directrix, and the eccentricity for [latex]r=\\frac{2}{3-\\cos \\text{ }\\theta }[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q600388\">Show Solution<\/span><\/p>\n<div id=\"q600388\" class=\"hidden-answer\" style=\"display: none\">\n<p>ellipse; [latex]e=\\frac{1}{3};x=-2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Graphing the Polar Equations of Conics<\/h2>\n<p>When graphing in Cartesian coordinates, each conic section has a unique equation. This is not the case when graphing in polar coordinates. We must use the eccentricity of a conic section to determine which type of curve to graph, and then determine its specific characteristics. The first step is to rewrite the conic in standard form as we have done in the previous example. In other words, we need to rewrite the equation so that the denominator begins with 1. This enables us to determine [latex]e[\/latex] and, therefore, the shape of the curve. The next step is to substitute values for [latex]\\theta[\/latex] and solve for [latex]r[\/latex] to plot a few key points. Setting [latex]\\theta[\/latex] equal to [latex]0,\\frac{\\pi }{2},\\pi[\/latex], and [latex]\\frac{3\\pi }{2}[\/latex] provides the vertices so we can create a rough sketch of the graph.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 2: Graphing a Parabola in Polar Form<\/h3>\n<p>Graph [latex]r=\\frac{5}{3+3\\text{ }\\cos \\text{ }\\theta }[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q944366\">Show Solution<\/span><\/p>\n<div id=\"q944366\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 3, which is [latex]\\frac{1}{3}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} r&=\\frac{5}{3+3 \\cos \\theta } \\\\ &=\\frac{5\\left(\\frac{1}{3}\\right)}{3\\left(\\frac{1}{3}\\right)+3\\left(\\frac{1}{3}\\right)\\cos \\theta } \\\\ r&=\\frac{\\frac{5}{3}}{1+\\cos \\theta } \\end{align}[\/latex]<\/p>\n<p>Because [latex]e=1[\/latex], we will graph a <strong>parabola<\/strong> with a focus at the origin. The function has a [latex]\\cos \\text{ }\\theta[\/latex], and there is an addition sign in the denominator, so the directrix is [latex]x=p[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{5}{3}&=ep\\\\ \\frac{5}{3}&=\\left(1\\right)p\\\\ \\frac{5}{3}&=p\\end{align}[\/latex]<\/p>\n<p>The directrix is [latex]x=\\frac{5}{3}[\/latex].<\/p>\n<p>Plotting a few key points as in the table below&nbsp;will enable us to see the vertices.<\/p>\n<table id=\"Table_10_05_01\" summary=\"..\">\n<thead>\n<tr>\n<th><\/th>\n<th>A<\/th>\n<th>B<\/th>\n<th>C<\/th>\n<th>D<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]\\theta[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td>[latex]\\pi[\/latex]<\/td>\n<td>[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]r=\\frac{5}{3+3\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/td>\n<td>[latex]\\frac{5}{6}\\approx 0.83[\/latex]<\/td>\n<td>[latex]\\frac{5}{3}\\approx 1.67[\/latex]<\/td>\n<td>undefined<\/td>\n<td>[latex]\\frac{5}{3}\\approx 1.67[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183125\/CNX_Precalc_Figure_10_05_0022.jpg\" alt=\"\" width=\"487\" height=\"376\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p>We can check our result with a graphing utility.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183127\/CNX_Precalc_Figure_10_05_0032.jpg\" alt=\"\" width=\"487\" height=\"376\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Graphing a Hyperbola in Polar Form<\/h3>\n<p>Graph [latex]r=\\frac{8}{2 - 3 \\sin \\theta }[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q930942\">Show Solution<\/span><\/p>\n<div id=\"q930942\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 2, which is [latex]\\frac{1}{2}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} r&=\\frac{8}{2 - 3\\sin \\theta } \\\\ &=\\frac{8\\left(\\frac{1}{2}\\right)}{2\\left(\\frac{1}{2}\\right)-3\\left(\\frac{1}{2}\\right)\\sin \\theta } \\\\ r&=\\frac{4}{1-\\frac{3}{2} \\sin \\theta } \\end{align}[\/latex]<\/p>\n<p>Because [latex]e=\\frac{3}{2},e>1[\/latex], so we will graph a <strong>hyperbola<\/strong> with a focus at the origin. The function has a [latex]\\sin \\text{ }\\theta[\/latex] term and there is a subtraction sign in the denominator, so the directrix is [latex]y=-p[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}4&=ep \\\\ 4&=\\left(\\frac{3}{2}\\right)p \\\\ 4\\left(\\frac{2}{3}\\right)&=p \\\\ \\frac{8}{3}&=p \\end{align}[\/latex]<\/p>\n<p>The directrix is [latex]y=-\\frac{8}{3}[\/latex].