{"id":2540,"date":"2025-08-13T17:54:12","date_gmt":"2025-08-13T17:54:12","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2540"},"modified":"2025-10-22T23:06:06","modified_gmt":"2025-10-22T23:06:06","slug":"hyperbolas-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/hyperbolas-learn-it-2\/","title":{"raw":"Hyperbolas: Learn It 2","rendered":"Hyperbolas: Learn It 2"},"content":{"raw":"
\r\n

hyperbola centered at (0,0)<\/h3>\r\nThe standard form of the equation of a hyperbola with center [latex]\\left(0,0\\right)[\/latex] and transverse axis on the x<\/em>-axis is\r\n

[latex]\\frac{{x}^{2}}{{a}^{2}}-\\frac{{y}^{2}}{{b}^{2}}=1[\/latex]<\/p>\r\nwhere\r\n

    \r\n \t
  • the length of the transverse axis is [latex]2a[\/latex]<\/li>\r\n \t
  • the coordinates of the vertices are [latex]\\left(\\pm a,0\\right)[\/latex]<\/li>\r\n \t
  • the length of the conjugate axis is [latex]2b[\/latex]<\/li>\r\n \t
  • the coordinates of the co-vertices are [latex]\\left(0,\\pm b\\right)[\/latex]<\/li>\r\n \t
  • the distance between the foci is [latex]2c[\/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[\/latex]<\/li>\r\n \t
  • the coordinates of the foci are [latex]\\left(\\pm c,0\\right)[\/latex]<\/li>\r\n \t
  • the equations of the asymptotes are [latex]y=\\pm \\frac{b}{a}x[\/latex]<\/li>\r\n<\/ul>\r\nThe standard form of the equation of a hyperbola with center [latex]\\left(0,0\\right)[\/latex] and transverse axis on the y<\/em>-axis is\r\n

    [latex]\\frac{{y}^{2}}{{a}^{2}}-\\frac{{x}^{2}}{{b}^{2}}=1[\/latex]<\/p>\r\nwhere\r\n

      \r\n \t
    • the length of the transverse axis is [latex]2a[\/latex]<\/li>\r\n \t
    • the coordinates of the vertices are [latex]\\left(0,\\pm a\\right)[\/latex]<\/li>\r\n \t
    • the length of the conjugate axis is [latex]2b[\/latex]<\/li>\r\n \t
    • the coordinates of the co-vertices are [latex]\\left(\\pm b,0\\right)[\/latex]<\/li>\r\n \t
    • the distance between the foci is [latex]2c[\/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[\/latex]<\/li>\r\n \t
    • the coordinates of the foci are [latex]\\left(0,\\pm c\\right)[\/latex]<\/li>\r\n \t
    • the equations of the asymptotes are [latex]y=\\pm \\frac{a}{b}x[\/latex]<\/li>\r\n<\/ul>\r\nNote that the vertices, co-vertices, and foci are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[\/latex]. When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci.\r\n\r\n\"\"\r\n\r\n<\/section>
      How To: Given the equation of a hyperbola in standard form, locate its vertices and foci.<\/strong>\r\n
        \r\n \t
      1. \r\n
          \r\n \t
        1. Determine whether the transverse axis lies on the x<\/em>- or y<\/em>-axis. Notice that [latex]{a}^{2}[\/latex] is always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n

          a. If the equation has the form [latex]\\frac{{x}^{2}}{{a}^{2}}-\\frac{{y}^{2}}{{b}^{2}}=1[\/latex], then the transverse axis lies on the x<\/em>-axis. The vertices are located at [latex]\\left(\\pm a,0\\right)[\/latex], and the foci are located at [latex]\\left(\\pm c,0\\right)[\/latex].<\/p>\r\n

          b. If the equation has the form [latex]\\frac{{y}^{2}}{{a}^{2}}-\\frac{{x}^{2}}{{b}^{2}}=1[\/latex], then the transverse axis lies on the y<\/em>-axis. The vertices are located at [latex]\\left(0,\\pm a\\right)[\/latex], and the foci are located at [latex]\\left(0,\\pm c\\right)[\/latex].<\/p>\r\n\r\n

            \r\n \t
          1. Solve for [latex]a[\/latex] using the equation [latex]a=\\sqrt{{a}^{2}}[\/latex].<\/li>\r\n \t
          2. Solve for [latex]c[\/latex] using the equation [latex]c=\\sqrt{{a}^{2}+{b}^{2}}[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>
            Identify the vertices and foci of the hyperbola<\/strong> with equation [latex]\\frac{{y}^{2}}{49}-\\frac{{x}^{2}}{32}=1[\/latex].[reveal-answer q=\"768741\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"768741\"]\r\n\r\nThe equation has the form [latex]\\frac{{y}^{2}}{{a}^{2}}-\\frac{{x}^{2}}{{b}^{2}}=1[\/latex], so the transverse axis lies on the y<\/em>-axis. The hyperbola is centered at the origin, so the vertices serve as the y<\/em>-intercepts of the graph. To find the vertices, set [latex]x=0[\/latex], and solve for [latex]y[\/latex].\r\n

