{"id":3300,"date":"2025-08-16T01:53:09","date_gmt":"2025-08-16T01:53:09","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=3300"},"modified":"2025-10-23T17:13:17","modified_gmt":"2025-10-23T17:13:17","slug":"conic-sections-in-polar-coordinates-apply-it","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/conic-sections-in-polar-coordinates-apply-it\/","title":{"raw":"Conic Sections in Polar Coordinates: Apply It","rendered":"Conic Sections in Polar Coordinates: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Identify a conic in polar form.<\/li>\r\n \t<li>Graph the polar equations of conics.<\/li>\r\n \t<li>De\ufb01ne conics in terms of a focus and a directrix.<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox recall\" aria-label=\"Recall\">\r\n<p class=\"whitespace-normal break-words\">A conic section in polar form with focus at the origin has the equation [latex]r = \\frac{ep}{1 \\pm e\\cos\\theta}[\/latex] or [latex]r = \\frac{ep}{1 \\pm e\\sin\\theta}[\/latex], where [latex]e[\/latex] is the eccentricity and [latex]p[\/latex] is the distance to the directrix (a fixed reference line). The eccentricity determines the shape:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">If [latex]0 \\leq e &lt; 1[\/latex]: ellipse<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]e = 1[\/latex]: parabola<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]e &gt; 1[\/latex]: hyperbola<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">The trigonometric function indicates directrix orientation: [latex]\\cos\\theta[\/latex] means a vertical directrix ([latex]x = \\pm p[\/latex]), while [latex]\\sin\\theta[\/latex] means a horizontal directrix ([latex]y = \\pm p[\/latex]).\r\n\r\n<\/section>\r\n<h2 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Orbital Mechanics<\/h2>\r\n<p class=\"whitespace-normal break-words\">Satellites, planets, and comets follow conic paths around celestial bodies. The shape of an orbit depends on the object's velocity and distance from the body it orbits. Understanding these polar equations helps scientists predict orbital behavior and plan space missions.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-normal break-words\">A satellite's orbital path is described by [latex]r = \\frac{6}{3 + 2\\sin\\theta}[\/latex], where [latex]r[\/latex] is measured in thousands of kilometers. Identify the type of orbit, the eccentricity, and the directrix.<\/p>\r\n<p class=\"whitespace-normal break-words\">\r\n[reveal-answer q=\"681883\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"681883\"]First, rewrite in standard form by multiplying numerator and denominator by [latex]\\frac{1}{3}[\/latex]: [latex]\\begin{aligned} r &amp;= \\frac{6}{3 + 2\\sin\\theta} \\cdot \\frac{\\frac{1}{3}}{\\frac{1}{3}} \\ &amp;= \\frac{6 \\cdot \\frac{1}{3}}{3 \\cdot \\frac{1}{3} + 2 \\cdot \\frac{1}{3}\\sin\\theta} \\ &amp;= \\frac{2}{1 + \\frac{2}{3}\\sin\\theta} \\end{aligned}[\/latex] Now identify the characteristics: The eccentricity is [latex]e = \\frac{2}{3}[\/latex]. Since [latex]e &lt; 1[\/latex], this is an ellipse. Since [latex]\\sin\\theta[\/latex] appears in the denominator with a plus sign, the directrix is [latex]y = p[\/latex] where [latex]p &gt; 0[\/latex]. To find [latex]p[\/latex], use [latex]ep = 2[\/latex]: [latex]\\begin{aligned} \\frac{2}{3} \\cdot p &amp;= 2 \\ p &amp;= 2 \\cdot \\frac{3}{2} \\ p &amp;= 3 \\end{aligned}[\/latex] The satellite follows an elliptical orbit with eccentricity [latex]e = \\frac{2}{3}[\/latex] and directrix [latex]y = 3[\/latex] (3,000 km above the origin).[\/hidden-answer]<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-normal break-words\">A space probe follows the path [latex]r = \\frac{7}{2 - 2\\sin\\theta}[\/latex]. Identify the orbit characteristics.<\/p>\r\n<p class=\"whitespace-normal break-words\">\r\n[reveal-answer q=\"109528\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"109528\"]Multiply numerator and denominator by [latex]\\frac{1}{2}[\/latex]: [latex]\\begin{aligned} r &amp;= \\frac{7}{2 - 2\\sin\\theta} \\cdot \\frac{\\frac{1}{2}}{\\frac{1}{2}} \\ &amp;= \\frac{\\frac{7}{2}}{1 - \\sin\\theta} \\end{aligned}[\/latex] The eccentricity is [latex]e = 1[\/latex], so this is a parabola. Since [latex]\\sin\\theta[\/latex] appears with a minus sign, the directrix is [latex]y = -p[\/latex] where [latex]p &gt; 0[\/latex]. From [latex]ep = \\frac{7}{2}[\/latex]: [latex]\\begin{aligned} (1) \\cdot p &amp;= \\frac{7}{2} \\ p &amp;= \\frac{7}{2} \\end{aligned}[\/latex] The directrix is [latex]y = -\\frac{7}{2} = -3.5[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/section><section class=\"textbox connectIt\" aria-label=\"Connect It\">A parabolic orbit ([latex]e = 1[\/latex]) represents the boundary between closed orbits (ellipses) and open orbits (hyperbolas). Objects at exactly escape velocity follow parabolic paths!