{"id":1989,"date":"2015-11-12T18:30:43","date_gmt":"2015-11-12T18:30:43","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1989"},"modified":"2015-11-12T18:30:43","modified_gmt":"2015-11-12T18:30:43","slug":"de%ef%ac%81ning-conics-in-terms-of-a-focus-and-a-directrix","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/de%ef%ac%81ning-conics-in-terms-of-a-focus-and-a-directrix\/","title":{"raw":"De\ufb01ning Conics in Terms of a Focus and a Directrix","rendered":"De\ufb01ning Conics in Terms of a Focus and a Directrix"},"content":{"raw":"

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.\n<\/p>

\n

How To: Given the focus, eccentricity, and directrix of a conic, determine the polar equation.\n<\/strong><\/h3>\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\t
  2. Determine the sign in the denominator. If [latex]p<0[\/latex], use subtraction. If [latex]p>0[\/latex], use addition.<\/li>\n\t
  3. Write the coefficient of the trigonometric function as the given eccentricity.<\/li>\n\t
  4. Write the absolute value of [latex]p[\/latex] in the numerator, and simplify the equation.<\/li>\n<\/ol><\/div>\n
    \n

    Example 5: Finding the Polar Form of a Vertical Conic Given a Focus at the Origin and the Eccentricity and Directrix<\/h3>\nFind 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].\n\n<\/div>\n
    \n

    Solution<\/h3>\nThe directrix is [latex]y=-p[\/latex], so we know the trigonometric function in the denominator is sine.\n\nBecause [latex]y=-2,-2<0[\/latex], so we know there is a subtraction sign in the denominator. We use the standard form of\n
    [latex]r=\\frac{ep}{1-e\\text{ }\\sin \\text{ }\\theta }[\/latex]<\/div>\nand [latex]e=3[\/latex] and [latex]|-2|=2=p[\/latex].\n\nTherefore,\n
    [latex]\\begin{array}{l}\\hfill \\\\ \\begin{array}{l}r=\\frac{\\left(3\\right)\\left(2\\right)}{1 - 3\\text{ }\\sin \\text{ }\\theta }\\hfill \\\\ r=\\frac{6}{1 - 3\\text{ }\\sin \\text{ }\\theta }\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
    \n

    Example 6: Finding the Polar Form of a Horizontal Conic Given a Focus at the Origin and the Eccentricity and Directrix<\/h3>\nFind 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].\n\n<\/div>\n
    \n

    Solution<\/h3>\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\n
    [latex]r=\\frac{ep}{1+e\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/div>\nand [latex]e=\\frac{3}{5}[\/latex] and [latex]|4|=4=p[\/latex].\n\nTherefore,\n
    [latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ r=\\frac{\\left(\\frac{3}{5}\\right)\\left(4\\right)}{1+\\frac{3}{5}\\cos \\theta }\\hfill \\end{array}\\hfill \\\\ r=\\frac{\\frac{12}{5}}{1+\\frac{3}{5}\\cos \\theta }\\hfill \\\\ r=\\frac{\\frac{12}{5}}{1\\left(\\frac{5}{5}\\right)+\\frac{3}{5}\\cos \\theta }\\hfill \\\\ r=\\frac{\\frac{12}{5}}{\\frac{5}{5}+\\frac{3}{5}\\cos \\theta }\\hfill \\\\ r=\\frac{12}{5}\\cdot \\frac{5}{5+3\\cos \\theta }\\hfill \\\\ r=\\frac{12}{5+3\\cos \\theta }\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
    \n

    Try It 3<\/h3>\nFind the polar form of the conic given a focus at the origin, [latex]e=1[\/latex], and directrix [latex]x=-1[\/latex].\n\nSolution<\/a>\n\n<\/div>\n
    \n

    Example 5: Converting a Conic in Polar Form to Rectangular Form<\/h3>\nConvert the conic [latex]r=\\frac{1}{5 - 5\\sin \\theta }[\/latex] to rectangular form.\n\n<\/div>\n
    \n

    Solution<\/h3>\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].\n
    [latex]\\begin{array}{ll}\\text{ }r=\\frac{1}{5 - 5\\sin \\theta }\\hfill & \\hfill \\\\ r\\cdot \\left(5 - 5\\sin \\theta \\right)=\\frac{1}{5 - 5\\sin \\theta }\\cdot \\left(5 - 5\\sin \\theta \\right)\\hfill & \\text{Eliminate the fraction}.\\hfill \\\\ \\text{ }5r - 5r\\sin \\theta =1\\hfill & \\text{Distribute}.\\hfill \\\\ \\text{ }5r=1+5r\\sin \\theta \\hfill & \\text{Isolate }5r.\\hfill \\\\ \\text{ }25{r}^{2}={\\left(1+5r\\sin \\theta \\right)}^{2}\\hfill & \\text{Square both sides}.\\hfill \\\\ \\text{ }25\\left({x}^{2}+{y}^{2}\\right)={\\left(1+5y\\right)}^{2}\\hfill & \\text{Substitute }r=\\sqrt{{x}^{2}+{y}^{2}}\\text{ and }y=r\\sin \\theta .\\hfill \\\\ \\text{ }25{x}^{2}+25{y}^{2}=1+10y+25{y}^{2}\\hfill & \\text{Distribute and use FOIL}.\\hfill \\\\ \\text{ }25{x}^{2}-10y=1\\hfill & \\text{Rearrange terms and set equal to 1}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
    \n

