{"id":2610,"date":"2025-08-13T18:11:40","date_gmt":"2025-08-13T18:11:40","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2610"},"modified":"2025-08-13T18:11:40","modified_gmt":"2025-08-13T18:11:40","slug":"conic-sections-in-polar-coordinates-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/conic-sections-in-polar-coordinates-learn-it-3\/","title":{"raw":"Conic Sections in Polar Coordinates: Learn It 3","rendered":"Conic Sections in Polar Coordinates: Learn It 3"},"content":{"raw":"

Defining Conics in Terms of a Focus and a Directrix<\/h2>\r\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.\r\n\r\n
How To: Given the focus, eccentricity, and directrix of a conic, determine the polar equation.\r\n<\/strong>\r\n
    \r\n \t
  1. 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>\r\n \t
  2. Determine the sign in the denominator. If [latex]p<0[\/latex], use subtraction. If [latex]p>0[\/latex], use addition.<\/li>\r\n \t
  3. Write the coefficient of the trigonometric function as the given eccentricity.<\/li>\r\n \t
  4. Write the absolute value of [latex]p[\/latex] in the numerator, and simplify the equation.<\/li>\r\n<\/ol>\r\n<\/section>
    Find the polar form of the conic<\/strong> given a focus<\/strong> at the origin, [latex]e=3[\/latex] and directrix<\/strong> [latex]y=-2[\/latex].\r\n\r\n[reveal-answer q=\"155638\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"155638\"]\r\n\r\nThe directrix is [latex]y=-p[\/latex], so we know the trigonometric function in the denominator is sine.\r\n\r\nBecause [latex]y=-2,-2<0[\/latex], so we know there is a subtraction sign in the denominator. We use the standard form of\r\n

    [latex]r=\\frac{ep}{1-e \\sin \\theta }[\/latex]<\/p>\r\nand [latex]e=3[\/latex] and [latex]|-2|=2=p[\/latex].\r\n\r\nTherefore,\r\n

    [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>\r\n[\/hidden-answer]\r\n\r\n<\/section>

    Find the polar form of a conic<\/strong> given a focus<\/strong> at the origin, [latex]e=\\frac{3}{5}[\/latex], and directrix<\/strong> [latex]x=4[\/latex].\r\n\r\n[reveal-answer q=\"323889\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"323889\"]\r\n\r\nBecause 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\r\n

    [latex]r=\\frac{ep}{1+e \\cos \\theta }[\/latex]<\/p>\r\nand [latex]e=\\frac{3}{5}[\/latex] and [latex]|4|=4=p[\/latex].\r\n\r\nTherefore,\r\n

    [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>\r\n[\/hidden-answer]\r\n\r\n<\/section>

    \r\n
    \r\n\r\nFind the polar form of the conic given a focus at the origin, [latex]e=1[\/latex], and directrix [latex]x=-1[\/latex].\r\n\r\n[reveal-answer q=\"733852\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"733852\"]\r\n\r\n[latex]r=\\frac{1}{1-\\cos \\theta }[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>
    Convert the conic [latex]r=\\frac{1}{5 - 5\\sin \\theta }[\/latex] to rectangular form.\r\n\r\n[reveal-answer q=\"462269\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"462269\"]\r\n\r\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].\r\n

    [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>\r\n[\/hidden-answer]\r\n\r\n<\/section>

    \r\n
    \r\n\r\nConvert the conic [latex]r=\\frac{2}{1+2\\text{ }\\cos \\text{ }\\theta }[\/latex] to rectangular form.\r\n\r\n[reveal-answer q=\"554425\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"554425\"]\r\n\r\n[latex]4 - 8x+3{x}^{2}-{y}^{2}=0[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>","rendered":"

    Defining Conics in Terms of a Focus and a Directrix<\/h2>\n

    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

    How To: Given the focus, eccentricity, and directrix of a conic, determine the polar equation.
    \n<\/strong><\/p>\n
      \n
    1. 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
    2. Determine the sign in the denominator. If [latex]p<0[\/latex], use subtraction. If [latex]p>0[\/latex], use addition.<\/li>\n
    3. Write the coefficient of the trigonometric function as the given eccentricity.<\/li>\n
    4. Write the absolute value of [latex]p[\/latex] in the numerator, and simplify the equation.<\/li>\n<\/ol>\n<\/section>\n
      Find the polar form of the conic<\/strong> given a focus<\/strong> at the origin, [latex]e=3[\/latex] and directrix<\/strong> [latex]y=-2[\/latex].<\/p>\n