{"id":1031,"date":"2025-06-20T17:29:49","date_gmt":"2025-06-20T17:29:49","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=1031"},"modified":"2025-09-10T18:55:46","modified_gmt":"2025-09-10T18:55:46","slug":"area-and-arc-length-in-polar-coordinates-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/area-and-arc-length-in-polar-coordinates-learn-it-4\/","title":{"raw":"Conic Sections: Learn It 5","rendered":"Conic Sections: Learn It 5"},"content":{"raw":"

Eccentricity and Directrix<\/h2>\r\nAn alternative way to describe a conic section involves the directrices, the foci, and a new property called eccentricity. The value of a conic's eccentricity can uniquely identify which type of conic section it is.\r\n\r\n
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eccentricity<\/h3>\r\n

The eccentricity<\/strong> [latex]e[\/latex] of a conic section equals the distance from any point on the conic to its focus, divided by the perpendicular distance from that point to the nearest directrix. This ratio remains constant for all points on the conic.<\/p>\r\n \r\n

The eccentricity value determines the conic type:<\/p>\r\n\r\n

    \r\n \t
  • If [latex]e=1[\/latex], the conic is a parabola.<\/li>\r\n \t
  • If [latex]e<1[\/latex], it is an ellipse.<\/li>\r\n \t
  • If [latex]e>1[\/latex], it is a hyperbola.<\/li>\r\n \t
  • If [latex]e = 0[\/latex], it is a circle (special case of ellipse)<\/li>\r\n<\/ul>\r\n
      <\/ol>\r\n<\/section>The directrix<\/strong> of a conic section is the line that, together with the focus, helps define the conic. Parabolas have one focus and one directrix, while ellipses and hyperbolas have two foci and two associated directrices. The three conic sections with their directrices appear in the following figure.\r\n
      <\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"965\"]\"This Figure 16. The three conic sections with their foci and directrices.[\/caption]<\/figure>\r\nLet's verify these eccentricity relationships by examining each conic type.\r\n\r\nFor parabolas, by definition, the distance from any point on a parabola to the focus equals the distance to the directrix. Therefore, [latex]e = \\frac{\\text{distance to focus}}{\\text{distance to directrix}} = 1[\/latex].\r\n

      For ellipses, consider a horizontal ellipse with directrices at [latex]x = \\pm \\frac{a^2}{c}[\/latex]. The right vertex is at [latex]\\left(a,0\\right)[\/latex] and the right focus is [latex]\\left(c,0\\right)[\/latex]. Therefore the distance from the vertex to the focus is [latex]a-c[\/latex] and the distance from the vertex to the right directrix is [latex]\\frac{{a}^{2}}{c}-c[\/latex]. This gives the eccentricity as:<\/p>\r\n

      [latex]e = \\frac{a-c}{\\frac{a^2}{c}-a} = \\frac{c(a-c)}{a^2-ac} = \\frac{c(a-c)}{a(a-c)} = \\frac{c}{a}[\/latex]<\/p>\r\n

      Since [latex]c < a[\/latex] for ellipses, we have [latex]e < 1[\/latex].<\/p>\r\nFor hyperbolas, the directrices are also at [latex]x = \\pm \\frac{a^2}{c}[\/latex], and similar calculations give [latex]e = \\frac{c}{a}[\/latex]. However, for hyperbolas [latex]c > a[\/latex], so [latex]e > 1[\/latex].\r\n\r\n

      \r\n
      \r\n

      Determine the eccentricity of the ellipse described by the equation<\/p>\r\n\r\n

      [latex]\\frac{{\\left(x - 3\\right)}^{2}}{16}+\\frac{{\\left(y+2\\right)}^{2}}{25}=1[\/latex].<\/div>\r\n<\/div>\r\n[reveal-answer q=\"44558890\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558890\"]\r\n
      \r\n

      From the equation we see that [latex]a=5[\/latex] and [latex]b=4[\/latex]. The value of c<\/em> can be calculated using the equation [latex]{a}^{2}={b}^{2}+{c}^{2}[\/latex] for an ellipse. Substituting the values of a<\/em> and b<\/em> and solving for c<\/em> gives [latex]c=3[\/latex]. Therefore the eccentricity of the ellipse is [latex]e=\\frac{c}{a}=\\frac{3}{5}=0.6[\/latex].<\/p>\r\n\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.