{"id":2527,"date":"2025-08-13T17:48:24","date_gmt":"2025-08-13T17:48:24","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2527"},"modified":"2025-08-13T17:48:36","modified_gmt":"2025-08-13T17:48:36","slug":"ellipses-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/ellipses-learn-it-4\/","title":{"raw":"Ellipses: Learn It 4","rendered":"Ellipses: Learn It 4"},"content":{"raw":"

Graphing Ellipses Not Centered at the Origin<\/h2>\r\nWhen an ellipse<\/strong> is not centered at the origin, we can still use the standard forms to find the key features of the graph. When the ellipse is centered at some point, [latex]\\left(h,k\\right)[\/latex], we use the standard forms [latex]\\frac{{\\left(x-h\\right)}^{2}}{{a}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1,\\text{ }a>b[\/latex] for horizontal ellipses and [latex]\\frac{{\\left(x-h\\right)}^{2}}{{b}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}=1,\\text{ }a>b[\/latex] for vertical ellipses. From these standard equations, we can easily determine the center, vertices, co-vertices, foci, and positions of the major and minor axes.\r\n\r\n
How To: Given the standard form of an equation for an ellipse centered at [latex]\\left(h,k\\right)[\/latex], sketch the graph.<\/strong>\r\n
    \r\n \t
  • Use the standard forms of the equations of an ellipse to determine the center, position of the major axis, vertices, co-vertices, and foci.\r\n
      \r\n \t
    • If the equation is in the form [latex]\\frac{{\\left(x-h\\right)}^{2}}{{a}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex], where [latex]a>b[\/latex], then\r\n
        \r\n \t
      • the center is [latex]\\left(h,k\\right)[\/latex]<\/li>\r\n \t
      • the major axis is parallel to the x<\/em>-axis<\/li>\r\n \t
      • the coordinates of the vertices are [latex]\\left(h\\pm a,k\\right)[\/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 coordinates of the foci are [latex]\\left(h\\pm c,k\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t
      • If the equation is in the form [latex]\\frac{{\\left(x-h\\right)}^{2}}{{b}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}=1[\/latex], where [latex]a>b[\/latex], then\r\n
          \r\n \t
        • the center is [latex]\\left(h,k\\right)[\/latex]<\/li>\r\n \t
        • the major axis is parallel to the y<\/em>-axis<\/li>\r\n \t
        • the coordinates of the vertices are [latex]\\left(h,k\\pm a\\right)[\/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 coordinates of the foci are [latex]\\left(h,k\\pm c\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t
        • Solve for [latex]c[\/latex] using the equation [latex]{c}^{2}={a}^{2}-{b}^{2}[\/latex].<\/li>\r\n \t
        • Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.<\/li>\r\n<\/ul>\r\n<\/section>
          Graph the ellipse given by the equation, [latex]\\frac{{\\left(x+2\\right)}^{2}}{4}+\\frac{{\\left(y - 5\\right)}^{2}}{9}=1[\/latex]. Identify and label the center, vertices, co-vertices, and foci.[reveal-answer q=\"754503\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"754503\"]\r\n\r\nFirst, we determine the position of the major axis. Because [latex]9>4[\/latex], the major axis is parallel to the y<\/em>-axis. Therefore, the equation is in the form [latex]\\frac{{\\left(x-h\\right)}^{2}}{{b}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}=1[\/latex], where [latex]{b}^{2}=4[\/latex] and [latex]{a}^{2}=9[\/latex]. It follows that:\r\n
            \r\n \t
          • the center of the ellipse is [latex]\\left(h,k\\right)=\\left(-2,\\text{5}\\right)[\/latex]<\/li>\r\n \t
          • the coordinates of the vertices are [latex]\\left(h,k\\pm a\\right)=\\left(-2,5\\pm \\sqrt{9}\\right)=\\left(-2,5\\pm 3\\right)[\/latex], or [latex]\\left(-2,\\text{2}\\right)[\/latex] and [latex]\\left(-2,\\text{8}\\right)[\/latex]<\/li>\r\n \t
          • the coordinates of the co-vertices are [latex]\\left(h\\pm b,k\\right)=\\left(-2\\pm \\sqrt{4},5\\right)=\\left(-2\\pm 2,5\\right)[\/latex], or [latex]\\left(-4,5\\right)[\/latex] and [latex]\\left(0,\\text{5}\\right)[\/latex]<\/li>\r\n \t
          • the coordinates of the foci are [latex]\\left(h,k\\pm c\\right)[\/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[\/latex]. Solving for [latex]c[\/latex], we have:<\/li>\r\n<\/ul>\r\n

            [latex]\\begin{align} c&=\\pm \\sqrt{{a}^{2}-{b}^{2}} \\\\ &=\\pm \\sqrt{9 - 4} \\\\ &=\\pm \\sqrt{5} \\end{align}[\/latex]<\/p>\r\nTherefore, the coordinates of the foci are [latex]\\left(-2,\\text{5}-\\sqrt{5}\\right)[\/latex] and [latex]\\left(-2,\\text{5+}\\sqrt{5}\\right)[\/latex].\r\n\r\nNext, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.\r\n\r\n\"\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

