{"id":2635,"date":"2024-08-08T23:05:47","date_gmt":"2024-08-08T23:05:47","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&p=2635"},"modified":"2025-08-15T16:58:56","modified_gmt":"2025-08-15T16:58:56","slug":"parabolas-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/parabolas-learn-it-4\/","title":{"raw":"Parabolas: Learn It 4","rendered":"Parabolas: Learn It 4"},"content":{"raw":"

Parabolas with Vertices Not at the Origin<\/h2>\r\nLike other graphs we\u2019ve worked with, the graph of a parabola can be translated. If a parabola is translated [latex]h[\/latex] units horizontally and [latex]k[\/latex] units vertically, the vertex 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\nTo graph parabolas with a vertex [latex]\\left(h,k\\right)[\/latex] other than the origin, we use the standard form [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex] for parabolas that have an axis of symmetry parallel to the x<\/em>-axis, and [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex] for parabolas that have an axis of symmetry parallel to the y<\/em>-axis. These standard forms are given below, along with their general graphs and key features.\r\n\r\n
\r\n

standard forms of parabolas with vertex [latex](h, k)[\/latex]<\/h3>\r\nThe table summarizes the standard features of parabolas with a vertex at a point [latex]\\left(h,k\\right)[\/latex].\r\n\r\n\r\n\r\n\r\n\r\n
Axis of Symmetry<\/strong><\/td>\r\nEquation<\/strong><\/td>\r\nFocus<\/strong><\/td>\r\nDirectrix<\/strong><\/td>\r\nEndpoints of focal diameter<\/strong><\/td>\r\n<\/tr>\r\n
[latex]y=k[\/latex]<\/td>\r\n[latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex]<\/td>\r\n[latex]\\left(h+p,\\text{ }k\\right)[\/latex]<\/td>\r\n[latex]x=h-p[\/latex]<\/td>\r\n[latex]\\left(h+p,\\text{ }k\\pm 2p\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]x=h[\/latex]<\/td>\r\n[latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex]<\/td>\r\n[latex]\\left(h,\\text{ }k+p\\right)[\/latex]<\/td>\r\n[latex]y=k-p[\/latex]<\/td>\r\n[latex]\\left(h\\pm 2p,\\text{ }k+p\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]\"\" (a) When [latex]p>0[\/latex], the parabola opens right. (b) When [latex]p<0[\/latex], the parabola opens left. (c) When [latex]p>0[\/latex], the parabola opens up. (d) When [latex]p<0[\/latex], the parabola opens down.[\/caption]<\/section>
How To: Given a standard form equation for a parabola centered at (h<\/em>, k<\/em>), sketch the graph.<\/strong>\r\n
    \r\n \t
  1. Determine which of the standard forms applies to the given equation: [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex] or [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex].<\/li>\r\n \t
  2. Use the standard form identified in Step 1 to determine the vertex, axis of symmetry, focus, equation of the directrix, and endpoints of the focal diameter.\r\n
      \r\n \t
    1. If the equation is in the form [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex], then:\r\n
        \r\n \t
      • use the given equation to identify [latex]h[\/latex] and [latex]k[\/latex] for the vertex, [latex]\\left(h,k\\right)[\/latex]<\/li>\r\n \t
      • use the value of [latex]k[\/latex] to determine the axis of symmetry, [latex]y=k[\/latex]<\/li>\r\n \t
      • set [latex]4p[\/latex] equal to the coefficient of [latex]\\left(x-h\\right)[\/latex] in the given equation to solve for [latex]p[\/latex]. If [latex]p>0[\/latex], the parabola opens right. If [latex]p<0[\/latex], the parabola opens left.<\/li>\r\n \t
      • use [latex]h,k[\/latex], and [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(h+p,\\text{ }k\\right)[\/latex]<\/li>\r\n \t
      • use [latex]h[\/latex] and [latex]p[\/latex] to find the equation of the directrix, [latex]x=h-p[\/latex]<\/li>\r\n \t
      • use [latex]h,k[\/latex], and [latex]p[\/latex] to find the endpoints of the focal diameter, [latex]\\left(h+p,k\\pm 2p\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t
      • If the equation is in the form [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex], then:\r\n
          \r\n \t
        • use the given equation to identify [latex]h[\/latex] and [latex]k[\/latex] for the vertex, [latex]\\left(h,k\\right)[\/latex]<\/li>\r\n \t
        • use the value of [latex]h[\/latex] to determine the axis of symmetry, [latex]x=h[\/latex]<\/li>\r\n \t
        • set [latex]4p[\/latex] equal to the coefficient of [latex]\\left(y-k\\right)[\/latex] in the given equation to solve for [latex]p[\/latex]. If [latex]p>0[\/latex], the parabola opens up. If [latex]p<0[\/latex], the parabola opens down.<\/li>\r\n \t
        • use [latex]h,k[\/latex], and [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(h,\\text{ }k+p\\right)[\/latex]<\/li>\r\n \t
        • use [latex]k[\/latex] and [latex]p[\/latex] to find the equation of the directrix, [latex]y=k-p[\/latex]<\/li>\r\n \t
        • use [latex]h,k[\/latex], and [latex]p[\/latex] to find the endpoints of the focal diameter, [latex]\\left(h\\pm 2p,\\text{ }k+p\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
        • Plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.<\/li>\r\n<\/ol>\r\n<\/section>
          Graph [latex]{\\left(y - 1\\right)}^{2}=-16\\left(x+3\\right)[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter.[reveal-answer q=\"943688\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"943688\"]The standard form that applies to the given equation is [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex]. Thus, the axis of symmetry is parallel to the [latex]x[\/latex]-axis.It follows that:\r\n
            \r\n \t
          • the vertex<\/strong> is [latex]\\left(h,k\\right)=\\left(-3,1\\right)[\/latex]<\/li>\r\n \t
          • the axis of symmetry<\/strong> is [latex]y=k=1[\/latex]<\/li>\r\n \t
          • [latex]-16=4p[\/latex], so [latex]p=-4[\/latex]. Since [latex]p<0[\/latex], the parabola opens left<\/strong>.<\/li>\r\n \t
          • the coordinates of the focus<\/strong> are [latex]\\left(h+p,k\\right)=\\left(-3+\\left(-4\\right),1\\right)=\\left(-7,1\\right)[\/latex]<\/li>\r\n \t
          • the equation of the directrix<\/strong> is [latex]x=h-p=-3-\\left(-4\\right)=1[\/latex]<\/li>\r\n \t
          • the endpoints of the focal diameter<\/strong> are [latex]\\left(h+p,k\\pm 2p\\right)=\\left(-3+\\left(-4\\right),1\\pm 2\\left(-4\\right)\\right)[\/latex], or [latex]\\left(-7,-7\\right)[\/latex] and [latex]\\left(-7,9\\right)[\/latex]<\/li>\r\n<\/ul>\r\nNext we plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"399\"]\"\" Parabola with key features labeled[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>Sometimes, our equations for parabolas might not initially appear in the standard form, which can make it challenging to identify their key features like the vertex, focus, and directrix. To address this, we often need to convert the given equation into the standard form by completing the square<\/strong>. This transformation allows us to rewrite the equation in a more familiar format, making it easier to analyze and graph the parabola. Completing the square helps us identify important components of the parabola and better understand its geometric properties.\r\n\r\n
            Graph [latex]{x}^{2}-8x - 28y - 208=0[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter.[reveal-answer q=\"335853\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"335853\"]Start by writing the equation of the parabola<\/strong> in standard form. The standard form that applies to the given equation is [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex]. Thus, the axis of symmetry is parallel to the [latex]y[\/latex]-axis.To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable [latex]x[\/latex] in order to complete the square.\r\n