<\/p>\n<p>Plotting a few key points as in the table below&nbsp;will enable us to see the vertices.<\/p>\n<table id=\"Table_10_05_02\" summary=\"..\">\n<thead>\n<tr>\n<th><\/th>\n<th>A<\/th>\n<th>B<\/th>\n<th>C<\/th>\n<th>D<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]\\theta[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td>[latex]\\pi[\/latex]<\/td>\n<td>[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]r=\\frac{8}{2 - 3\\sin \\theta }[\/latex]<\/td>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]-8[\/latex]<\/td>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]\\frac{8}{5}=1.6[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183130\/CNX_Precalc_Figure_10_05_0042.jpg\" alt=\"\" width=\"975\" height=\"810\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 4: Graphing an Ellipse in Polar Form<\/h3>\n<p>Graph [latex]r=\\frac{10}{5 - 4 \\cos \\theta }[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q252925\">Show Solution<\/span><\/p>\n<div id=\"q252925\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 5, which is [latex]\\frac{1}{5}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}r&=\\frac{10}{5 - 4\\cos \\theta } \\\\ &=\\frac{10\\left(\\frac{1}{5}\\right)}{5\\left(\\frac{1}{5}\\right)-4\\left(\\frac{1}{5}\\right)\\cos \\theta } \\\\ r&=\\frac{2}{1-\\frac{4}{5}\\cos \\theta } \\end{align}[\/latex]<\/p>\n<p>Because [latex]e=\\frac{4}{5},e<1[\/latex], so we will graph an <strong>ellipse<\/strong> with a <strong>focus<\/strong> at the origin. The function has a [latex]\\text{cos}\\theta[\/latex], and there is a subtraction sign in the denominator, so the <strong>directrix<\/strong> is [latex]x=-p[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2&=ep \\\\ 2&=\\left(\\frac{4}{5}\\right)p \\\\ 2\\left(\\frac{5}{4}\\right)&=p \\\\ \\frac{5}{2}&=p \\end{align}[\/latex]<\/p>\n<p>The directrix is [latex]x=-\\frac{5}{2}[\/latex].<\/p>\n<p>Plotting a few key points as in the table below&nbsp;will enable us to see the vertices.<\/p>\n<table id=\"Table_10_05_03\" summary=\"..\">\n<thead>\n<tr>\n<th><\/th>\n<th>A<\/th>\n<th>B<\/th>\n<th>C<\/th>\n<th>D<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]\\theta[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n<td>[latex]\\pi[\/latex]<\/td>\n<td>[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]r=\\frac{10}{5 - 4\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/td>\n<td>[latex]10[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]\\frac{10}{9}\\approx 1.1[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183132\/CNX_Precalc_Figure_10_05_0062.jpg\" alt=\"\" width=\"487\" height=\"431\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p>We can check our result using a graphing utility.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183134\/CNX_Precalc_Figure_10_05_0072.jpg\" alt=\"\" width=\"487\" height=\"431\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 7.<\/b> [latex]r=\\frac{10}{5 - 4 \\cos \\theta }[\/latex] graphed on a viewing window of [latex]\\left[-3,12,1\\right][\/latex] by [latex]\\left[-4,4,1\\right],\\theta \\text{min =}0[\/latex] and [latex]\\theta \\text{max =}2\\pi [\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Graph [latex]r=\\frac{2}{4-\\cos \\theta }[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q275168\">Show Solution<\/span><\/p>\n<div id=\"q275168\" class=\"hidden-answer\" style=\"display: none\">\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183136\/CNX_Precalc_Figure_10_05_0092.jpg\" alt=\"\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Defining Conics in Terms of a Focus and a Directrix<\/h2>\n<p>So far we have been using polar equations of conics to describe and graph the curve. Now we will work in reverse; we will use information about the origin, eccentricity, and directrix to determine the polar equation.