{"id":1042,"date":"2025-06-20T17:30:20","date_gmt":"2025-06-20T17:30:20","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=1042"},"modified":"2025-08-27T15:26:56","modified_gmt":"2025-08-27T15:26:56","slug":"conic-sections-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/conic-sections-learn-it-3\/","title":{"raw":"Conic Sections: Learn It 3","rendered":"Conic Sections: Learn It 3"},"content":{"raw":"

Ellipses<\/h2>\r\nAn ellipse has two foci (plural of focus) and two directrices (plural of directrix). We'll examine the directrices in detail later in this section.\r\n\r\n
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ellipse<\/h3>\r\nAn ellipse<\/strong> is the set of all points for which the sum of their distances from two fixed points (the foci) is constant.\r\n\r\n<\/section>\r\n
\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"567\"]\"An Figure 8. A typical ellipse in which the sum of the distances from any point on the ellipse to the foci is constant.[\/caption]<\/figure>\r\n

Figure 8 shows a typical ellipse with its essential components labeled. Understanding these parts will help you work with ellipse equations and solve related problems. Let's examine each component and its coordinates in this standard position.<\/p>\r\n\r\n

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  • The Foci<\/strong>: Points [latex]F[\/latex] and [latex]F'[\/latex] are both the same distance [latex]c[\/latex] from the origin. Their coordinates are [latex]F(c,0)[\/latex] and [latex]F'(-c,0)[\/latex].<\/li>\r\n \t
  • The Major Axis<\/strong>: The longest distance across the ellipse. Points [latex]P[\/latex] and [latex]P'[\/latex] mark the ends of the major axis with coordinates [latex]\\left(a,0\\right)[\/latex] and \u00a0[latex]\\left(-a,0\\right)[\/latex]. The major axis length is [latex]2a[\/latex].<\/li>\r\n \t
  • The Vertices<\/strong>: Points [latex]P[\/latex] and [latex]P'[\/latex] are called the vertices of the ellipse\u2014the endpoints of the major axis.<\/li>\r\n \t
  • The Minor Axis<\/strong>: The shortest distance across the ellipse, perpendicular to the major axis. Points [latex]Q[\/latex] and [latex]Q'[\/latex] mark its ends with coordinates [latex]\\left(0,b\\right)[\/latex] and [latex]\\left(0,-b\\right)[\/latex].<\/li>\r\n<\/ul>\r\n
    The major axis can be horizontal or vertical\u2014whichever is longer. The minor axis is always perpendicular to the major axis and represents the shorter dimension.<\/section>Now we'll use these components to derive the standard form equation of an ellipse. According to the ellipse definition, the sum of distances from any point on the ellipse to the two foci remains constant. Let's use point [latex]P[\/latex] to find this constant sum.\r\n

    Since [latex]P[\/latex] has coordinates [latex]\\left(a,0\\right)[\/latex], the sum of the distances is<\/p>\r\n\r\n

    [latex]d\\left(P,F\\right)+d\\left(P,{F}^{\\prime }\\right)=\\left(a-c\\right)+\\left(a+c\\right)=2a[\/latex].<\/div>\r\nThis means for any arbitrary point [latex]A(x,y)[\/latex] on the ellipse, the sum also equals [latex]2a[\/latex]. Using the distance formula:\r\n
    [latex]\\begin{array}{ccc}\\hfill d\\left(A,F\\right)+d\\left(A,{F}^{\\prime }\\right)& =\\hfill & 2a\\hfill \\\\ \\hfill \\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}+\\sqrt{{\\left(x+c\\right)}^{2}+{y}^{2}}& =\\hfill & 2a.\\hfill \\end{array}[\/latex]<\/div>\r\n

    To solve for the ellipse equation, we'll eliminate the radicals through careful algebra. Subtract the second radical from both sides and square:<\/p>\r\n\r\n