\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<p class=\"whitespace-normal break-words\">A comet's path is given by [latex]r = \\frac{12}{4 + 5\\cos\\theta}[\/latex]. Identify the type of conic, the eccentricity, and the directrix.<\/p>\r\n<p class=\"whitespace-normal break-words\">Type of conic: [response area]<\/p>\r\n<p class=\"whitespace-normal break-words\">Eccentricity [latex]e =[\/latex] [response area]<\/p>\r\n<p class=\"whitespace-normal break-words\">Directrix: [response area]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Correct answer:<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Type: hyperbola<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Eccentricity: [latex]\\frac{5}{4}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Directrix: [latex]x = \\frac{12}{5}[\/latex] or [latex]x = 2.4[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Feedback for correct answer:<\/strong> Excellent! You rewrote the equation as [latex]r = \\frac{3}{1 + \\frac{5}{4}\\cos\\theta}[\/latex], identified [latex]e = \\frac{5}{4} &gt; 1[\/latex] (hyperbola), and solved [latex]\\frac{5}{4}p = 3[\/latex] to get [latex]p = \\frac{12}{5}[\/latex]. Since [latex]\\cos\\theta[\/latex] appears with a plus sign, the directrix is [latex]x = \\frac{12}{5}[\/latex]. This hyperbolic path means the comet will escape the gravitational pull!<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Feedback for incorrect answer:<\/strong> Start by multiplying numerator and denominator by [latex]\\frac{1}{4}[\/latex] to get standard form. The coefficient of [latex]\\cos\\theta[\/latex] in the denominator is your eccentricity [latex]e[\/latex]. Compare [latex]e[\/latex] to 1 to determine the conic type. Since [latex]\\cos\\theta[\/latex] appears, the directrix has form [latex]x = p[\/latex].<\/p>\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<p class=\"whitespace-normal break-words\">Identify the conic type and eccentricity for [latex]r = \\frac{2}{3 - \\cos\\theta}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Type: [response area]<\/p>\r\n<p class=\"whitespace-normal break-words\">Eccentricity [latex]e =[\/latex] [response area]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Correct answer:<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Type: ellipse<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Eccentricity: [latex]\\frac{1}{3}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Feedback for correct answer:<\/strong> Perfect! You converted to standard form [latex]r = \\frac{\\frac{2}{3}}{1 - \\frac{1}{3}\\cos\\theta}[\/latex] and identified [latex]e = \\frac{1}{3} &lt; 1[\/latex], which means this is an ellipse. The relatively small eccentricity indicates a nearly circular orbit.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Feedback for incorrect answer:<\/strong> Multiply both numerator and denominator by [latex]\\frac{1}{3}[\/latex] to rewrite with 1 in the denominator. The coefficient of [latex]\\cos\\theta[\/latex] after this conversion is your eccentricity. Since this value is less than 1, what type of conic is it?<\/p>\r\n\r\n<\/section><section class=\"textbox recall\" aria-label=\"Recall\">The general forms are [latex]r = \\frac{ep}{1 \\pm e\\cos\\theta}[\/latex] or [latex]r = \\frac{ep}{1 \\pm e\\sin\\theta}[\/latex]. Choose [latex]\\cos\\theta[\/latex] for a vertical directrix ([latex]x = \\pm p[\/latex]) and [latex]\\sin\\theta[\/latex] for a horizontal directrix ([latex]y = \\pm p[\/latex]). Use [latex]+[\/latex] when [latex]p &gt; 0[\/latex] and [latex]-[\/latex] when [latex]p &lt; 0[\/latex].\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-normal break-words\">Mission control needs to place a satellite in an elliptical orbit with focus at Earth's center, eccentricity [latex]e = \\frac{3}{5}[\/latex], and directrix at [latex]x = 4[\/latex] (in thousands of km). Write the polar equation for this orbit.