{"id":2541,"date":"2025-08-13T17:54:20","date_gmt":"2025-08-13T17:54:20","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2541"},"modified":"2025-10-22T23:06:58","modified_gmt":"2025-10-22T23:06:58","slug":"hyperbolas-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/hyperbolas-learn-it-3\/","title":{"raw":"Hyperbolas: Learn It 3","rendered":"Hyperbolas: Learn It 3"},"content":{"raw":"

Hyperbolas Not Centered at the Origin<\/h2>\r\nLike the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translated [latex]h[\/latex] units horizontally and [latex]k[\/latex] units vertically, the center of the hyperbola<\/strong> will be [latex]\\left(h,k\\right)[\/latex]. This translation results in the standard form of the equation we saw previously, with [latex]x[\/latex] replaced by [latex]\\left(x-h\\right)[\/latex] and [latex]y[\/latex] replaced by [latex]\\left(y-k\\right)[\/latex].\r\n\r\n
\r\n

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

[latex]\\frac{{\\left(x-h\\right)}^{2}}{{a}^{2}}-\\frac{{\\left(y-k\\right)}^{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(h\\pm a,k\\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(h,k\\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(h\\pm c,k\\right)[\/latex]<\/li>\r\n<\/ul>\r\nThe asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle is [latex]2a[\/latex] and its width is [latex]2b[\/latex]. The slopes of the diagonals are [latex]\\pm \\frac{b}{a}[\/latex], and each diagonal passes through the center [latex]\\left(h,k\\right)[\/latex]. Using the point-slope formula<\/strong>, it is simple to show that the equations of the asymptotes are [latex]y=\\pm \\frac{b}{a}\\left(x-h\\right)+k[\/latex].\r\n\r\nThe standard form of the equation of a hyperbola with center [latex]\\left(h,k\\right)[\/latex] and transverse axis parallel to the y<\/em>-axis is\r\n

    [latex]\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}-\\frac{{\\left(x-h\\right)}^{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(h,k\\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(h\\pm b,k\\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(h,k\\pm c\\right)[\/latex]<\/li>\r\n<\/ul>\r\nUsing the reasoning above, the equations of the asymptotes are [latex]y=\\pm \\frac{a}{b}\\left(x-h\\right)+k[\/latex].\r\n\r\n\"\"\r\n\r\n<\/section>Like hyperbolas centered at the origin, hyperbolas centered at a point [latex]\\left(h,k\\right)[\/latex] have vertices, co-vertices, and foci that are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[\/latex]. We can use this relationship along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and foci are given.\r\n\r\n
      How To: Given the vertices and foci of a hyperbola centered at [latex]\\left(h,k\\right)[\/latex], write its equation in standard form.<\/strong>\r\n
        \r\n \t
      1. Determine whether the transverse axis is parallel to the x<\/em>- or y<\/em>-axis.\r\n
          \r\n \t
        1. If the y<\/em>-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the x<\/em>-axis. Use the standard form [latex]\\frac{{\\left(x-h\\right)}^{2}}{{a}^{2}}-\\frac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex].<\/li>\r\n \t
        2. If the x<\/em>-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the y<\/em>-axis. Use the standard form [latex]\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}-\\frac{{\\left(x-h\\right)}^{2}}{{b}^{2}}=1[\/latex].<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
        3. Identify the center of the hyperbola, [latex]\\left(h,k\\right)[\/latex], using the midpoint formula and the given coordinates for the vertices.<\/li>\r\n \t
        4. Find [latex]{a}^{2}[\/latex] by solving for the length of the transverse axis, [latex]2a[\/latex] , which is the distance between the given vertices.<\/li>\r\n \t
        5. Find [latex]{c}^{2}[\/latex] using [latex]h[\/latex] and [latex]k[\/latex] found in Step 2 along with the given coordinates for the foci.<\/li>\r\n \t
        6. Solve for [latex]{b}^{2}[\/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}[\/latex].<\/li>\r\n \t
        7. Substitute the values for [latex]h,k,{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>
          \r\n