            [latex]\\begin{gathered}{x}^{2}-8x - 28y - 208=0 \\\\ {x}^{2}-8x=28y+208 \\\\ {x}^{2}-8x+16=28y+208+16 \\\\ {\\left(x - 4\\right)}^{2}=28y+224 \\\\ {\\left(x - 4\\right)}^{2}=28\\left(y+8\\right) \\\\ {\\left(x - 4\\right)}^{2}=4\\cdot 7\\cdot \\left(y+8\\right) \\end{gathered}[\/latex]<\/p>\r\nIt follows that:\r\n

              \r\n \t
            • the vertex<\/strong> is [latex]\\left(h,k\\right)=\\left(4,-8\\right)[\/latex]<\/li>\r\n \t
            • the axis of symmetry<\/strong> is [latex]x=h=4[\/latex]<\/li>\r\n \t
            • since [latex]p=7,p>0[\/latex] and so the parabola opens up<\/strong><\/li>\r\n \t
            • the coordinates of the focus<\/strong> are [latex]\\left(h,k+p\\right)=\\left(4,-8+7\\right)=\\left(4,-1\\right)[\/latex]<\/li>\r\n \t
            • the equation of the directrix is [latex]y=k-p=-8 - 7=-15[\/latex]<\/li>\r\n \t
            • the endpoints of the focal diameter<\/strong> are [latex]\\left(h\\pm 2p,k+p\\right)=\\left(4\\pm 2\\left(7\\right),-8+7\\right)[\/latex], or [latex]\\left(-10,-1\\right)[\/latex] and [latex]\\left(18,-1\\right)[\/latex]<\/li>\r\n<\/ul>\r\nNext we plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"400\"]\"\" Parabola with key features labeled[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
              [ohm2_question hide_question_numbers=1]24913[\/ohm2_question]<\/section>
              [ohm2_question hide_question_numbers=1]24914[\/ohm2_question]<\/section>","rendered":"

              Parabolas with Vertices Not at the Origin<\/h2>\n

              Like other graphs we\u2019ve worked with, the graph of a parabola can be translated. If a parabola is translated [latex]h[\/latex] units horizontally and [latex]k[\/latex] units vertically, the vertex 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