            [latex]\\begin{align}&1=\\frac{{y}^{2}}{49}-\\frac{{x}^{2}}{32} \\\\ &1=\\frac{{y}^{2}}{49}-\\frac{{0}^{2}}{32} \\\\ &1=\\frac{{y}^{2}}{49} \\\\ &{y}^{2}=49 \\\\ &y=\\pm \\sqrt{49}=\\pm 7 \\end{align}[\/latex]<\/p>\r\nThe foci are located at [latex]\\left(0,\\pm c\\right)[\/latex]. Solving for [latex]c[\/latex],\r\n

            [latex]c=\\sqrt{{a}^{2}+{b}^{2}}=\\sqrt{49+32}=\\sqrt{81}=9[\/latex]<\/p>\r\nTherefore, the vertices are located at [latex]\\left(0,\\pm 7\\right)[\/latex], and the foci are located at [latex]\\left(0,9\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

            Identify the vertices and foci of the hyperbola with equation [latex]\\frac{{x}^{2}}{9}-\\frac{{y}^{2}}{25}=1[\/latex].[reveal-answer q=\"303710\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"303710\"]Vertices: [latex]\\left(\\pm 3,0\\right)[\/latex]; Foci: [latex]\\left(\\pm \\sqrt{34},0\\right)[\/latex][\/hidden-answer]\r\n\r\n<\/section>
            [ohm_question hide_question_numbers=1]173983[\/ohm_question]<\/section>\r\n

            Writing Equations of Hyperbolas in Standard Form<\/h3>\r\nJust as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverse and conjugate axes. Conversely, an equation for a hyperbola can be found given its key features. We begin by finding standard equations for hyperbolas centered at the origin. Then we will turn our attention to finding standard equations for hyperbolas centered at some point other than the origin.\r\n

            Hyperbolas Centered at the Origin<\/h3>\r\nReviewing the standard forms given for hyperbolas centered at [latex]\\left(0,0\\right)[\/latex], we see that the vertices, co-vertices, and foci are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[\/latex]. Note that this equation can also be rewritten as [latex]{b}^{2}={c}^{2}-{a}^{2}[\/latex]. This relationship is used to write the equation for a hyperbola when given the coordinates of its foci and vertices.\r\n\r\n
            How To: Given the vertices and foci of a hyperbola centered at [latex]\\left(0,\\text{0}\\right)[\/latex], write its equation in standard form.<\/strong>\r\n
              \r\n \t
            1. Determine whether the transverse axis lies on the x<\/em>- or y<\/em>-axis.\r\n
                \r\n \t
              1. If the given coordinates of the vertices and foci have the form [latex]\\left(\\pm a,0\\right)[\/latex] and [latex]\\left(\\pm c,0\\right)[\/latex], respectively, then the transverse axis is the x<\/em>-axis. Use the standard form [latex]\\frac{{x}^{2}}{{a}^{2}}-\\frac{{y}^{2}}{{b}^{2}}=1[\/latex].<\/li>\r\n \t
              2. If the given coordinates of the vertices and foci have the form [latex]\\left(0,\\pm a\\right)[\/latex] and [latex]\\left(0,\\pm c\\right)[\/latex], respectively, then the transverse axis is the y<\/em>-axis. Use the standard form [latex]\\frac{{y}^{2}}{{a}^{2}}-\\frac{{x}^{2}}{{b}^{2}}=1[\/latex].<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
              3. Find [latex]{b}^{2}[\/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}[\/latex].<\/li>\r\n \t
              4. Substitute the values for [latex]{a}^{2}[\/latex] and [latex]{b}^{2}[\/latex] into the standard form of the equation determined in Step 1.<\/li>\r\n<\/ol>\r\n<\/section>
                What is the standard form equation of the hyperbola<\/strong> that has vertices [latex]\\left(\\pm 6,0\\right)[\/latex] and foci [latex]\\left(\\pm 2\\sqrt{10},0\\right)?[\/latex][reveal-answer q=\"377345\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"377345\"]\r\n\r\nThe vertices and foci are on the x<\/em>-axis. Thus, the equation for the hyperbola will have the form [latex]\\frac{{x}^{2}}{{a}^{2}}-\\frac{{y}^{2}}{{b}^{2}}=1[\/latex].\r\n\r\nThe vertices are [latex]\\left(\\pm 6,0\\right)[\/latex], so [latex]a=6[\/latex] and [latex]{a}^{2}=36[\/latex].\r\n\r\nThe foci are [latex]\\left(\\pm 2\\sqrt{10},0\\right)[\/latex], so [latex]c=2\\sqrt{10}[\/latex] and [latex]{c}^{2}=40[\/latex].\r\n\r\nSolving for [latex]{b}^{2}[\/latex], we have\r\n

