{"id":2526,"date":"2025-08-13T17:48:21","date_gmt":"2025-08-13T17:48:21","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2526"},"modified":"2025-08-13T17:49:04","modified_gmt":"2025-08-13T17:49:04","slug":"ellipses-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/ellipses-learn-it-3\/","title":{"raw":"Ellipses: Learn It 3","rendered":"Ellipses: Learn It 3"},"content":{"raw":"

Graphing Ellipses Centered at the Origin<\/h2>\r\nJust as we can write the equation for an ellipse given its graph, we can graph an ellipse given its equation. To graph ellipses centered at the origin, we use the standard form [latex]\\frac{{x}^{2}}{{a}^{2}}+\\frac{{y}^{2}}{{b}^{2}}=1,\\text{ }a>b[\/latex] for horizontal ellipses and [latex]\\frac{{x}^{2}}{{b}^{2}}+\\frac{{y}^{2}}{{a}^{2}}=1,\\text{ }a>b[\/latex] for vertical ellipses.\r\n\r\n
How To: Given the standard form of an equation for an ellipse centered at [latex]\\left(0,0\\right)[\/latex], sketch the graph.<\/strong>\r\n
    \r\n \t
  • Use the standard forms of the equations of an ellipse to determine the major axis, vertices, co-vertices, and foci.\r\n
      \r\n \t
    • If the equation is in the form [latex]\\frac{{x}^{2}}{{a}^{2}}+\\frac{{y}^{2}}{{b}^{2}}=1[\/latex], where [latex]a>b[\/latex], then\r\n
        \r\n \t
      • the major axis is the x<\/em>-axis<\/li>\r\n \t
      • the coordinates of the vertices are [latex]\\left(\\pm a,0\\right)[\/latex]<\/li>\r\n \t
      • the coordinates of the co-vertices are [latex]\\left(0,\\pm b\\right)[\/latex]<\/li>\r\n \t
      • the coordinates of the foci are [latex]\\left(\\pm c,0\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t
      • If the equation is in the form [latex]\\frac{{x}^{2}}{{b}^{2}}+\\frac{{y}^{2}}{{a}^{2}}=1[\/latex], where [latex]a>b[\/latex], then\r\n
          \r\n \t
        • the major axis is the y<\/em>-axis<\/li>\r\n \t
        • the coordinates of the vertices are [latex]\\left(0,\\pm a\\right)[\/latex]<\/li>\r\n \t
        • the coordinates of the co-vertices are [latex]\\left(\\pm b,0\\right)[\/latex]<\/li>\r\n \t
        • the coordinates of the foci are [latex]\\left(0,\\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{{x}^{2}}{9}+\\frac{{y}^{2}}{25}=1[\/latex]. Identify and label the center, vertices, co-vertices, and foci.[reveal-answer q=\"322489\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"322489\"]\r\n\r\nFirst, we determine the position of the major axis. Because [latex]25>9[\/latex], the major axis is on the y<\/em>-axis. Therefore, the equation is in the form [latex]\\frac{{x}^{2}}{{b}^{2}}+\\frac{{y}^{2}}{{a}^{2}}=1[\/latex], where [latex]{b}^{2}=9[\/latex] and [latex]{a}^{2}=25[\/latex]. It follows that:\r\n
          \r\n
            \r\n \t
          • the center of the ellipse is [latex]\\left(0,0\\right)[\/latex]<\/li>\r\n \t
          • the coordinates of the vertices are [latex]\\left(0,\\pm a\\right)=\\left(0,\\pm \\sqrt{25}\\right)=\\left(0,\\pm 5\\right)[\/latex]<\/li>\r\n \t
          • the coordinates of the co-vertices are [latex]\\left(\\pm b,0\\right)=\\left(\\pm \\sqrt{9},0\\right)=\\left(\\pm 3,0\\right)[\/latex]<\/li>\r\n \t
          • the coordinates of the foci are [latex]\\left(0,\\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{25 - 9} \\\\ &=\\pm \\sqrt{16} \\\\ &=\\pm 4 \\end{align}[\/latex]<\/p>\r\n\r\n<\/div>\r\nTherefore, the coordinates of the foci are [latex]\\left(0,\\pm 4\\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\nGraph the ellipse given by the equation [latex]\\frac{{x}^{2}}{36}+\\frac{{y}^{2}}{4}=1[\/latex]. Identify and label the center, vertices, co-vertices, and foci.\r\n\r\n[reveal-answer q=\"363169\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"363169\"]\r\n\r\ncenter: [latex]\\left(0,0\\right)[\/latex]; vertices: [latex]\\left(\\pm 6,0\\right)[\/latex]; co-vertices: [latex]\\left(0,\\pm 2\\right)[\/latex]; foci: [latex]\\left(\\pm 4\\sqrt{2},0\\right)[\/latex]\r\n\"\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>
            [ohm_question hide_question_numbers=1]173953[\/ohm_question]<\/section>
            Graph the ellipse given by the equation [latex]4{x}^{2}+25{y}^{2}=100[\/latex]. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.[reveal-answer q=\"829376\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"829376\"]\r\n\r\nFirst, use algebra to rewrite the equation in standard form.\r\n