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given the focus, eccentricity, and directrix of a conic, determine the polar equation.<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Determine whether the directrix is horizontal or vertical. If the directrix is given in terms of [latex]y[\/latex], we use the general polar form in terms of sine. If the directrix is given in terms of [latex]x[\/latex], we use the general polar form in terms of cosine.<\/li>\n<li>Determine the sign in the denominator. If [latex]p<0[\/latex], use subtraction. If [latex]p>0[\/latex], use addition.<\/li>\n<li>Write the coefficient of the trigonometric function as the given eccentricity.<\/li>\n<li>Write the absolute value of [latex]p[\/latex] in the numerator, and simplify the equation.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Finding the Polar Form of a Vertical Conic Given a Focus at the Origin and the Eccentricity and Directrix<\/h3>\n<p>Find the polar form of the <strong>conic<\/strong> given a <strong>focus<\/strong> at the origin, [latex]e=3[\/latex] and <strong>directrix<\/strong> [latex]y=-2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q155638\">Show Solution<\/span><\/p>\n<div id=\"q155638\" class=\"hidden-answer\" style=\"display: none\">\n<p>The directrix is [latex]y=-p[\/latex], so we know the trigonometric function in the denominator is sine.<\/p>\n<p>Because [latex]y=-2,-2<0[\/latex], so we know there is a subtraction sign in the denominator. We use the standard form of\n\n\n<p style=\"text-align: center;\">[latex]r=\\frac{ep}{1-e \\sin \\theta }[\/latex]<\/p>\n<p>and [latex]e=3[\/latex] and [latex]|-2|=2=p[\/latex].<\/p>\n<p>Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}r&=\\frac{\\left(3\\right)\\left(2\\right)}{1 - 3 \\sin \\theta } \\\\ r&=\\frac{6}{1 - 3 \\sin \\theta } \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 6: Finding the Polar Form of a Horizontal Conic Given a Focus at the Origin and the Eccentricity and Directrix<\/h3>\n<p>Find the <strong>polar form of a conic<\/strong> given a <strong>focus<\/strong> at the origin, [latex]e=\\frac{3}{5}[\/latex], and <strong>directrix<\/strong> [latex]x=4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q323889\">Show Solution<\/span><\/p>\n<div id=\"q323889\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because the directrix is [latex]x=p[\/latex], we know the function in the denominator is cosine. Because [latex]x=4,4>0[\/latex], so we know there is an addition sign in the denominator. We use the standard form of<\/p>\n<p style=\"text-align: center;\">[latex]r=\\frac{ep}{1+e \\cos \\theta }[\/latex]<\/p>\n<p>and [latex]e=\\frac{3}{5}[\/latex] and [latex]|4|=4=p[\/latex].<\/p>\n<p>Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} r&=\\frac{\\left(\\frac{3}{5}\\right)\\left(4\\right)}{1+\\frac{3}{5}\\cos \\theta } \\\\ &=\\frac{\\frac{12}{5}}{1+\\frac{3}{5}\\cos \\theta } \\\\ &=\\frac{\\frac{12}{5}}{1\\left(\\frac{5}{5}\\right)+\\frac{3}{5}\\cos \\theta } \\\\ &=\\frac{\\frac{12}{5}}{\\frac{5}{5}+\\frac{3}{5}\\cos \\theta } \\\\ &=\\frac{12}{5}\\cdot \\frac{5}{5+3\\cos \\theta } \\\\ r&=\\frac{12}{5+3\\cos \\theta } \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find the polar form of the conic given a focus at the origin, [latex]e=1[\/latex], and directrix [latex]x=-1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q733852\">Show Solution<\/span><\/p>\n<div id=\"q733852\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]r=\\frac{1}{1-\\cos \\theta }[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 7: Converting a Conic in Polar Form to Rectangular Form<\/h3>\n<p>Convert the conic [latex]r=\\frac{1}{5 - 5\\sin \\theta }[\/latex] to