                [latex]\\begin{align}&{b}^{2}={c}^{2}-{a}^{2} \\\\ &{b}^{2}=40 - 36 && \\text{Substitute for }{c}^{2}\\text{ and }{a}^{2}. \\\\ &{b}^{2}=4 && \\text{Subtract}. \\end{align}[\/latex]<\/p>\r\nFinally, we substitute [latex]{a}^{2}=36[\/latex] and [latex]{b}^{2}=4[\/latex] into the standard form of the equation, [latex]\\frac{{x}^{2}}{{a}^{2}}-\\frac{{y}^{2}}{{b}^{2}}=1[\/latex]. The equation of the hyperbola is [latex]\\frac{{x}^{2}}{36}-\\frac{{y}^{2}}{4}=1[\/latex].\r\n\r\n\"\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

                What is the standard form equation of the hyperbola that has vertices [latex]\\left(0,\\pm 2\\right)[\/latex] and foci [latex]\\left(0,\\pm 2\\sqrt{5}\\right)?[\/latex][reveal-answer q=\"409322\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"409322\"][latex]\\frac{{y}^{2}}{4}-\\frac{{x}^{2}}{16}=1[\/latex][\/hidden-answer]\r\n\r\n<\/section>
                [ohm_question hide_question_numbers=1]146722[\/ohm_question]<\/section>","rendered":"
                \n

                hyperbola centered at (0,0)<\/h3>\n

                The standard form of the equation of a hyperbola with center [latex]\\left(0,0\\right)[\/latex] and transverse axis on the x<\/em>-axis is<\/p>\n

                [latex]\\frac{{x}^{2}}{{a}^{2}}-\\frac{{y}^{2}}{{b}^{2}}=1[\/latex]<\/p>\n

                where<\/p>\n

                  \n
                • the length of the transverse axis is [latex]2a[\/latex]<\/li>\n
                • the coordinates of the vertices are [latex]\\left(\\pm a,0\\right)[\/latex]<\/li>\n
                • the length of the conjugate axis is [latex]2b[\/latex]<\/li>\n
                • the coordinates of the co-vertices are [latex]\\left(0,\\pm b\\right)[\/latex]<\/li>\n
                • the distance between the foci is [latex]2c[\/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[\/latex]<\/li>\n
                • the coordinates of the foci are [latex]\\left(\\pm c,0\\right)[\/latex]<\/li>\n
                • the equations of the asymptotes are [latex]y=\\pm \\frac{b}{a}x[\/latex]<\/li>\n<\/ul>\n

                  The standard form of the equation of a hyperbola with center [latex]\\left(0,0\\right)[\/latex] and transverse axis on the y<\/em>-axis is<\/p>\n

                  [latex]\\frac{{y}^{2}}{{a}^{2}}-\\frac{{x}^{2}}{{b}^{2}}=1[\/latex]<\/p>\n

                  where<\/p>\n

                    \n
                  • the length of the transverse axis is [latex]2a[\/latex]<\/li>\n
                  • the coordinates of the vertices are [latex]\\left(0,\\pm a\\right)[\/latex]<\/li>\n
                  • the length of the conjugate axis is [latex]2b[\/latex]<\/li>\n
                  • the coordinates of the co-vertices are [latex]\\left(\\pm b,0\\right)[\/latex]<\/li>\n
                  • the distance between the foci is [latex]2c[\/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[\/latex]<\/li>\n
                  • the coordinates of the foci are [latex]\\left(0,\\pm c\\right)[\/latex]<\/li>\n
                  • the equations of the asymptotes are [latex]y=\\pm \\frac{a}{b}x[\/latex]<\/li>\n<\/ul>\n

                    Note that the vertices, co-vertices, and foci are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[\/latex]. When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci.<\/p>\n

                    \"\"<\/p>\n<\/section>\n

                    How To: Given the equation of a hyperbola in standard form, locate its vertices and foci.<\/strong><\/p>\n
                      \n
                    1. \n
                        \n
                      1. Determine whether the transverse axis lies on the x<\/em>– or y<\/em>-axis. Notice that [latex]{a}^{2}[\/latex] is always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                        a. If the equation has the form [latex]\\frac{{x}^{2}}{{a}^{2}}-\\frac{{y}^{2}}{{b}^{2}}=1[\/latex], then the transverse axis lies on the x<\/em>-axis. The vertices are located at [latex]\\left(\\pm a,0\\right)[\/latex], and the foci are located at [latex]\\left(\\pm c,0\\right)[\/latex].<\/p>\n

                        b. If the equation has the form [latex]\\frac{{y}^{2}}{{a}^{2}}-\\frac{{x}^{2}}{{b}^{2}}=1[\/latex], then the transverse axis lies on the y<\/em>-axis. The vertices are located at [latex]\\left(0,\\pm a\\right)[\/latex], and the foci are located at [latex]\\left(0,\\pm c\\right)[\/latex].<\/p>\n

                          \n
                        1. Solve for [latex]a[\/latex] using the equation [latex]a=\\sqrt{{a}^{2}}[\/latex].<\/li>\n
                        2. Solve for [latex]c[\/latex] using the equation [latex]c=\\sqrt{{a}^{2}+{b}^{2}}[\/latex].<\/li>\n<\/ol>\n<\/section>\n
                          Identify the vertices and foci of the hyperbola<\/strong> with equation [latex]\\frac{{y}^{2}}{49}-\\frac{{x}^{2}}{32}=1[\/latex].<\/p>\n