{"id":1988,"date":"2015-11-12T18:30:43","date_gmt":"2015-11-12T18:30:43","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1988"},"modified":"2015-11-12T18:30:43","modified_gmt":"2015-11-12T18:30:43","slug":"graphing-the-polar-equations-of-conics","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/graphing-the-polar-equations-of-conics\/","title":{"raw":"Graphing the Polar Equations of Conics","rendered":"Graphing the Polar Equations of Conics"},"content":{"raw":"

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.\n<\/p>

\n

Example 2: Graphing a Parabola in Polar Form<\/h3>\nGraph [latex]r=\\frac{5}{3+3\\text{ }\\cos \\text{ }\\theta }[\/latex].\n\n<\/div>\n
\n

Solution<\/h3>\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
[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ r=\\frac{5}{3+3\\text{ }\\cos \\text{ }\\theta }=\\frac{5\\left(\\frac{1}{3}\\right)}{3\\left(\\frac{1}{3}\\right)+3\\left(\\frac{1}{3}\\right)\\cos \\text{ }\\theta }\\hfill \\end{array}\\hfill \\\\ r=\\frac{\\frac{5}{3}}{1+\\cos \\text{ }\\theta }\\hfill \\end{array}[\/latex]<\/div>\nBecause [latex]e=1[\/latex], we will graph a 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
[latex]\\begin{array}{l}\\frac{5}{3}=ep\\\\ \\frac{5}{3}=\\left(1\\right)p\\\\ \\frac{5}{3}=p\\end{array}[\/latex]<\/div>\nThe directrix is [latex]x=\\frac{5}{3}[\/latex].\n\nPlotting a few key points as in the table below\u00a0will enable us to see the vertices.\n\n
A<\/th>\nB<\/th>\nC<\/th>\nD<\/th>\n<\/tr><\/thead>
[latex]\\theta [\/latex]<\/td>\n[latex]0[\/latex]<\/td>\n[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n[latex]\\pi [\/latex]<\/td>\n[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<\/tr>
[latex]r=\\frac{5}{3+3\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/td>\n[latex]\\frac{5}{6}\\approx 0.83[\/latex]<\/td>\n[latex]\\frac{5}{3}\\approx 1.67[\/latex]<\/td>\nundefined<\/td>\n[latex]\\frac{5}{3}\\approx 1.67[\/latex]<\/td>\n<\/tr><\/tbody><\/table>\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"\"Figure 3<\/b>[\/caption]\n\n<\/div>\n
\n

Analysis of the Solution<\/h3>\nWe can check our result with a graphing utility.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"\"Figure 4<\/b>[\/caption]\n\n<\/div>\n
\n

Example 3: Graphing a Hyperbola in Polar Form<\/h3>\nGraph [latex]r=\\frac{8}{2 - 3\\text{ }\\sin \\text{ }\\theta }[\/latex].\n\n<\/div>\n
\n

Solution<\/h3>\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
[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ r=\\frac{8}{2 - 3\\sin \\text{ }\\theta }=\\frac{8\\left(\\frac{1}{2}\\right)}{2\\left(\\frac{1}{2}\\right)-3\\left(\\frac{1}{2}\\right)\\sin \\text{ }\\theta }\\hfill \\end{array}\\hfill \\\\ r=\\frac{4}{1-\\frac{3}{2}\\text{ }\\sin \\text{ }\\theta }\\hfill \\end{array}[\/latex]<\/div>\nBecause [latex]e=\\frac{3}{2},e>1[\/latex], so we will graph a 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
[latex]\\begin{array}{l}\\text{ }4=ep\\hfill \\\\ \\text{ }4=\\left(\\frac{3}{2}\\right)p\\hfill \\\\ 4\\left(\\frac{2}{3}\\right)=p\\hfill \\\\ \\text{ }\\frac{8}{3}=p\\hfill \\end{array}[\/latex]<\/div>\nThe directrix is [latex]y=-\\frac{8}{3}[\/latex].\n\nPlotting a few key points as in the table below\u00a0will enable us to see the vertices.\n\n
A<\/th>\nB<\/th>\nC<\/th>\nD<\/th>\n<\/tr><\/thead>
[latex]\\theta [\/latex]<\/td>\n[latex]0[\/latex]<\/td>\n[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n[latex]\\pi [\/latex]<\/td>\n[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<\/tr>
[latex]r=\\frac{8}{2 - 3\\sin \\theta }[\/latex]<\/td>\n[latex]4[\/latex]<\/td>\n[latex]-8[\/latex]<\/td>\n[latex]4[\/latex]<\/td>\n[latex]\\frac{8}{5}=1.6[\/latex]<\/td>\n<\/tr><\/tbody><\/table>\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]\"\"Figure 5<\/b>[\/caption]\n\n<\/div>\n
\n

Example 4: Graphing an Ellipse in Polar Form<\/h3>\nGraph [latex]r=\\frac{10}{5 - 4\\text{ }\\cos \\text{ }\\theta }[\/latex].\n\n<\/div>\n
\n