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Solution:<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The directrix [latex]x = 4[\/latex] tells us:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Use [latex]\\cos\\theta[\/latex] in the denominator (vertical directrix)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Since [latex]4 &gt; 0[\/latex], use the [latex]+[\/latex] sign<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]p = 4[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">The standard form is:<\/p>\r\n<p class=\"whitespace-normal break-words\">[latex]r = \\frac{ep}{1 + e\\cos\\theta}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Substituting [latex]e = \\frac{3}{5}[\/latex] and [latex]p = 4[\/latex]:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">[latex]\\begin{aligned} r &amp;= \\frac{\\frac{3}{5} \\cdot 4}{1 + \\frac{3}{5}\\cos\\theta} \\ &amp;= \\frac{\\frac{12}{5}}{1 + \\frac{3}{5}\\cos\\theta} \\end{aligned}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">To eliminate the fraction in the numerator, multiply by [latex]\\frac{5}{5}[\/latex]:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">[latex]\\begin{aligned} r &amp;= \\frac{\\frac{12}{5}}{\\frac{5}{5} + \\frac{3}{5}\\cos\\theta} \\ &amp;= \\frac{12}{5} \\cdot \\frac{5}{5 + 3\\cos\\theta} \\ &amp;= \\frac{12}{5 + 3\\cos\\theta} \\end{aligned}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">The orbital equation is [latex]r = \\frac{12}{5 + 3\\cos\\theta}[\/latex].<\/p>\r\n\r\n<\/section><section class=\"textbox youChoose\" aria-label=\"You Choose\">\r\n<p class=\"whitespace-normal break-words\">You are a mission planner designing trajectories for three different space missions. Each mission requires a specific orbital path with a focus at Earth's center. Choose one mission and write its polar equation.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Mission A: Communications Satellite<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Orbit type: Elliptical (for stable, repeating coverage)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Eccentricity: [latex]e = \\frac{2}{3}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Directrix: [latex]y = 6[\/latex] thousand km<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Mission B: Interstellar Probe<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Orbit type: Hyperbolic (to escape the solar system)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Eccentricity: [latex]e = \\frac{5}{2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Directrix: [latex]x = -4[\/latex] thousand km<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Mission C: Solar Observation Probe<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Orbit type: Parabolic (at exact escape velocity)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Eccentricity: [latex]e = 1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Directrix: [latex]y = -3[\/latex] thousand km<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Choose your mission and write the polar equation: [latex]r =[\/latex] [response area]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Correct answer for Mission A:<\/strong> [latex]r = \\frac{4}{1 + \\frac{2}{3}\\sin\\theta}[\/latex] or [latex]r = \\frac{12}{3 + 2\\sin\\theta}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Feedback for Mission A correct:<\/strong> Excellent mission planning! You identified that directrix [latex]y = 6[\/latex] requires [latex]\\sin\\theta[\/latex] with a plus sign. With [latex]e = \\frac{2}{3}[\/latex] and [latex]p = 6[\/latex], you calculated [latex]ep = 4[\/latex]. Your equation [latex]r = \\frac{4}{1 + \\frac{2}{3}\\sin\\theta}[\/latex] creates a stable elliptical orbit perfect for consistent satellite coverage. The eccentricity of [latex]\\frac{2}{3}[\/latex] provides an elongated but closed orbit.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Feedback for Mission A incorrect:<\/strong> For directrix [latex]y = 6[\/latex], use the form [latex]r = \\frac{ep}{1 + e\\sin\\theta}[\/latex] (positive [latex]p[\/latex] means plus sign). Calculate [latex]ep = \\frac{2}{3} \\cdot 6 = 4[\/latex] for the numerator.