    [latex]\\begin{array}{ccc}\\hfill \\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}& =\\hfill & 2a-\\sqrt{{\\left(x+c\\right)}^{2}+{y}^{2}}\\hfill \\\\ \\hfill {\\left(x-c\\right)}^{2}+{y}^{2}& =\\hfill & 4{a}^{2}-4a\\sqrt{{\\left(x+c\\right)}^{2}+{y}^{2}}+{\\left(x+c\\right)}^{2}+{y}^{2}\\hfill \\\\ \\hfill {x}^{2}-2cx+{c}^{2}+{y}^{2}& =\\hfill & 4{a}^{2}-4a\\sqrt{{\\left(x+c\\right)}^{2}+{y}^{2}}+{x}^{2}+2cx+{c}^{2}+{y}^{2}\\hfill \\\\ \\hfill -2cx& =\\hfill & 4{a}^{2}-4a\\sqrt{{\\left(x+c\\right)}^{2}+{y}^{2}}+2cx.\\hfill \\end{array}[\/latex]<\/div>\r\nIsolate the radical and square again:\r\n
    [latex]\\begin{array}{ccc}\\hfill -2cx& =\\hfill & 4{a}^{2}-4a\\sqrt{{\\left(x+c\\right)}^{2}+{y}^{2}}+2cx\\hfill \\\\ \\hfill 4a\\sqrt{{\\left(x+c\\right)}^{2}+{y}^{2}}& =\\hfill & 4{a}^{2}+4cx\\hfill \\\\ \\hfill \\sqrt{{\\left(x+c\\right)}^{2}+{y}^{2}}& =\\hfill & a+\\frac{cx}{a}\\hfill \\\\ \\hfill {\\left(x+c\\right)}^{2}+{y}^{2}& =\\hfill & {a}^{2}+2cx+\\frac{{c}^{2}{x}^{2}}{{a}^{2}}\\hfill \\\\ \\hfill {x}^{2}+2cx+{c}^{2}+{y}^{2}& =\\hfill & {a}^{2}+2cx+\\frac{{c}^{2}{x}^{2}}{{a}^{2}}\\hfill \\\\ \\hfill {x}^{2}+{c}^{2}+{y}^{2}& =\\hfill & {a}^{2}+\\frac{{c}^{2}{x}^{2}}{{a}^{2}}.\\hfill \\end{array}[\/latex]<\/div>\r\nRearrange the variables on the left-hand side of the equation and the constants on the right-hand side:\r\n
    [latex]\\begin{array}{}\\\\ \\hfill {x}^{2}-\\frac{{c}^{2}{x}^{2}}{{a}^{2}}+{y}^{2}& =\\hfill & {a}^{2}-{c}^{2}\\hfill \\\\ \\hfill \\frac{\\left({a}^{2}-{c}^{2}\\right){x}^{2}}{{a}^{2}}+{y}^{2}& =\\hfill & {a}^{2}-{c}^{2}.\\hfill \\end{array}[\/latex]<\/div>\r\n

    Divide both sides by [latex]{a}^{2}-{c}^{2}[\/latex]. This gives the equation:<\/p>\r\n\r\n

    [latex]\\frac{{x}^{2}}{{a}^{2}}+\\frac{{y}^{2}}{{a}^{2}-{c}^{2}}=1[\/latex].<\/div>\r\n
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    Looking back at Figure 8, each green line segment from point [latex]Q[\/latex] to a focus has length [latex]a[\/latex]. This happens because the sum of distances from [latex]Q[\/latex] to both foci equals [latex]2a[\/latex], and these segments have equal length. These segments form a right triangle with hypotenuse [latex]a[\/latex] and legs of length [latex]b[\/latex] and [latex]c[\/latex]. Using the Pythagorean theorem: [latex]a^2 = b^2 + c^2[\/latex], which gives us [latex]b^2 = a^2 - c^2[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    \r\n

    Substituting this relationship into our equation [latex]\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1[\/latex],we get:<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    [latex]\\frac{{x}^{2}}{{a}^{2}}+\\frac{{y}^{2}}{{b}^{2}}=1[\/latex].<\/div>\r\n \r\n

    When the center moves from the origin to point [latex]\\left(h,k\\right)[\/latex], we get standard form of an ellipse<\/strong>.<\/p>\r\n\r\n

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    standard form of an ellipse<\/strong><\/h3>\r\n

    The standard form depends on whether the major axis is horizontal or vertical.<\/p>\r\n \r\n\r\nEllipse with Horizontal Major Axis<\/strong>: Center at [latex]\\left(h,k\\right)[\/latex], horizontal major axis of length [latex]2a[\/latex], vertical minor axis of length [latex]2b[\/latex]:\r\n

