{"id":2237,"date":"2025-08-11T17:30:27","date_gmt":"2025-08-11T17:30:27","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2237"},"modified":"2025-08-13T16:59:53","modified_gmt":"2025-08-13T16:59:53","slug":"polar-coordinates-learn-it-5","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/polar-coordinates-learn-it-5\/","title":{"raw":"Polar Coordinates: Learn It 5","rendered":"Polar Coordinates: Learn It 5"},"content":{"raw":"

Identify and Graph Polar Equations by Converting to Rectangular Equations<\/h2>\r\nWe have learned how to convert rectangular coordinates to polar coordinates, and we have seen that the points are indeed the same. We have also transformed polar equations to rectangular equations and vice versa. Now we will demonstrate that their graphs, while drawn on different grids, are identical.\r\n\r\n
Convert the polar equation [latex]r=2\\sec \\theta [\/latex] to a rectangular equation, and draw its corresponding graph.[reveal-answer q=\"584712\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"584712\"]The conversion is\r\n

[latex]\\begin{gathered}r=2\\sec \\theta \\\\ r=\\frac{2}{\\cos \\theta } \\\\ r\\cos \\theta =2 \\\\ x=2 \\end{gathered}[\/latex]<\/p>\r\nNotice that the equation [latex]r=2\\sec \\theta [\/latex] drawn on the polar grid is clearly the same as the vertical line [latex]x=2[\/latex] drawn on the rectangular grid. Just as [latex]x=c[\/latex] is the standard form for a vertical line in rectangular form, [latex]r=c\\sec \\theta [\/latex] is the standard form for a vertical line in polar form.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]\"Plots (a) Polar grid (b) Rectangular coordinate system[\/caption]\r\n\r\nA similar discussion would demonstrate that the graph of the function [latex]r=2\\csc \\theta [\/latex] will be the horizontal line [latex]y=2[\/latex]. In fact, [latex]r=c\\csc \\theta [\/latex] is the standard form for a horizontal line in polar form, corresponding to the rectangular form [latex]y=c[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

Rewrite the polar equation [latex]r=\\frac{3}{1 - 2\\cos \\theta }[\/latex] as a Cartesian equation.[reveal-answer q=\"165840\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"165840\"]The goal is to eliminate [latex]\\theta [\/latex] and [latex]r[\/latex], and introduce [latex]x[\/latex] and [latex]y[\/latex]. We clear the fraction, and then use substitution. In order to replace [latex]r[\/latex] with [latex]x[\/latex] and [latex]y[\/latex], we must use the expression [latex]{x}^{2}+{y}^{2}={r}^{2}[\/latex].\r\n

[latex]\\begin{align}&r=\\frac{3}{1 - 2\\cos \\theta } \\\\ &r\\left(1 - 2\\cos \\theta \\right)=3 \\\\ &r\\left(1 - 2\\left(\\frac{x}{r}\\right)\\right)=3&& \\text{Use }\\cos \\theta =\\frac{x}{r}\\text{ to eliminate }\\theta . \\\\ &r - 2x=3 \\\\ &r=3+2x&& \\text{Isolate }r. \\\\ &{r}^{2}={\\left(3+2x\\right)}^{2}&& \\text{Square both sides}. \\\\ &{x}^{2}+{y}^{2}={\\left(3+2x\\right)}^{2}&& \\text{Use }{x}^{2}+{y}^{2}={r}^{2}. \\end{align}[\/latex]<\/p>\r\nThe Cartesian equation is [latex]{x}^{2}+{y}^{2}={\\left(3+2x\\right)}^{2}[\/latex]. However, to graph it, especially using a graphing calculator or computer program, we want to isolate [latex]y[\/latex].\r\n

[latex]\\begin{gathered}{x}^{2}+{y}^{2}={\\left(3+2x\\right)}^{2} \\\\ {y}^{2}={\\left(3+2x\\right)}^{2}-{x}^{2} \\\\ y=\\pm \\sqrt{{\\left(3+2x\\right)}^{2}-{x}^{2}} \\end{gathered}[\/latex]<\/p>\r\nWhen our entire equation has been changed from [latex]r[\/latex] and [latex]\\theta [\/latex] to [latex]x[\/latex] and [latex]y[\/latex], we can stop, unless asked to solve for [latex]y[\/latex] or simplify.\r\n\r\n\"Plots\r\n\r\nThe \"hour-glass\" shape of the graph is called a hyperbola<\/em>. Hyperbolas have many interesting geometric features and applications, which we will investigate further in Analytic Geometry<\/a>.\r\n