    Try It 4<\/h3>\nConvert the conic [latex]r=\\frac{2}{1+2\\text{ }\\cos \\text{ }\\theta }[\/latex] to rectangular form.\n\nSolution<\/a>\n\n<\/div>","rendered":"

    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.\n<\/p>\n

    \n

    How To: Given the focus, eccentricity, and directrix of a conic, determine the polar equation.
    \n<\/strong><\/h3>\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<\/div>\n
      \n

      Example 5: Finding the Polar Form of a Vertical Conic Given a Focus at the Origin and the Eccentricity and Directrix<\/h3>\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<\/div>\n

      \n

      Solution<\/h3>\n

      The directrix is [latex]y=-p[\/latex], so we know the trigonometric function in the denominator is sine.<\/p>\n

      Because [latex]y=-2,-2<0[\/latex], so we know there is a subtraction sign in the denominator. We use the standard form of\n\n\n

      [latex]r=\\frac{ep}{1-e\\text{ }\\sin \\text{ }\\theta }[\/latex]<\/div>\n

      and [latex]e=3[\/latex] and [latex]|-2|=2=p[\/latex].<\/p>\n

      Therefore,<\/p>\n

      [latex]\\begin{array}{l}\\hfill \\\\ \\begin{array}{l}r=\\frac{\\left(3\\right)\\left(2\\right)}{1 - 3\\text{ }\\sin \\text{ }\\theta }\\hfill \\\\ r=\\frac{6}{1 - 3\\text{ }\\sin \\text{ }\\theta }\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
      \n

      Example 6: Finding the Polar Form of a Horizontal Conic Given a Focus at the Origin and the Eccentricity and Directrix<\/h3>\n

      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].<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

      Because 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<\/p>\n

      [latex]r=\\frac{ep}{1+e\\text{ }\\cos \\text{ }\\theta }[\/latex]<\/div>\n

      and [latex]e=\\frac{3}{5}[\/latex] and [latex]|4|=4=p[\/latex].<\/p>\n

      Therefore,<\/p>\n

      [latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ r=\\frac{\\left(\\frac{3}{5}\\right)\\left(4\\right)}{1+\\frac{3}{5}\\cos \\theta }\\hfill \\end{array}\\hfill \\\\ r=\\frac{\\frac{12}{5}}{1+\\frac{3}{5}\\cos \\theta }\\hfill \\\\ r=\\frac{\\frac{12}{5}}{1\\left(\\frac{5}{5}\\right)+\\frac{3}{5}\\cos \\theta }\\hfill \\\\ r=\\frac{\\frac{12}{5}}{\\frac{5}{5}+\\frac{3}{5}\\cos \\theta }\\hfill \\\\ r=\\frac{12}{5}\\cdot \\frac{5}{5+3\\cos \\theta }\\hfill \\\\ r=\\frac{12}{5+3\\cos \\theta }\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
      \n

      Try It 3<\/h3>\n

      Find the polar form of the conic given a focus at the origin, [latex]e=1[\/latex], and directrix [latex]x=-1[\/latex].<\/p>\n

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

      \n

      Example 5: Converting a Conic in Polar Form to Rectangular Form<\/h3>\n

      Convert the conic [latex]r=\\frac{1}{5 - 5\\sin \\theta }[\/latex] to rectangular form.<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

      We 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].<\/p>\n

      [latex]\\begin{array}{ll}\\text{ }r=\\frac{1}{5 - 5\\sin \\theta }\\hfill & \\hfill \\\\ r\\cdot \\left(5 - 5\\sin \\theta \\right)=\\frac{1}{5 - 5\\sin \\theta }\\cdot \\left(5 - 5\\sin \\theta \\right)\\hfill & \\text{Eliminate the fraction}.\\hfill \\\\ \\text{ }5r - 5r\\sin \\theta =1\\hfill & \\text{Distribute}.\\hfill \\\\ \\text{ }5r=1+5r\\sin \\theta \\hfill & \\text{Isolate }5r.\\hfill \\\\ \\text{ }25{r}^{2}={\\left(1+5r\\sin \\theta \\right)}^{2}\\hfill & \\text{Square both sides}.\\hfill \\\\ \\text{ }25\\left({x}^{2}+{y}^{2}\\right)={\\left(1+5y\\right)}^{2}\\hfill & \\text{Substitute }r=\\sqrt{{x}^{2}+{y}^{2}}\\text{ and }y=r\\sin \\theta .\\hfill \\\\ \\text{ }25{x}^{2}+25{y}^{2}=1+10y+25{y}^{2}\\hfill & \\text{Distribute and use FOIL}.\\hfill \\\\ \\text{ }25{x}^{2}-10y=1\\hfill & \\text{Rearrange terms and set equal to 1}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
      \n

      Try It 4<\/h3>\n

      Convert the conic [latex]r=\\frac{2}{1+2\\text{ }\\cos \\text{ }\\theta }[\/latex] to rectangular form.<\/p>\n

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

      \n\t\t\t

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