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Correct answer for Mission B:<\/strong> [latex]r = \\frac{10}{1 - \\frac{5}{2}\\cos\\theta}[\/latex] or [latex]r = \\frac{20}{2 - 5\\cos\\theta}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Feedback for Mission B correct:<\/strong> Outstanding! You recognized that directrix [latex]x = -4[\/latex] requires [latex]\\cos\\theta[\/latex] with a minus sign. With [latex]e = \\frac{5}{2}[\/latex] and [latex]p = 4[\/latex], you found [latex]ep = 10[\/latex]. Your hyperbolic trajectory [latex]r = \\frac{10}{1 - \\frac{5}{2}\\cos\\theta}[\/latex] will successfully launch the probe out of the solar system. Since [latex]e &gt; 1[\/latex], this open orbit provides the escape path needed!<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Feedback for Mission B incorrect:<\/strong> For directrix [latex]x = -4[\/latex], use the form [latex]r = \\frac{ep}{1 - e\\cos\\theta}[\/latex] (negative [latex]p[\/latex] means minus sign, but use [latex]|p| = 4[\/latex]). Calculate [latex]ep = \\frac{5}{2} \\cdot 4 = 10[\/latex] for the numerator.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Correct answer for Mission C:<\/strong> [latex]r = \\frac{3}{1 - \\sin\\theta}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Feedback for Mission C correct:<\/strong> Perfect trajectory design! You correctly identified that directrix [latex]y = -3[\/latex] requires [latex]\\sin\\theta[\/latex] with a minus sign. For a parabola, [latex]e = 1[\/latex], so [latex]ep = (1)(3) = 3[\/latex]. Your equation [latex]r = \\frac{3}{1 - \\sin\\theta}[\/latex] places the probe at exactly escape velocity\u2014the boundary between returning to Earth and escaping forever. This is ideal for a solar observation mission that needs to break free of Earth's orbit!<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Feedback for Mission C incorrect:<\/strong> A parabola has [latex]e = 1[\/latex]. For directrix [latex]y = -3[\/latex], use [latex]r = \\frac{ep}{1 - e\\sin\\theta}[\/latex]. With [latex]e = 1[\/latex] and [latex]|p| = 3[\/latex], calculate [latex]ep[\/latex] for the numerator.<\/p>\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\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<\/section>\n<section class=\"textbox recall\" aria-label=\"Recall\">\n<p class=\"whitespace-normal break-words\">A conic section in polar form with focus at the origin has the equation [latex]r = \\frac{ep}{1 \\pm e\\cos\\theta}[\/latex] or [latex]r = \\frac{ep}{1 \\pm e\\sin\\theta}[\/latex], where [latex]e[\/latex] is the eccentricity and [latex]p[\/latex] is the distance to the directrix (a fixed reference line). The eccentricity determines the shape:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">If [latex]0 \\leq e < 1[\/latex]: ellipse<\/li>\n<li class=\"whitespace-normal break-words\">If [latex]e = 1[\/latex]: parabola<\/li>\n<li class=\"whitespace-normal break-words\">If [latex]e > 1[\/latex]: hyperbola<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">The trigonometric function indicates directrix orientation: [latex]\\cos\\theta[\/latex] means a vertical directrix ([latex]x = \\pm p[\/latex]), while [latex]\\sin\\theta[\/latex] means a horizontal directrix ([latex]y = \\pm p[\/latex]).<\/p>\n<\/section>\n<h2 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Orbital Mechanics<\/h2>\n<p class=\"whitespace-normal break-words\">Satellites, planets, and comets follow conic paths around celestial bodies. The shape of an orbit depends on the object&#8217;s velocity and distance from the body it orbits. Understanding these polar equations helps scientists predict orbital behavior and plan space missions.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-normal break-words\">A satellite&#8217;s orbital path is described by [latex]r = \\frac{6}{3 + 2\\sin\\theta}[\/latex], where [latex]r[\/latex] is measured in thousands of kilometers. Identify the type of orbit, the eccentricity, and the directrix.