Solution<\/h3>\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
[latex]\\begin{array}{l}\\hfill \\\\ \\begin{array}{l}r=\\frac{10}{5 - 4\\cos \\text{ }\\theta }=\\frac{10\\left(\\frac{1}{5}\\right)}{5\\left(\\frac{1}{5}\\right)-4\\left(\\frac{1}{5}\\right)\\cos \\text{ }\\theta }\\hfill \\\\ r=\\frac{2}{1-\\frac{4}{5}\\text{ }\\cos \\text{ }\\theta }\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\nBecause [latex]e=\\frac{4}{5},e<1[\/latex], so we will graph an ellipse<\/strong> with a 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 directrix<\/strong> is [latex]x=-p[\/latex].\n
[latex]\\begin{array}{l}\\text{ }2=ep\\hfill \\\\ \\text{ }2=\\left(\\frac{4}{5}\\right)p\\hfill \\\\ 2\\left(\\frac{5}{4}\\right)=p\\hfill \\\\ \\text{ }\\frac{5}{2}=p\\hfill \\end{array}[\/latex]<\/div>\nThe directrix is [latex]x=-\\frac{5}{2}[\/latex].\n\nPlotting a few key points as in the table below\u00a0will enable us to see the vertices.\n\n
A<\/th>\nB<\/th>\nC<\/th>\nD<\/th>\n<\/tr><\/thead>
[latex]\\theta [\/latex]<\/td>\n[latex]0[\/latex]<\/td>\n[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n[latex]\\pi [\/latex]<\/td>\n[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<\/tr>
[latex]r=\\frac{10}{5 - 4\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/td>\n[latex]10[\/latex]<\/td>\n[latex]2[\/latex]<\/td>\n[latex]\\frac{10}{9}\\approx 1.1[\/latex]<\/td>\n[latex]2[\/latex]<\/td>\n<\/tr><\/tbody><\/table>\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"\"Figure 6<\/b>[\/caption]\n\n<\/div>\n

Analysis of the Solution<\/h3>\nWe can check our result using a graphing utility.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"\"Figure 7.<\/b> [latex]r=\\frac{10}{5 - 4\\text{ }\\cos \\text{ }\\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]\n
\n

Try It 2<\/h3>\nGraph [latex]r=\\frac{2}{4-\\cos \\text{ }\\theta }[\/latex].\n\nSolution<\/a>\n\n<\/div>","rendered":"

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.\n<\/p>\n

\n

Example 2: Graphing a Parabola in Polar Form<\/h3>\n

Graph [latex]r=\\frac{5}{3+3\\text{ }\\cos \\text{ }\\theta }[\/latex].<\/p>\n<\/div>\n

\n

Solution<\/h3>\n

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

[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ r=\\frac{5}{3+3\\text{ }\\cos \\text{ }\\theta }=\\frac{5\\left(\\frac{1}{3}\\right)}{3\\left(\\frac{1}{3}\\right)+3\\left(\\frac{1}{3}\\right)\\cos \\text{ }\\theta }\\hfill \\end{array}\\hfill \\\\ r=\\frac{\\frac{5}{3}}{1+\\cos \\text{ }\\theta }\\hfill \\end{array}[\/latex]<\/div>\n

Because [latex]e=1[\/latex], we will graph a 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

[latex]\\begin{array}{l}\\frac{5}{3}=ep\\\\ \\frac{5}{3}=\\left(1\\right)p\\\\ \\frac{5}{3}=p\\end{array}[\/latex]<\/div>\n

The directrix is [latex]x=\\frac{5}{3}[\/latex].<\/p>\n

Plotting a few key points as in the table below\u00a0will enable us to see the vertices.<\/p>\n\n\n\n\n\n\n
\n<\/th>\nA<\/th>\nB<\/th>\nC<\/th>\nD<\/th>\n<\/tr>\n<\/thead>\n
[latex]\\theta[\/latex]<\/td>\n[latex]0[\/latex]<\/td>\n[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n[latex]\\pi[\/latex]<\/td>\n[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<\/tr>\n
[latex]r=\\frac{5}{3+3\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/td>\n[latex]\\frac{5}{6}\\approx 0.83[\/latex]<\/td>\n[latex]\\frac{5}{3}\\approx 1.67[\/latex]<\/td>\nundefined<\/td>\n[latex]\\frac{5}{3}\\approx 1.67[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\"\"<\/p>\n

Figure 3<\/b><\/p>\n<\/div>\n<\/div>\n

\n

Analysis of the Solution<\/h3>\n

We can check our result with a graphing utility.<\/p>\n

\"\"<\/p>\n

Figure 4<\/b><\/p>\n<\/div>\n<\/div>\n

\n

Example 3: Graphing a Hyperbola in Polar Form<\/h3>\n

Graph [latex]r=\\frac{8}{2 - 3\\text{ }\\sin \\text{ }\\theta }[\/latex].<\/p>\n<\/div>\n