          What is the standard form equation of the <\/span>hyperbola<\/strong> that has vertices at [latex]\\left(0,-2\\right)[\/latex] and [latex]\\left(6,-2\\right)[\/latex] and foci at [latex]\\left(-2,-2\\right)[\/latex] and [latex]\\left(8,-2\\right)?[\/latex]<\/span><\/h3>\r\n[reveal-answer q=\"833629\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"833629\"]\r\n\r\nThe y<\/em>-coordinates of the vertices and foci are the same, so the transverse axis is parallel to the x<\/em>-axis. Thus, the equation of the hyperbola will have the form\r\n

          [latex]\\frac{{\\left(x-h\\right)}^{2}}{{a}^{2}}-\\frac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex]<\/p>\r\nFirst, we identify the center, [latex]\\left(h,k\\right)[\/latex]. The center is halfway between the vertices [latex]\\left(0,-2\\right)[\/latex] and [latex]\\left(6,-2\\right)[\/latex]. Applying the midpoint formula, we have\r\n

          [latex]\\left(h,k\\right)=\\left(\\frac{0+6}{2},\\frac{-2+\\left(-2\\right)}{2}\\right)=\\left(3,-2\\right)[\/latex]<\/p>\r\nNext, we find [latex]{a}^{2}[\/latex]. The length of the transverse axis, [latex]2a[\/latex], is bounded by the vertices. So, we can find [latex]{a}^{2}[\/latex] by finding the distance between the x<\/em>-coordinates of the vertices.\r\n

          [latex]\\begin{gathered}2a=|0 - 6| \\\\ 2a=6 \\\\ a=3 \\\\ {a}^{2}=9 \\end{gathered}[\/latex]<\/p>\r\nNow we need to find [latex]{c}^{2}[\/latex]. The coordinates of the foci are [latex]\\left(h\\pm c,k\\right)[\/latex]. So [latex]\\left(h-c,k\\right)=\\left(-2,-2\\right)[\/latex] and [latex]\\left(h+c,k\\right)=\\left(8,-2\\right)[\/latex]. We can use the x<\/em>-coordinate from either of these points to solve for [latex]c[\/latex]. Using the point [latex]\\left(8,-2\\right)[\/latex], and substituting [latex]h=3[\/latex],\r\n

          [latex]\\begin{gathered}h+c=8 \\\\ 3+c=8 \\\\ c=5 \\\\ {c}^{2}=25 \\end{gathered}[\/latex]<\/p>\r\nNext, solve for [latex]{b}^{2}[\/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}:[\/latex]\r\n

          [latex]\\begin{align}{b}^{2}&={c}^{2}-{a}^{2} \\\\ &=25 - 9 \\\\ &=16 \\end{align}[\/latex]<\/p>\r\nFinally, substitute the values found for [latex]h,k,{a}^{2}[\/latex], and [latex]{b}^{2}[\/latex] into the standard form of the equation.\r\n

          [latex]\\frac{{\\left(x - 3\\right)}^{2}}{9}-\\frac{{\\left(y+2\\right)}^{2}}{16}=1[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

          What is the standard form equation of the hyperbola that has vertices [latex]\\left(1,-2\\right)[\/latex] and [latex]\\left(1,\\text{8}\\right)[\/latex] and foci [latex]\\left(1,-10\\right)[\/latex] and [latex]\\left(1,16\\right)?[\/latex][reveal-answer q=\"56207\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"56207\"][latex]\\frac{{\\left(y - 3\\right)}^{2}}{25}+\\frac{{\\left(x - 1\\right)}^{2}}{144}=1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
          [ohm_question hide_question_numbers=1]174002[\/ohm_question]<\/section>","rendered":"