<\/p>\n<p class=\"whitespace-normal break-words\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q681883\">Show Solution<\/button><\/p>\n<div id=\"q681883\" class=\"hidden-answer\" style=\"display: none\">First, rewrite in standard form by multiplying numerator and denominator by [latex]\\frac{1}{3}[\/latex]: [latex]\\begin{aligned} r &= \\frac{6}{3 + 2\\sin\\theta} \\cdot \\frac{\\frac{1}{3}}{\\frac{1}{3}} \\ &= \\frac{6 \\cdot \\frac{1}{3}}{3 \\cdot \\frac{1}{3} + 2 \\cdot \\frac{1}{3}\\sin\\theta} \\ &= \\frac{2}{1 + \\frac{2}{3}\\sin\\theta} \\end{aligned}[\/latex] Now identify the characteristics: The eccentricity is [latex]e = \\frac{2}{3}[\/latex]. Since [latex]e < 1[\/latex], this is an ellipse. Since [latex]\\sin\\theta[\/latex] appears in the denominator with a plus sign, the directrix is [latex]y = p[\/latex] where [latex]p > 0[\/latex]. To find [latex]p[\/latex], use [latex]ep = 2[\/latex]: [latex]\\begin{aligned} \\frac{2}{3} \\cdot p &= 2 \\ p &= 2 \\cdot \\frac{3}{2} \\ p &= 3 \\end{aligned}[\/latex] The satellite follows an elliptical orbit with eccentricity [latex]e = \\frac{2}{3}[\/latex] and directrix [latex]y = 3[\/latex] (3,000 km above the origin).<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-normal break-words\">A space probe follows the path [latex]r = \\frac{7}{2 - 2\\sin\\theta}[\/latex]. Identify the orbit characteristics.<\/p>\n<p class=\"whitespace-normal break-words\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q109528\">Show Solution<\/button><\/p>\n<div id=\"q109528\" class=\"hidden-answer\" style=\"display: none\">Multiply numerator and denominator by [latex]\\frac{1}{2}[\/latex]: [latex]\\begin{aligned} r &= \\frac{7}{2 - 2\\sin\\theta} \\cdot \\frac{\\frac{1}{2}}{\\frac{1}{2}} \\ &= \\frac{\\frac{7}{2}}{1 - \\sin\\theta} \\end{aligned}[\/latex] The eccentricity is [latex]e = 1[\/latex], so this is a parabola. Since [latex]\\sin\\theta[\/latex] appears with a minus sign, the directrix is [latex]y = -p[\/latex] where [latex]p > 0[\/latex]. From [latex]ep = \\frac{7}{2}[\/latex]: [latex]\\begin{aligned} (1) \\cdot p &= \\frac{7}{2} \\ p &= \\frac{7}{2} \\end{aligned}[\/latex] The directrix is [latex]y = -\\frac{7}{2} = -3.5[\/latex].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox connectIt\" aria-label=\"Connect It\">A parabolic orbit ([latex]e = 1[\/latex]) represents the boundary between closed orbits (ellipses) and open orbits (hyperbolas). Objects at exactly escape velocity follow parabolic paths!<\/p>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<p class=\"whitespace-normal break-words\">A comet&#8217;s path is given by [latex]r = \\frac{12}{4 + 5\\cos\\theta}[\/latex]. Identify the type of conic, the eccentricity, and the directrix.<\/p>\n<p class=\"whitespace-normal break-words\">Type of conic: [response area]<\/p>\n<p class=\"whitespace-normal break-words\">Eccentricity [latex]e =[\/latex] [response area]<\/p>\n<p class=\"whitespace-normal break-words\">Directrix: [response area]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Correct answer:<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Type: hyperbola<\/li>\n<li class=\"whitespace-normal break-words\">Eccentricity: [latex]\\frac{5}{4}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Directrix: [latex]x = \\frac{12}{5}[\/latex] or [latex]x = 2.4[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Feedback for correct answer:<\/strong> Excellent! You rewrote the equation as [latex]r = \\frac{3}{1 + \\frac{5}{4}\\cos\\theta}[\/latex], identified [latex]e = \\frac{5}{4} > 1[\/latex] (hyperbola), and solved [latex]\\frac{5}{4}p = 3[\/latex] to get [latex]p = \\frac{12}{5}[\/latex]. Since [latex]\\cos\\theta[\/latex] appears with a plus sign, the directrix is [latex]x = \\frac{12}{5}[\/latex]. This hyperbolic path means the comet will escape the gravitational pull!<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Feedback for incorrect answer:<\/strong> Start by multiplying numerator and denominator by [latex]\\frac{1}{4}[\/latex] to get standard form. The coefficient of [latex]\\cos\\theta[\/latex] in the denominator is your eccentricity [latex]e[\/latex]. Compare [latex]e[\/latex] to 1 to determine the conic type. Since [latex]\\cos\\theta[\/latex] appears, the directrix has form [latex]x = p[\/latex].<\/p>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<p class=\"whitespace-normal break-words\">Identify the conic type and eccentricity for [latex]r = \\frac{2}{3 - \\cos\\theta}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Type: [response area]<\/p>\n<p class=\"whitespace-normal break-words\">Eccentricity [latex]e =[\/latex] [response area]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Correct answer:<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Type: ellipse<\/li>\n<li class=\"whitespace-normal break-words\">Eccentricity: [latex]\\frac{1}{3}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Feedback for correct answer:<\/strong> Perfect! You converted to standard form [latex]r = \\frac{\\frac{2}{3}}{1 - \\frac{1}{3}\\cos\\theta}[\/latex] and identified [latex]e = \\frac{1}{3} < 1[\/latex], which means this is an ellipse. The relatively small eccentricity indicates a nearly circular orbit.