Analysis of the Solution<\/h4>\r\nIn this example, the right side of the equation can be expanded and the equation simplified further, as shown above. However, the equation cannot be written as a single function in Cartesian form. We may wish to write the rectangular equation in the hyperbola\u2019s standard form. To do this, we can start with the initial equation.\r\n

[latex]\\begin{align}&{x}^{2}+{y}^{2}={\\left(3+2x\\right)}^{2} \\\\ &{x}^{2}+{y}^{2}-{\\left(3+2x\\right)}^{2}=0 \\\\ &{x}^{2}+{y}^{2}-\\left(9+12x+4{x}^{2}\\right)=0 \\\\ &{x}^{2}+{y}^{2}-9 - 12x - 4{x}^{2}=0 \\\\ &-3{x}^{2}-12x+{y}^{2}=9&& \\text{Multiply through by }-1. \\\\ &3{x}^{2}+12x-{y}^{2}=-9 \\\\ &3\\left({x}^{2}+4x+\\right)-{y}^{2}=-9&& \\text{Organize terms to complete the square for}x. \\\\ &3\\left({x}^{2}+4x+4\\right)-{y}^{2}=-9+12 \\\\ &3{\\left(x+2\\right)}^{2}-{y}^{2}=3 \\\\ &{\\left(x+2\\right)}^{2}-\\frac{{y}^{2}}{3}=1 \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

Rewrite the polar equation [latex]r=2\\sin \\theta [\/latex] in Cartesian form.[reveal-answer q=\"407542\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"407542\"][latex]{x}^{2}+{y}^{2}=2y[\/latex] or, in the standard form for a circle, [latex]{x}^{2}+{\\left(y - 1\\right)}^{2}=1[\/latex][\/hidden-answer]<\/section>
[ohm_question hide_question_numbers=1]173797[\/ohm_question]<\/section>\r\n
Rewrite the polar equation [latex]r=\\sin \\left(2\\theta \\right)[\/latex] in Cartesian form.[reveal-answer q=\"810563\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"810563\"]\r\n

[latex]\\begin{align}&r=\\sin \\left(2\\theta \\right)&& \\text{Use the double angle identity for sine}. \\\\ &r=2\\sin \\theta \\cos \\theta&& \\text{Use }\\cos \\theta =\\frac{x}{r}\\text{ and }\\sin \\theta =\\frac{y}{r}. \\\\ &r=2\\left(\\frac{x}{r}\\right)\\left(\\frac{y}{r}\\right)&& \\text{Simplify}. \\\\ &r=\\frac{2xy}{{r}^{2}}&& \\text{ Multiply both sides by }{r}^{2}. \\\\ &{r}^{3}=2xy \\\\ &{\\left(\\sqrt{{x}^{2}+{y}^{2}}\\right)}^{3}=2xy&& \\text{As }{x}^{2}+{y}^{2}={r}^{2},r=\\sqrt{{x}^{2}+{y}^{2}}. \\end{align}[\/latex]<\/p>\r\nThis equation can also be written as\r\n

[latex]{\\left({x}^{2}+{y}^{2}\\right)}^{\\frac{3}{2}}=2xy\\text{ or }{x}^{2}+{y}^{2}={\\left(2xy\\right)}^{\\frac{2}{3}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><\/div>","rendered":"

Identify and Graph Polar Equations by Converting to Rectangular Equations<\/h2>\n

We have learned how to convert rectangular coordinates to polar coordinates, and we have seen that the points are indeed the same. We have also transformed polar equations to rectangular equations and vice versa. Now we will demonstrate that their graphs, while drawn on different grids, are identical.<\/p>\n

Convert the polar equation [latex]r=2\\sec \\theta[\/latex] to a rectangular equation, and draw its corresponding graph.<\/p>\n