              To graph parabolas with a vertex [latex]\\left(h,k\\right)[\/latex] other than the origin, we use the standard form [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex] for parabolas that have an axis of symmetry parallel to the x<\/em>-axis, and [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex] for parabolas that have an axis of symmetry parallel to the y<\/em>-axis. These standard forms are given below, along with their general graphs and key features.<\/p>\n

              \n

              standard forms of parabolas with vertex [latex](h, k)[\/latex]<\/h3>\n

              The table summarizes the standard features of parabolas with a vertex at a point [latex]\\left(h,k\\right)[\/latex].<\/p>\n\n\n\n\n\n
              Axis of Symmetry<\/strong><\/td>\nEquation<\/strong><\/td>\nFocus<\/strong><\/td>\nDirectrix<\/strong><\/td>\nEndpoints of focal diameter<\/strong><\/td>\n<\/tr>\n
              [latex]y=k[\/latex]<\/td>\n[latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex]<\/td>\n[latex]\\left(h+p,\\text{ }k\\right)[\/latex]<\/td>\n[latex]x=h-p[\/latex]<\/td>\n[latex]\\left(h+p,\\text{ }k\\pm 2p\\right)[\/latex]<\/td>\n<\/tr>\n
              [latex]x=h[\/latex]<\/td>\n[latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex]<\/td>\n[latex]\\left(h,\\text{ }k+p\\right)[\/latex]<\/td>\n[latex]y=k-p[\/latex]<\/td>\n[latex]\\left(h\\pm 2p,\\text{ }k+p\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
              \"\"
              (a) When [latex]p>0[\/latex], the parabola opens right. (b) When [latex]p<0[\/latex], the parabola opens left. (c) When [latex]p>0[\/latex], the parabola opens up. (d) When [latex]p<0[\/latex], the parabola opens down.<\/figcaption><\/figure>\n<\/section>\n
              How To: Given a standard form equation for a parabola centered at (h<\/em>, k<\/em>), sketch the graph.<\/strong><\/p>\n
                \n
              1. Determine which of the standard forms applies to the given equation: [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex] or [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex].<\/li>\n
              2. Use the standard form identified in Step 1 to determine the vertex, axis of symmetry, focus, equation of the directrix, and endpoints of the focal diameter.\n
                  \n
                1. If the equation is in the form [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex], then:\n
                    \n
                  • use the given equation to identify [latex]h[\/latex] and [latex]k[\/latex] for the vertex, [latex]\\left(h,k\\right)[\/latex]<\/li>\n
                  • use the value of [latex]k[\/latex] to determine the axis of symmetry, [latex]y=k[\/latex]<\/li>\n
                  • set [latex]4p[\/latex] equal to the coefficient of [latex]\\left(x-h\\right)[\/latex] in the given equation to solve for [latex]p[\/latex]. If [latex]p>0[\/latex], the parabola opens right. If [latex]p<0[\/latex], the parabola opens left.<\/li>\n
                  • use [latex]h,k[\/latex], and [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(h+p,\\text{ }k\\right)[\/latex]<\/li>\n
                  • use [latex]h[\/latex] and [latex]p[\/latex] to find the equation of the directrix, [latex]x=h-p[\/latex]<\/li>\n
                  • use [latex]h,k[\/latex], and [latex]p[\/latex] to find the endpoints of the focal diameter, [latex]\\left(h+p,k\\pm 2p\\right)[\/latex]<\/li>\n<\/ul>\n<\/li>\n
                  • If the equation is in the form [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex], then:\n
                      \n
                    • use the given equation to identify [latex]h[\/latex] and [latex]k[\/latex] for the vertex, [latex]\\left(h,k\\right)[\/latex]<\/li>\n
                    • use the value of [latex]h[\/latex] to determine the axis of symmetry, [latex]x=h[\/latex]<\/li>\n
                    • set [latex]4p[\/latex] equal to the coefficient of [latex]\\left(y-k\\right)[\/latex] in the given equation to solve for [latex]p[\/latex]. If [latex]p>0[\/latex], the parabola opens up. If [latex]p<0[\/latex], the parabola opens down.<\/li>\n
                    • use [latex]h,k[\/latex], and [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(h,\\text{ }k+p\\right)[\/latex]<\/li>\n
                    • use [latex]k[\/latex] and [latex]p[\/latex] to find the equation of the directrix, [latex]y=k-p[\/latex]<\/li>\n
                    • use [latex]h,k[\/latex], and [latex]p[\/latex] to find the endpoints of the focal diameter, [latex]\\left(h\\pm 2p,\\text{ }k+p\\right)[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/li>\n
                    • Plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.<\/li>\n<\/ol>\n<\/section>\n
                      Graph [latex]{\\left(y - 1\\right)}^{2}=-16\\left(x+3\\right)[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter.<\/p>\n