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Feedback for incorrect answer:<\/strong> Multiply both numerator and denominator by [latex]\\frac{1}{3}[\/latex] to rewrite with 1 in the denominator. The coefficient of [latex]\\cos\\theta[\/latex] after this conversion is your eccentricity. Since this value is less than 1, what type of conic is it?<\/p>\n<\/section>\n<section class=\"textbox recall\" aria-label=\"Recall\">The general forms are [latex]r = \\frac{ep}{1 \\pm e\\cos\\theta}[\/latex] or [latex]r = \\frac{ep}{1 \\pm e\\sin\\theta}[\/latex]. Choose [latex]\\cos\\theta[\/latex] for a vertical directrix ([latex]x = \\pm p[\/latex]) and [latex]\\sin\\theta[\/latex] for a horizontal directrix ([latex]y = \\pm p[\/latex]). Use [latex]+[\/latex] when [latex]p > 0[\/latex] and [latex]-[\/latex] when [latex]p < 0[\/latex].\n\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-normal break-words\">Mission control needs to place a satellite in an elliptical orbit with focus at Earth&#8217;s center, eccentricity [latex]e = \\frac{3}{5}[\/latex], and directrix at [latex]x = 4[\/latex] (in thousands of km). Write the polar equation for this orbit.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Solution:<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The directrix [latex]x = 4[\/latex] tells us:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Use [latex]\\cos\\theta[\/latex] in the denominator (vertical directrix)<\/li>\n<li class=\"whitespace-normal break-words\">Since [latex]4 > 0[\/latex], use the [latex]+[\/latex] sign<\/li>\n<li class=\"whitespace-normal break-words\">[latex]p = 4[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">The standard form is:<\/p>\n<p class=\"whitespace-normal break-words\">[latex]r = \\frac{ep}{1 + e\\cos\\theta}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Substituting [latex]e = \\frac{3}{5}[\/latex] and [latex]p = 4[\/latex]:<\/p>\n<p class=\"whitespace-pre-wrap break-words\">[latex]\\begin{aligned} r &= \\frac{\\frac{3}{5} \\cdot 4}{1 + \\frac{3}{5}\\cos\\theta} \\ &= \\frac{\\frac{12}{5}}{1 + \\frac{3}{5}\\cos\\theta} \\end{aligned}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">To eliminate the fraction in the numerator, multiply by [latex]\\frac{5}{5}[\/latex]:<\/p>\n<p class=\"whitespace-pre-wrap break-words\">[latex]\\begin{aligned} r &= \\frac{\\frac{12}{5}}{\\frac{5}{5} + \\frac{3}{5}\\cos\\theta} \\ &= \\frac{12}{5} \\cdot \\frac{5}{5 + 3\\cos\\theta} \\ &= \\frac{12}{5 + 3\\cos\\theta} \\end{aligned}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">The orbital equation is [latex]r = \\frac{12}{5 + 3\\cos\\theta}[\/latex].<\/p>\n<\/section>\n<section class=\"textbox youChoose\" aria-label=\"You Choose\">\n<p class=\"whitespace-normal break-words\">You are a mission planner designing trajectories for three different space missions. Each mission requires a specific orbital path with a focus at Earth&#8217;s center. Choose one mission and write its polar equation.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Mission A: Communications Satellite<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Orbit type: Elliptical (for stable, repeating coverage)<\/li>\n<li class=\"whitespace-normal break-words\">Eccentricity: [latex]e = \\frac{2}{3}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Directrix: [latex]y = 6[\/latex] thousand km<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Mission B: Interstellar Probe<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Orbit type: Hyperbolic (to escape the solar system)<\/li>\n<li class=\"whitespace-normal break-words\">Eccentricity: [latex]e = \\frac{5}{2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Directrix: [latex]x = -4[\/latex] thousand km<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Mission C: Solar Observation Probe<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Orbit type: Parabolic (at exact escape