rectangular form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q462269\">Show Solution<\/span><\/p>\n<div id=\"q462269\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will rearrange the formula to use the identities [latex]r=\\sqrt{{x}^{2}+{y}^{2}},x=r\\cos \\theta ,\\text{and }y=r\\sin \\theta[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&r=\\frac{1}{5 - 5\\sin \\theta } \\\\ &r\\cdot \\left(5 - 5\\sin \\theta \\right)=\\frac{1}{5 - 5\\sin \\theta }\\cdot \\left(5 - 5\\sin \\theta \\right)&& \\text{Eliminate the fraction}. \\\\ &5r - 5r\\sin \\theta =1&& \\text{Distribute}. \\\\ &5r=1+5r\\sin \\theta && \\text{Isolate }5r. \\\\ &25{r}^{2}={\\left(1+5r\\sin \\theta \\right)}^{2}&& \\text{Square both sides}. \\\\ &25\\left({x}^{2}+{y}^{2}\\right)={\\left(1+5y\\right)}^{2}&& \\text{Substitute }r=\\sqrt{{x}^{2}+{y}^{2}}\\text{ and }y=r\\sin \\theta . \\\\ &25{x}^{2}+25{y}^{2}=1+10y+25{y}^{2}&& \\text{Distribute and use FOIL}. \\\\ &25{x}^{2}-10y=1&& \\text{Rearrange terms and set equal to 1}. \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Convert the conic [latex]r=\\frac{2}{1+2\\text{ }\\cos \\text{ }\\theta }[\/latex] to rectangular form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q554425\">Show Solution<\/span><\/p>\n<div id=\"q554425\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]4 - 8x+3{x}^{2}-{y}^{2}=0[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>Any conic may be determined by a single focus, the corresponding eccentricity, and the directrix. We can also define a conic in terms of a fixed point, the focus [latex]P\\left(r,\\theta \\right)[\/latex] at the pole, and a line, the directrix, which is perpendicular to the polar axis.<\/li>\n<li>A conic is the set of all points [latex]e=\\frac{PF}{PD}[\/latex], where eccentricity [latex]e[\/latex] is a positive real number. Each conic may be written in terms of its polar equation.<\/li>\n<li>The polar equations of conics can be graphed.<\/li>\n<li>Conics can be defined in terms of a focus, a directrix, and eccentricity.<\/li>\n<li>We can use the identities [latex]r=\\sqrt{{x}^{2}+{y}^{2}},x=r\\text{ }\\cos \\text{ }\\theta[\/latex], and [latex]y=r\\text{ }\\sin \\text{ }\\theta[\/latex] to convert the equation for a conic from polar to rectangular form.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id2172786\" class=\"definition\">\n<dt>eccentricity<\/dt>\n<dd id=\"fs-id2172791\">the ratio of the distances from a point [latex]P[\/latex] on the graph to the focus [latex]F[\/latex] and to the directrix [latex]D[\/latex] represented by [latex]e=\\frac{PF}{PD}[\/latex], where [latex]e[\/latex] is a positive real number<\/dd>\n<\/dl>\n<dl id=\"fs-id2172885\" class=\"definition\">\n<dt>polar equation<\/dt>\n<dd id=\"fs-id1271646\">an equation of a curve in polar coordinates [latex]r[\/latex] and [latex]\\theta[\/latex]<\/dd>\n<\/dl>\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-1434\">\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":503070,"menu_order":5,"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-1434","chapter","type-chapter","status-publish","hentry"],"part":1429,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/pressbooks\/v2\/chapters\/1434","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/wp\/v2\/users\/503070"}],"version-history":[{"count":0,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/pressbooks\/v2\/chapters\/1434\/revisions"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/pressbooks\/v2\/parts\/1429"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/pressbooks\/v2\/chapters\/1434\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/wp\/v2\/media?parent=1434"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/pressbooks\/v2\/chapter-type?post=1434"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/wp\/v2\/contributor?post=1434"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-math1613\/wp-json\/wp\/v2\/license?post=1434"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}