            [latex]\\begin{gathered} 4{x}^{2}+25{y}^{2}=100 \\\\ \\frac{4{x}^{2}}{100}+\\frac{25{y}^{2}}{100}=\\frac{100}{100} \\\\ \\frac{{x}^{2}}{25}+\\frac{{y}^{2}}{4}=1 \\end{gathered}[\/latex]<\/p>\r\nNext, we determine the position of the major axis. Because [latex]25>4[\/latex], the major axis is on the x<\/em>-axis. Therefore, the equation is in the form [latex]\\frac{{x}^{2}}{{a}^{2}}+\\frac{{y}^{2}}{{b}^{2}}=1[\/latex], where [latex]{a}^{2}=25[\/latex] and [latex]{b}^{2}=4[\/latex]. It follows that:\r\n

            \r\n
              \r\n \t
            • the center of the ellipse is [latex]\\left(0,0\\right)[\/latex]<\/li>\r\n \t
            • the coordinates of the vertices are [latex]\\left(\\pm a,0\\right)=\\left(\\pm \\sqrt{25},0\\right)=\\left(\\pm 5,0\\right)[\/latex]<\/li>\r\n \t
            • the coordinates of the co-vertices are [latex]\\left(0,\\pm b\\right)=\\left(0,\\pm \\sqrt{4}\\right)=\\left(0,\\pm 2\\right)[\/latex]<\/li>\r\n \t
            • the coordinates of the foci are [latex]\\left(\\pm c,0\\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{25 - 4} \\\\ &=\\pm \\sqrt{21} \\end{align}[\/latex]<\/p>\r\n\r\n<\/div>\r\nTherefore the coordinates of the foci are [latex]\\left(\\pm \\sqrt{21},0\\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]49{x}^{2}+16{y}^{2}=784[\/latex]. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.[reveal-answer q=\"205282\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"205282\"]Standard form: [latex]\\frac{{x}^{2}}{16}+\\frac{{y}^{2}}{49}=1[\/latex]; center: [latex]\\left(0,0\\right)[\/latex]; vertices: [latex]\\left(0,\\pm 7\\right)[\/latex]; co-vertices: [latex]\\left(\\pm 4,0\\right)[\/latex]; foci: [latex]\\left(0,\\pm \\sqrt{33}\\right)[\/latex]\r\n\"\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

              Graphing Ellipses Centered at the Origin<\/h2>\n

              Just as we can write the equation for an ellipse given its graph, we can graph an ellipse given its equation. To graph ellipses centered at the origin, we use the standard form [latex]\\frac{{x}^{2}}{{a}^{2}}+\\frac{{y}^{2}}{{b}^{2}}=1,\\text{ }a>b[\/latex] for horizontal ellipses and [latex]\\frac{{x}^{2}}{{b}^{2}}+\\frac{{y}^{2}}{{a}^{2}}=1,\\text{ }a>b[\/latex] for vertical ellipses.<\/p>\n

              How To: Given the standard form of an equation for an ellipse centered at [latex]\\left(0,0\\right)[\/latex], sketch the graph.<\/strong><\/p>\n
                \n
              • Use the standard forms of the equations of an ellipse to determine the major axis, vertices, co-vertices, and foci.\n
                  \n
                • If the equation is in the form [latex]\\frac{{x}^{2}}{{a}^{2}}+\\frac{{y}^{2}}{{b}^{2}}=1[\/latex], where [latex]a>b[\/latex], then\n
                    \n
                  • the major axis is the x<\/em>-axis<\/li>\n
                  • the coordinates of the vertices are [latex]\\left(\\pm a,0\\right)[\/latex]<\/li>\n
                  • the coordinates of the co-vertices are [latex]\\left(0,\\pm b\\right)[\/latex]<\/li>\n
                  • the coordinates of the foci are [latex]\\left(\\pm c,0\\right)[\/latex]<\/li>\n<\/ul>\n<\/li>\n
                  • If the equation is in the form [latex]\\frac{{x}^{2}}{{b}^{2}}+\\frac{{y}^{2}}{{a}^{2}}=1[\/latex], where [latex]a>b[\/latex], then\n
                      \n
                    • the major axis is the y<\/em>-axis<\/li>\n
                    • the coordinates of the vertices are [latex]\\left(0,\\pm a\\right)[\/latex]<\/li>\n
                    • the coordinates of the co-vertices are [latex]\\left(\\pm b,0\\right)[\/latex]<\/li>\n
                    • the coordinates of the foci are [latex]\\left(0,\\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{{x}^{2}}{9}+\\frac{{y}^{2}}{25}=1[\/latex]. Identify and label the center, vertices, co-vertices, and foci.<\/p>\n