velocity)<\/li>\n<li class=\"whitespace-normal break-words\">Eccentricity: [latex]e = 1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Directrix: [latex]y = -3[\/latex] thousand km<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Choose your mission and write the polar equation: [latex]r =[\/latex] [response area]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Correct answer for Mission A:<\/strong> [latex]r = \\frac{4}{1 + \\frac{2}{3}\\sin\\theta}[\/latex] or [latex]r = \\frac{12}{3 + 2\\sin\\theta}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Feedback for Mission A correct:<\/strong> Excellent mission planning! You identified that directrix [latex]y = 6[\/latex] requires [latex]\\sin\\theta[\/latex] with a plus sign. With [latex]e = \\frac{2}{3}[\/latex] and [latex]p = 6[\/latex], you calculated [latex]ep = 4[\/latex]. Your equation [latex]r = \\frac{4}{1 + \\frac{2}{3}\\sin\\theta}[\/latex] creates a stable elliptical orbit perfect for consistent satellite coverage. The eccentricity of [latex]\\frac{2}{3}[\/latex] provides an elongated but closed orbit.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Feedback for Mission A incorrect:<\/strong> For directrix [latex]y = 6[\/latex], use the form [latex]r = \\frac{ep}{1 + e\\sin\\theta}[\/latex] (positive [latex]p[\/latex] means plus sign). Calculate [latex]ep = \\frac{2}{3} \\cdot 6 = 4[\/latex] for the numerator.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Correct answer for Mission B:<\/strong> [latex]r = \\frac{10}{1 - \\frac{5}{2}\\cos\\theta}[\/latex] or [latex]r = \\frac{20}{2 - 5\\cos\\theta}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Feedback for Mission B correct:<\/strong> Outstanding! You recognized that directrix [latex]x = -4[\/latex] requires [latex]\\cos\\theta[\/latex] with a minus sign. With [latex]e = \\frac{5}{2}[\/latex] and [latex]p = 4[\/latex], you found [latex]ep = 10[\/latex]. Your hyperbolic trajectory [latex]r = \\frac{10}{1 - \\frac{5}{2}\\cos\\theta}[\/latex] will successfully launch the probe out of the solar system. Since [latex]e > 1[\/latex], this open orbit provides the escape path needed!<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Feedback for Mission B incorrect:<\/strong> For directrix [latex]x = -4[\/latex], use the form [latex]r = \\frac{ep}{1 - e\\cos\\theta}[\/latex] (negative [latex]p[\/latex] means minus sign, but use [latex]|p| = 4[\/latex]). Calculate [latex]ep = \\frac{5}{2} \\cdot 4 = 10[\/latex] for the numerator.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Correct answer for Mission C:<\/strong> [latex]r = \\frac{3}{1 - \\sin\\theta}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Feedback for Mission C correct:<\/strong> Perfect trajectory design! You correctly identified that directrix [latex]y = -3[\/latex] requires [latex]\\sin\\theta[\/latex] with a minus sign. For a parabola, [latex]e = 1[\/latex], so [latex]ep = (1)(3) = 3[\/latex]. Your equation [latex]r = \\frac{3}{1 - \\sin\\theta}[\/latex] places the probe at exactly escape velocity\u2014the boundary between returning to Earth and escaping forever. This is ideal for a solar observation mission that needs to break free of Earth&#8217;s orbit!<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Feedback for Mission C incorrect:<\/strong> A parabola has [latex]e = 1[\/latex]. For directrix [latex]y = -3[\/latex], use [latex]r = \\frac{ep}{1 - e\\sin\\theta}[\/latex]. With [latex]e = 1[\/latex] and [latex]|p| = 3[\/latex], calculate [latex]ep[\/latex] for the numerator.<\/p>\n<\/section>\n","protected":false},"author":67,"menu_order":31,"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":"apply_it","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\/3300"}],"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\/67"}],"version-history":[{"count":3,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3300\/revisions"}],"predecessor-version":[{"id":4859,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3300\/revisions\/4859"}],"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\/3300\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=3300"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=3300"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=3300"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=3300"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}