{"id":1030,"date":"2025-06-20T17:29:46","date_gmt":"2025-06-20T17:29:46","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=1030"},"modified":"2025-09-10T18:53:20","modified_gmt":"2025-09-10T18:53:20","slug":"area-and-arc-length-in-polar-coordinates-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/area-and-arc-length-in-polar-coordinates-learn-it-3\/","title":{"raw":"Conic Sections: Learn It 6","rendered":"Conic Sections: Learn It 6"},"content":{"raw":"

Polar Equations of Conic Sections<\/h2>\r\n

Sometimes it is useful to write or identify the equation of a conic section in polar form. To do this, we need the concept of the focal parameter<\/strong>.<\/p>\r\n\r\n

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focal parameter<\/h3>\r\nThe focal parameter [latex]p[\/latex] of a conic section is the distance from a focus to the nearest directrix.\r\n\r\n<\/section>The following table shows the focal parameters for different types of conics, where [latex]a[\/latex] is the length of the semi-major axis (half the length of the major axis), [latex]c[\/latex] is the distance from the origin to the focus, and [latex]e[\/latex] is the eccentricity. For parabolas, [latex]a[\/latex] represents the distance from the vertex to the focus.\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Eccentricities and Focal Parameters of the Conic Sections<\/span><\/caption>\r\n
Conic<\/th>\r\n[latex]e[\/latex]<\/th>\r\n[latex]p[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
Ellipse<\/td>\r\n[latex]0 < e < 1[\/latex]<\/td>\r\n[latex]\\frac{{a}^{2}-{c}^{2}}{c}=\\frac{a\\left(1-{e}^{2}\\right)}{c}[\/latex]<\/td>\r\n<\/tr>\r\n
Parabola<\/td>\r\n[latex]e=1[\/latex]<\/td>\r\n[latex]2a[\/latex]<\/td>\r\n<\/tr>\r\n
Hyperbola<\/td>\r\n[latex]e > 1[\/latex]<\/td>\r\n[latex]\\frac{{c}^{2}-{a}^{2}}{c}=\\frac{a\\left({e}^{2}-1\\right)}{e}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

Using the definitions of the focal parameter and eccentricity of the conic section, we can derive an equation for any conic section in polar coordinates. In particular, we assume that one of the foci of a given conic section lies at the pole. Then using the definition of the various conic sections in terms of distances, it is possible to prove the following theorem.<\/p>\r\n\r\n

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polar equation of conic sections<\/h3>\r\n

The polar equation of a conic section with focal parameter [latex]p[\/latex] is:<\/p>\r\n

[latex]r = \\frac{ep}{1 \\pm e\\cos\\theta}[\/latex] or [latex]r = \\frac{ep}{1 \\pm e\\sin\\theta}[\/latex]<\/p>\r\n\r\n

    \r\n \t
  • Left equation: horizontal major axis (cosine term)<\/li>\r\n \t
  • Right equation: vertical major axis (sine term)<\/li>\r\n<\/ul>\r\n<\/section>\r\n

    To analyze a conic section written in polar form, follow these steps:<\/p>\r\n\r\n

      \r\n \t
    • Step 1<\/strong>: Make the constant term in the denominator equal to [latex]1[\/latex]. Divide both numerator and denominator by the constant that appears in front of the [latex]\\pm[\/latex] sign.<\/li>\r\n \t
    • Step 2<\/strong>: Identify the eccentricity [latex]e[\/latex] as the coefficient of the trigonometric function in the denominator.<\/li>\r\n \t
    • Step 3<\/strong>: Determine the conic type using the eccentricity value and orientation using the trigonometric function:\r\n
        \r\n \t
      • If cosine appears: horizontal orientation<\/li>\r\n \t
      • If sine appears: vertical orientation<\/li>\r\n \t
      • If both appear: the axes are rotated<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n
        The center of the conic is not necessarily at the origin. The center is at the origin only when the conic is a circle (when [latex]e = 0[\/latex]).<\/section>
        \r\n
        \r\n

        Identify and create a graph of the conic section described by the equation<\/p>\r\n\r\n

        [latex]r=\\dfrac{3}{1+2\\cos\\theta }[\/latex].<\/div>\r\n \r\n\r\n<\/div>\r\n[reveal-answer q=\"44558869\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558869\"]\r\n
        \r\n

        The constant term in the denominator is 1, so the eccentricity of the conic is 2. This is a hyperbola. The focal parameter p<\/em> can be calculated by using the equation [latex]ep=3[\/latex]. Since [latex]e=2[\/latex], this gives [latex]p=\\frac{3}{2}[\/latex]. The cosine function appears in the denominator, so the hyperbola is horizontal. Pick a few values for [latex]\\theta [\/latex] and create a table of values. Then we can graph the hyperbola (Figure 17).<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        [latex]\\theta [\/latex]<\/th>\r\n[latex]r[\/latex]<\/th>\r\n[latex]\\theta [\/latex]<\/th>\r\n[latex]r[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
        0<\/td>\r\n1<\/td>\r\n[latex]\\pi [\/latex]<\/td>\r\n\u22123<\/td>\r\n<\/tr>\r\n
        [latex]\\frac{\\pi }{4}[\/latex]<\/td>\r\n[latex]\\frac{3}{1+\\sqrt{2}}\\approx 1.2426[\/latex]<\/td>\r\n[latex]\\frac{5\\pi }{4}[\/latex]<\/td>\r\n[latex]\\frac{3}{1-\\sqrt{2}}\\approx -7.2426[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]\\frac{\\pi }{2}[\/latex]<\/td>\r\n3<\/td>\r\n[latex]\\frac{3\\pi }{2}[\/latex]<\/td>\r\n3<\/td>\r\n<\/tr>\r\n
        [latex]\\frac{3\\pi }{4}[\/latex]<\/td>\r\n[latex]\\frac{3}{1-\\sqrt{2}}\\approx -7.2426[\/latex]<\/td>\r\n[latex]\\frac{7\\pi }{4}[\/latex]<\/td>\r\n[latex]\\frac{3}{1+\\sqrt{2}}\\approx 1.2426[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
        [caption id=\"\" align=\"aligncenter\" width=\"417\"]\"Graph Figure 17. Graph of the hyperbola described in [link].[\/caption]<\/figure>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
        Watch the following video to see the worked solution to the example above.