\n

Solution<\/h3>\n

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

[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ r=\\frac{8}{2 - 3\\sin \\text{ }\\theta }=\\frac{8\\left(\\frac{1}{2}\\right)}{2\\left(\\frac{1}{2}\\right)-3\\left(\\frac{1}{2}\\right)\\sin \\text{ }\\theta }\\hfill \\end{array}\\hfill \\\\ r=\\frac{4}{1-\\frac{3}{2}\\text{ }\\sin \\text{ }\\theta }\\hfill \\end{array}[\/latex]<\/div>\n

Because [latex]e=\\frac{3}{2},e>1[\/latex], so we will graph a 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

[latex]\\begin{array}{l}\\text{ }4=ep\\hfill \\\\ \\text{ }4=\\left(\\frac{3}{2}\\right)p\\hfill \\\\ 4\\left(\\frac{2}{3}\\right)=p\\hfill \\\\ \\text{ }\\frac{8}{3}=p\\hfill \\end{array}[\/latex]<\/div>\n

The directrix is [latex]y=-\\frac{8}{3}[\/latex].<\/p>\n

Plotting a few key points as in the table below\u00a0will enable us to see the vertices.<\/p>\n\n\n\n\n\n\n
\n<\/th>\nA<\/th>\nB<\/th>\nC<\/th>\nD<\/th>\n<\/tr>\n<\/thead>\n
[latex]\\theta[\/latex]<\/td>\n[latex]0[\/latex]<\/td>\n[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n[latex]\\pi[\/latex]<\/td>\n[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<\/tr>\n
[latex]r=\\frac{8}{2 - 3\\sin \\theta }[\/latex]<\/td>\n[latex]4[\/latex]<\/td>\n[latex]-8[\/latex]<\/td>\n[latex]4[\/latex]<\/td>\n[latex]\\frac{8}{5}=1.6[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\"\"<\/p>\n

Figure 5<\/b><\/p>\n<\/div>\n<\/div>\n

\n

Example 4: Graphing an Ellipse in Polar Form<\/h3>\n

Graph [latex]r=\\frac{10}{5 - 4\\text{ }\\cos \\text{ }\\theta }[\/latex].<\/p>\n<\/div>\n

\n

Solution<\/h3>\n

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

[latex]\\begin{array}{l}\\hfill \\\\ \\begin{array}{l}r=\\frac{10}{5 - 4\\cos \\text{ }\\theta }=\\frac{10\\left(\\frac{1}{5}\\right)}{5\\left(\\frac{1}{5}\\right)-4\\left(\\frac{1}{5}\\right)\\cos \\text{ }\\theta }\\hfill \\\\ r=\\frac{2}{1-\\frac{4}{5}\\text{ }\\cos \\text{ }\\theta }\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n

Because [latex]e=\\frac{4}{5},e<1[\/latex], so we will graph an ellipse<\/strong> with a 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 directrix<\/strong> is [latex]x=-p[\/latex].<\/p>\n

[latex]\\begin{array}{l}\\text{ }2=ep\\hfill \\\\ \\text{ }2=\\left(\\frac{4}{5}\\right)p\\hfill \\\\ 2\\left(\\frac{5}{4}\\right)=p\\hfill \\\\ \\text{ }\\frac{5}{2}=p\\hfill \\end{array}[\/latex]<\/div>\n

The directrix is [latex]x=-\\frac{5}{2}[\/latex].<\/p>\n

Plotting a few key points as in the table below\u00a0will enable us to see the vertices.<\/p>\n\n\n\n\n\n\n
\n<\/th>\nA<\/th>\nB<\/th>\nC<\/th>\nD<\/th>\n<\/tr>\n<\/thead>\n
[latex]\\theta[\/latex]<\/td>\n[latex]0[\/latex]<\/td>\n[latex]\\frac{\\pi }{2}[\/latex]<\/td>\n[latex]\\pi[\/latex]<\/td>\n[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\n<\/tr>\n
[latex]r=\\frac{10}{5 - 4\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/td>\n[latex]10[\/latex]<\/td>\n[latex]2[\/latex]<\/td>\n[latex]\\frac{10}{9}\\approx 1.1[\/latex]<\/td>\n[latex]2[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\"\"<\/p>\n

Figure 6<\/b><\/p>\n<\/div>\n<\/div>\n

Analysis of the Solution<\/h3>\n

We can check our result using a graphing utility.<\/p>\n

\"\"<\/p>\n

Figure 7.<\/b> [latex]r=\\frac{10}{5 - 4\\text{ }\\cos \\text{ }\\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

\n

Try It 2<\/h3>\n

Graph [latex]r=\\frac{2}{4-\\cos \\text{ }\\theta }[\/latex].<\/p>\n

Solution<\/a><\/p>\n<\/div>\n\n\t\t\t

\n\t\t\t

Candela Citations<\/h3>\n\t\t\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t\t\t\t
CC licensed content, Specific attribution<\/div>