    [latex]\\frac{(x-h)^2}{a^2}+\\frac{(y-k)^2}{b^2}=1[\/latex]<\/p>\r\n\r\n

      \r\n \t
    • Foci: [latex](h \\pm c, k)[\/latex] where [latex]c^2 = a^2 - b^2[\/latex]<\/li>\r\n \t
    • Directrices: [latex]x = h \\pm \\frac{a^2}{c}[\/latex]<\/li>\r\n<\/ul>\r\nEllipse with Vertical Major Axis<\/strong>: Center at [latex]\\left(h,k\\right)[\/latex], vertical major axis of length [latex]2a[\/latex], horizontal minor axis of length [latex]2b[\/latex]:\r\n

      [latex]\\frac{(x-h)^2}{b^2}+\\frac{(y-k)^2}{a^2}=1[\/latex]<\/p>\r\n\r\n

        \r\n \t
      • Foci: [latex](h, k \\pm c)[\/latex] where [latex]c^2 = a^2 - b^2[\/latex]<\/li>\r\n \t
      • Directrices: [latex]y = k \\pm \\frac{a^2}{c}[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>An ellipse is called horizontal<\/strong> when its major axis is horizontal, and vertical<\/strong> when its major axis is vertical.\r\n\r\n
        In standard form, the larger denominator indicates the major axis direction. If [latex]a^2[\/latex] is under the [latex]x[\/latex]-term, the major axis is horizontal. If [latex]a^2[\/latex] is under the [latex]y[\/latex]-term, the major axis is vertical.<\/section>The general form<\/strong> of an ellipse is [latex]Ax^2 + By^2 + Cx + Dy + E = 0[\/latex], where [latex]A[\/latex] and [latex]B[\/latex] are both positive or both negative. To convert from general to standard form, use completing the square.\r\n\r\n
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        Put the equation [latex]9{x}^{2}+4{y}^{2}-36x+24y+36=0[\/latex] into standard form and graph the resulting ellipse.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n

        \r\n

        First subtract 36 from both sides of the equation:<\/p>\r\n\r\n

        [latex]9{x}^{2}+4{y}^{2}-36x+24y=-36[\/latex].<\/div>\r\n \r\n

        Next group the x<\/em> terms together and the y<\/em> terms together, and factor out the common factor:<\/p>\r\n\r\n

        [latex]\\begin{array}{ccc}\\hfill \\left(9{x}^{2}-36x\\right)+\\left(4{y}^{2}+24y\\right)& =\\hfill & -36\\hfill \\\\ \\hfill 9\\left({x}^{2}-4x\\right)+4\\left({y}^{2}+6y\\right)& =\\hfill & -36.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

        We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of x<\/em> and square it. This gives [latex]{\\left(\\frac{-4}{2}\\right)}^{2}=4[\/latex]. In the second set of parentheses, take half the coefficient of y<\/em> and square it. This gives [latex]{\\left(\\frac{6}{2}\\right)}^{2}=9[\/latex]. Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are adding 36 to the second set as well. Therefore the equation becomes<\/p>\r\n\r\n

        [latex]\\begin{array}{}\\\\ 9\\left({x}^{2}-4x+4\\right)+4\\left({y}^{2}+6y+9\\right)=-36+36+36\\hfill \\\\ 9\\left({x}^{2}-4x+4\\right)+4\\left({y}^{2}+6y+9\\right)=36.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

        Now factor both sets of parentheses and divide by 36:<\/p>\r\n\r\n

        [latex]\\begin{array}{}\\\\ \\hfill 9{\\left(x - 2\\right)}^{2}+4{\\left(y+3\\right)}^{2}& =\\hfill & 36\\hfill \\\\ \\hfill \\frac{9{\\left(x - 2\\right)}^{2}}{36}+\\frac{4{\\left(y+3\\right)}^{2}}{36}& =\\hfill & 1\\hfill \\\\ \\hfill \\frac{{\\left(x - 2\\right)}^{2}}{4}+\\frac{{\\left(y+3\\right)}^{2}}{9}& =\\hfill & 1.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

        The equation is now in standard form. Comparing this to the theorem equation gives [latex]h=2[\/latex], [latex]k=-3[\/latex], [latex]a=3[\/latex], and [latex]b=2[\/latex]. This is a vertical ellipse with center at [latex]\\left(2,-3\\right)[\/latex], major axis 6, and minor axis 4. The graph of this ellipse appears as follows.<\/p>\r\n\r\n

        <\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"356\"]\"An Figure 9. The ellipse 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.