          Hyperbolas Not Centered at the Origin<\/h2>\n

          Like the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translated [latex]h[\/latex] units horizontally and [latex]k[\/latex] units vertically, the center of the hyperbola<\/strong> will be [latex]\\left(h,k\\right)[\/latex]. This translation results in the standard form of the equation we saw previously, with [latex]x[\/latex] replaced by [latex]\\left(x-h\\right)[\/latex] and [latex]y[\/latex] replaced by [latex]\\left(y-k\\right)[\/latex].<\/p>\n

          \n

          hyperbolas centered at (h,k)<\/h3>\n

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

          [latex]\\frac{{\\left(x-h\\right)}^{2}}{{a}^{2}}-\\frac{{\\left(y-k\\right)}^{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(h\\pm a,k\\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(h,k\\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(h\\pm c,k\\right)[\/latex]<\/li>\n<\/ul>\n

            The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle is [latex]2a[\/latex] and its width is [latex]2b[\/latex]. The slopes of the diagonals are [latex]\\pm \\frac{b}{a}[\/latex], and each diagonal passes through the center [latex]\\left(h,k\\right)[\/latex]. Using the point-slope formula<\/strong>, it is simple to show that the equations of the asymptotes are [latex]y=\\pm \\frac{b}{a}\\left(x-h\\right)+k[\/latex].<\/p>\n

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

            [latex]\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}-\\frac{{\\left(x-h\\right)}^{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(h,k\\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(h\\pm b,k\\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(h,k\\pm c\\right)[\/latex]<\/li>\n<\/ul>\n

              Using the reasoning above, the equations of the asymptotes are [latex]y=\\pm \\frac{a}{b}\\left(x-h\\right)+k[\/latex].<\/p>\n

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

              Like hyperbolas centered at the origin, hyperbolas centered at a point [latex]\\left(h,k\\right)[\/latex] have vertices, co-vertices, and foci that are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[\/latex]. We can use this relationship along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and foci are given.<\/p>\n

              How To: Given the vertices and foci of a hyperbola centered at [latex]\\left(h,k\\right)[\/latex], write its equation in standard form.<\/strong><\/p>\n
                \n
              1. Determine whether the transverse axis is parallel to the x<\/em>– or y<\/em>-axis.\n
                  \n
                1. If the y<\/em>-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the x<\/em>-axis. Use the standard form [latex]\\frac{{\\left(x-h\\right)}^{2}}{{a}^{2}}-\\frac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex].<\/li>\n
                2. If the x<\/em>-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the y<\/em>-axis. Use the standard form [latex]\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}-\\frac{{\\left(x-h\\right)}^{2}}{{b}^{2}}=1[\/latex].<\/li>\n<\/ol>\n<\/li>\n
                3. Identify the center of the hyperbola, [latex]\\left(h,k\\right)[\/latex], using the midpoint formula and the given coordinates for the vertices.<\/li>\n
                4. Find [latex]{a}^{2}[\/latex] by solving for the length of the transverse axis, [latex]2a[\/latex] , which is the distance between the given vertices.<\/li>\n
                5. Find [latex]{c}^{2}[\/latex] using [latex]h[\/latex] and [latex]k[\/latex] found in Step 2 along with the given coordinates for the foci.<\/li>\n
                6. Solve for [latex]{b}^{2}[\/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}[\/latex].<\/li>\n
                7. Substitute the values for [latex]h,k,{a}^{2}[\/latex], and [latex]{b}^{2}[\/latex] into the standard form of the equation determined in Step 1.<\/li>\n<\/ol>\n<\/section>\n
                  \n

                  What is the standard form equation of the <\/span>hyperbola<\/strong> that has vertices at [latex]\\left(0,-2\\right)[\/latex] and [latex]\\left(6,-2\\right)[\/latex] and foci at [latex]\\left(-2,-2\\right)[\/latex] and [latex]\\left(8,-2\\right)?[\/latex]<\/span><\/h3>\n