            Graph the ellipse given by the equation [latex]\\frac{{\\left(x - 4\\right)}^{2}}{36}+\\frac{{\\left(y - 2\\right)}^{2}}{20}=1[\/latex]. Identify and label the center, vertices, co-vertices, and foci.[reveal-answer q=\"470748\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"470748\"]Center: [latex]\\left(4,2\\right)[\/latex]; vertices: [latex]\\left(-2,2\\right)[\/latex] and [latex]\\left(10,2\\right)[\/latex]; co-vertices: [latex]\\left(4,2 - 2\\sqrt{5}\\right)[\/latex] and [latex]\\left(4,2+2\\sqrt{5}\\right)[\/latex]; foci: [latex]\\left(0,2\\right)[\/latex] and [latex]\\left(8,2\\right)[\/latex]\r\n\"\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
            [ohm_question hide_question_numbers=1]173948[\/ohm_question]<\/section>
            How To: Given the general form of an equation for an ellipse centered at (h<\/em>, k<\/em>), express the equation in standard form.<\/strong>\r\n
              \r\n \t
            • Recognize that an ellipse described by an equation in the form [latex]a{x}^{2}+b{y}^{2}+cx+dy+e=0[\/latex] is in general form.<\/li>\r\n \t
            • Rearrange the equation by grouping terms that contain the same variable. Move the constant term to the opposite side of the equation.<\/li>\r\n \t
            • Factor out the coefficients of the [latex]{x}^{2}[\/latex] and [latex]{y}^{2}[\/latex] terms in preparation for completing the square.<\/li>\r\n \t
            • Complete the square for each variable to rewrite the equation in the form of the sum of multiples of two binomials squared set equal to a constant, [latex]{m}_{1}{\\left(x-h\\right)}^{2}+{m}_{2}{\\left(y-k\\right)}^{2}={m}_{3}[\/latex], where [latex]{m}_{1},{m}_{2}[\/latex], and [latex]{m}_{3}[\/latex] are constants.<\/li>\r\n \t
            • Divide both sides of the equation by the constant term to express the equation in standard form.<\/li>\r\n<\/ul>\r\n<\/section>
              Graph the ellipse given by the equation [latex]4{x}^{2}+9{y}^{2}-40x+36y+100=0[\/latex]. Identify and label the center, vertices, co-vertices, and foci.[reveal-answer q=\"499064\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"499064\"]\r\n\r\nWe must begin by rewriting the equation in standard form.\r\n
              [latex]4{x}^{2}+9{y}^{2}-40x+36y+100=0[\/latex]<\/div>\r\nGroup terms that contain the same variable, and move the constant to the opposite side of the equation.\r\n
              [latex]\\left(4{x}^{2}-40x\\right)+\\left(9{y}^{2}+36y\\right)=-100[\/latex]<\/div>\r\nFactor out the coefficients of the squared terms.\r\n
              [latex]4\\left({x}^{2}-10x\\right)+9\\left({y}^{2}+4y\\right)=-100[\/latex]<\/div>\r\nComplete the square twice. Remember to balance the equation by adding the same constants to each side.\r\n
              [latex]4\\left({x}^{2}-10x+25\\right)+9\\left({y}^{2}+4y+4\\right)=-100+100+36[\/latex]<\/div>\r\nRewrite as perfect squares.\r\n
              [latex]4{\\left(x - 5\\right)}^{2}+9{\\left(y+2\\right)}^{2}=36[\/latex]<\/div>\r\nDivide both sides by the constant term to place the equation in standard form.\r\n
              [latex]\\frac{{\\left(x - 5\\right)}^{2}}{9}+\\frac{{\\left(y+2\\right)}^{2}}{4}=1[\/latex]<\/div>\r\nNow that the equation is in standard form, we can determine the position of the major axis. Because [latex]9>4[\/latex], the major axis is parallel to the x<\/em>-axis. Therefore, the equation is in the form [latex]\\frac{{\\left(x-h\\right)}^{2}}{{a}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex], where [latex]{a}^{2}=9[\/latex] and [latex]{b}^{2}=4[\/latex]. It follows that:\r\n
                \r\n \t
              • the center of the ellipse is [latex]\\left(h,k\\right)=\\left(5,-2\\right)[\/latex]<\/li>\r\n \t
              • the coordinates of the vertices are [latex]\\left(h\\pm a,k\\right)=\\left(5\\pm \\sqrt{9},-2\\right)=\\left(5\\pm 3,-2\\right)[\/latex], or [latex]\\left(2,-2\\right)[\/latex] and [latex]\\left(8,-2\\right)[\/latex]<\/li>\r\n \t
              • the coordinates of the co-vertices are [latex]\\left(h,k\\pm b\\right)=\\left(\\text{5},-2\\pm \\sqrt{4}\\right)=\\left(\\text{5},-2\\pm 2\\right)[\/latex], or [latex]\\left(5,-4\\right)[\/latex] and [latex]\\left(5,\\text{0}\\right)[\/latex]<\/li>\r\n \t
              • the coordinates of the foci are [latex]\\left(h\\pm c,k\\right)[\/latex], where [latex]{c}^{2}={a}^{2}-{b}^{2}[\/latex]. Solving for [latex]c[\/latex], we have:<\/li>\r\n<\/ul>\r\n

