{"id":1043,"date":"2025-06-20T17:30:23","date_gmt":"2025-06-20T17:30:23","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=1043"},"modified":"2025-09-10T18:46:29","modified_gmt":"2025-09-10T18:46:29","slug":"conic-sections-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/conic-sections-learn-it-4\/","title":{"raw":"Conic Sections: Learn It 4","rendered":"Conic Sections: Learn It 4"},"content":{"raw":"

Hyperbolas<\/h2>\r\nLike ellipses, hyperbolas have two foci and two directrices. However, hyperbolas also feature two asymptotes\u2014lines the curve approaches but never touches.\r\n\r\n
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hyperbola<\/h3>\r\nA hyperbola<\/strong> is the set of all points where the difference between their distances from two fixed points (the foci) is constant.\r\n\r\n<\/section>
Notice the key difference from an ellipse: hyperbolas use the difference of distances, while ellipses use the sum of distances.<\/section>\r\n

Figure 12 shows the essential parts of a hyperbola that you'll need to understand for working with hyperbola equations.<\/p>\r\n\r\n

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  • The Transverse Axis<\/strong>: Also called the major axis, this passes through both foci and vertices. The hyperbola intersects this axis at two points called vertices.<\/li>\r\n \t
  • The Conjugate Axis<\/strong>: Also called the minor axis, this is perpendicular to the transverse axis and passes through the center. Unlike ellipses, the hyperbola doesn't intersect the conjugate axis.<\/li>\r\n \t
  • The Asymptotes<\/strong>: Two diagonal lines that the hyperbola approaches but never touches. These lines help determine the hyperbola's shape and orientation.<\/li>\r\n<\/ul>\r\n
    <\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"642\"]\"A Figure 12. A typical hyperbola in which the difference of the distances from any point on the ellipse to the foci is constant. The transverse axis is also called the major axis, and the conjugate axis is also called the minor axis.[\/caption]<\/figure>\r\n
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    The derivation of a hyperbola's equation follows the same process as for an ellipse, with one important difference in the definition. For a hyperbola, we consider the difference between two distances rather than the sum. Since this difference could be positive or negative, we use absolute value:<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    [latex]|d\\left(P,{F}_{1}\\right)-d\\left(P,{F}_{2}\\right)|=\\text{constant}[\/latex]<\/p>\r\nTo simplify the derivation, let's assume point [latex]P[\/latex] is on the right branch of the hyperbola, which allows us to drop the absolute value bars. The vertex of the right branch has coordinates [latex]\\left(a,0\\right)[\/latex], giving us:\r\n

    [latex]d\\left(P,{F}_{1}\\right)-d\\left(P,{F}_{2}\\right)=\\left(c+a\\right)-\\left(c-a\\right)=2a[\/latex].<\/div>\r\n

    For any point [latex]P(x,y)[\/latex] on the hyperbola:<\/p>\r\n\r\n

    [latex]\\begin{array}{ccc}\\hfill d\\left(P,{F}_{1}\\right)-d\\left(P,{F}_{2}\\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\nAdd the second radical to both sides and square:\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

    Dividing both sides by [latex]a^2 - c^2[\/latex]:<\/p>\r\n\r\n

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

    For hyperbolas, [latex]c > a[\/latex] (unlike ellipses where [latex]a > c[\/latex]). This makes [latex]a^2 - c^2[\/latex] negative. We define [latex]b^2 = c^2 - a^2[\/latex], so [latex]a^2 - c^2 = -b^2[\/latex].<\/p>\r\n

    Substituting this relationship:<\/p>\r\n\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 the standard form of a hyperbola.<\/p>\r\n\r\n

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    standard form of a hyperbola<\/h3>\r\nThe standard form depends on whether the transverse axis (major axis) is horizontal or vertical.\r\n\r\n \r\n\r\nHyperbola with Horizontal Transverse Axis<\/strong>: Center at [latex]\\left(h,k\\right)[\/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
    • Asymptotes: [latex]y = k \\pm \\frac{b}{a}(x-h)[\/latex]<\/li>\r\n \t
    • Directrices: [latex]x = h \\pm \\frac{a^2}{c}[\/latex]<\/li>\r\n<\/ul>\r\n

      Hyperbola with Vertical Transverse Axis<\/strong>: Center at [latex]\\left(h,k\\right)[\/latex]:<\/p>\r\n

      [latex]\\frac{(y-k)^2}{a^2}-\\frac{(x-h)^2}{b^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
      • Asymptotes: [latex]y = k \\pm \\frac{a}{b}(x-h)[\/latex]<\/li>\r\n \t
      • Directrices: [latex]y = k \\pm \\frac{a^2}{c}[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>A hyperbola is called horizontal<\/strong> when its transverse axis is horizontal, and vertical<\/strong> when its transverse axis is vertical.\r\n\r\n
        For hyperbolas, [latex]c^2 = a^2 + b^2[\/latex]. This differs from ellipses, where [latex]c^2 = a^2 - b^2[\/latex]. The relationship [latex]c > a[\/latex] always holds for hyperbolas.<\/section>The general form<\/strong> of a hyperbola is [latex]Ax^2 + By^2 + Cx + Dy + E = 0[\/latex], where [latex]A[\/latex] and [latex]B[\/latex] have opposite signs. 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}-16{y}^{2}+36x+32y - 124=0[\/latex] into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n

        \r\n

        First add [latex]124[\/latex] to both sides of the equation:<\/p>\r\n\r\n

        [latex]9{x}^{2}-16{y}^{2}+36x+32y=124[\/latex].<\/div>\r\n \r\n

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

        [latex]\\begin{array}{ccc}\\hfill \\left(9{x}^{2}+36x\\right)-\\left(16{y}^{2}-32y\\right)& =\\hfill & 124\\hfill \\\\ \\hfill 9\\left({x}^{2}+4x\\right)-16\\left({y}^{2}-2y\\right)& =\\hfill & 124.\\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 [latex]x[\/latex] 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{-2}{2}\\right)}^{2}=1[\/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 subtracting 16 from the second set of parentheses. Therefore the equation becomes<\/p>\r\n\r\n

        [latex]\\begin{array}{}\\\\ 9\\left({x}^{2}+4x+4\\right)-16\\left({y}^{2}-2y+1\\right)=124+36 - 16\\hfill \\\\ 9\\left({x}^{2}+4x+4\\right)-16\\left({y}^{2}-2y+1\\right)=144.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

        Next factor both sets of parentheses and divide by 144:<\/p>\r\n\r\n

        [latex]\\begin{array}{}\\\\ \\hfill 9{\\left(x+2\\right)}^{2}-16{\\left(y - 1\\right)}^{2}& =\\hfill & 144\\hfill \\\\ \\hfill \\frac{9{\\left(x+2\\right)}^{2}}{144}-\\frac{16{\\left(y - 1\\right)}^{2}}{144}& =\\hfill & 1\\hfill \\\\ \\hfill \\frac{{\\left(x+2\\right)}^{2}}{16}-\\frac{{\\left(y - 1\\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 gives [latex]h=-2[\/latex], [latex]k=1[\/latex], [latex]a=4[\/latex], and [latex]b=3[\/latex]. This is a horizontal hyperbola with center at [latex]\\left(-2,1\\right)[\/latex] and asymptotes given by the equations [latex]y=1\\pm \\frac{3}{4}\\left(x+2\\right)[\/latex]. The graph of this hyperbola appears in the following figure.<\/p>\r\n\r\n

        <\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"567\"]\"A Figure 13. Graph of the hyperbola 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.