                [latex]\\begin{align}c&=\\pm \\sqrt{{a}^{2}-{b}^{2}} \\\\ &=\\pm \\sqrt{9 - 4} \\\\ &=\\pm \\sqrt{5} \\end{align}[\/latex]<\/p>\r\nTherefore, the coordinates of the foci are [latex]\\left(\\text{5}-\\sqrt{5},-2\\right)[\/latex] and [latex]\\left(\\text{5+}\\sqrt{5},-2\\right)[\/latex].\r\n\r\nNext we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.\r\n\r\n\"\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

                \r\n
                \r\n\r\nExpress the equation of the ellipse given in standard form. Identify the center, vertices, co-vertices, and foci of the ellipse.\r\n

                [latex]4{x}^{2}+{y}^{2}-24x+2y+21=0[\/latex]<\/p>\r\n

                [reveal-answer q=\"8892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"8892\"]<\/p>\r\n

                [latex]\\frac{{\\left(x - 3\\right)}^{2}}{4}+\\frac{{\\left(y+1\\right)}^{2}}{16}=1[\/latex]; center: [latex]\\left(3,-1\\right)[\/latex]; vertices: [latex]\\left(3,-\\text{5}\\right)[\/latex] and [latex]\\left(3,\\text{3}\\right)[\/latex]; co-vertices: [latex]\\left(1,-1\\right)[\/latex] and [latex]\\left(5,-1\\right)[\/latex]; foci: [latex]\\left(3,-\\text{1}-2\\sqrt{3}\\right)[\/latex] and [latex]\\left(3,-\\text{1+}2\\sqrt{3}\\right)[\/latex]<\/p>\r\n

                [\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/section>","rendered":"

                Graphing Ellipses Not Centered at the Origin<\/h2>\n

                When an ellipse<\/strong> is not centered at the origin, we can still use the standard forms to find the key features of the graph. When the ellipse is centered at some point, [latex]\\left(h,k\\right)[\/latex], we use the standard forms [latex]\\frac{{\\left(x-h\\right)}^{2}}{{a}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1,\\text{ }a>b[\/latex] for horizontal ellipses and [latex]\\frac{{\\left(x-h\\right)}^{2}}{{b}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}=1,\\text{ }a>b[\/latex] for vertical ellipses. From these standard equations, we can easily determine the center, vertices, co-vertices, foci, and positions of the major and minor axes.<\/p>\n

                How To: Given the standard form of an equation for an ellipse centered at [latex]\\left(h,k\\right)[\/latex], sketch the graph.<\/strong><\/p>\n
                  \n
                • Use the standard forms of the equations of an ellipse to determine the center, position of the major axis, vertices, co-vertices, and foci.\n
                    \n
                  • If the equation is in the form [latex]\\frac{{\\left(x-h\\right)}^{2}}{{a}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex], where [latex]a>b[\/latex], then\n
                      \n
                    • the center is [latex]\\left(h,k\\right)[\/latex]<\/li>\n
                    • the major axis is parallel to the x<\/em>-axis<\/li>\n
                    • the coordinates of the vertices are [latex]\\left(h\\pm a,k\\right)[\/latex]<\/li>\n
                    • the coordinates of the co-vertices are [latex]\\left(h,k\\pm b\\right)[\/latex]<\/li>\n
                    • the coordinates of the foci are [latex]\\left(h\\pm c,k\\right)[\/latex]<\/li>\n<\/ul>\n<\/li>\n
                    • If the equation is in the form [latex]\\frac{{\\left(x-h\\right)}^{2}}{{b}^{2}}+\\frac{{\\left(y-k\\right)}^{2}}{{a}^{2}}=1[\/latex], where [latex]a>b[\/latex], then\n
                        \n
                      • the center is [latex]\\left(h,k\\right)[\/latex]<\/li>\n
                      • the major axis is parallel to the y<\/em>-axis<\/li>\n
                      • the coordinates of the vertices are [latex]\\left(h,k\\pm a\\right)[\/latex]<\/li>\n
                      • the coordinates of the co-vertices are [latex]\\left(h\\pm b,k\\right)[\/latex]<\/li>\n
                      • the coordinates of the foci are [latex]\\left(h,k\\pm c\\right)[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n
                      • Solve for [latex]c[\/latex] using the equation [latex]{c}^{2}={a}^{2}-{b}^{2}[\/latex].<\/li>\n
                      • Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.<\/li>\n<\/ul>\n<\/section>\n
                        Graph the ellipse given by the equation, [latex]\\frac{{\\left(x+2\\right)}^{2}}{4}+\\frac{{\\left(y - 5\\right)}^{2}}{9}=1[\/latex]. Identify and label the center, vertices, co-vertices, and foci.<\/p>\n