{"id":424,"date":"2019-07-15T22:44:55","date_gmt":"2019-07-15T22:44:55","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/chapter\/graph-hyperbolas\/"},"modified":"2019-07-15T22:44:55","modified_gmt":"2019-07-15T22:44:55","slug":"graph-hyperbolas","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/chapter\/graph-hyperbolas\/","title":{"raw":"Graph Hyperbolas","rendered":"Graph Hyperbolas"},"content":{"raw":"\n<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n \t<li>Graph a hyperbola centered at the origin.<\/li>\n \t<li>Graph a hyperbola not centered at the origin.<\/li>\n<\/ul>\n<\/div>\nWhen we have an equation in standard form for a hyperbola centered at the origin, we can interpret its parts to identify the key features of its graph: the center, vertices, co-vertices, asymptotes, foci, and lengths and positions of the transverse and conjugate axes. To graph hyperbolas centered at the origin, we use the standard form [latex]\\dfrac{{x}^{2}}{{a}^{2}}-\\dfrac{{y}^{2}}{{b}^{2}}=1[\/latex] for horizontal hyperbolas and the standard form [latex]\\dfrac{{y}^{2}}{{a}^{2}}-\\dfrac{{x}^{2}}{{b}^{2}}=1[\/latex] for vertical hyperbolas.\n<div class=\"textbox\">\n<h3>How To: Given a standard form equation for a hyperbola centered at [latex]\\left(0,0\\right)[\/latex], sketch the graph.<\/h3>\n<ul>\n \t<li>Determine which of the standard forms applies to the given equation.<\/li>\n \t<li>Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the vertices, co-vertices, and foci; and the equations for the asymptotes.\n<ul>\n \t<li>If the equation is in the form [latex]\\dfrac{{x}^{2}}{{a}^{2}}-\\dfrac{{y}^{2}}{{b}^{2}}=1[\/latex], then\n<ul>\n \t<li>the transverse axis is on the <em>x<\/em>-axis<\/li>\n \t<li>the coordinates of the vertices are [latex]\\left(\\pm a,0\\right)[\/latex]<\/li>\n \t<li>the coordinates of the co-vertices are [latex]\\left(0,\\pm b\\right)[\/latex]<\/li>\n \t<li>the coordinates of the foci are [latex]\\left(\\pm c,0\\right)[\/latex]<\/li>\n \t<li>the equations of the asymptotes are [latex]y=\\pm \\frac{b}{a}x[\/latex]<\/li>\n<\/ul>\n<\/li>\n \t<li>If the equation is in the form [latex]\\dfrac{{y}^{2}}{{a}^{2}}-\\dfrac{{x}^{2}}{{b}^{2}}=1[\/latex], then\n<ul>\n \t<li>the transverse axis is on the <em>y<\/em>-axis<\/li>\n \t<li>the coordinates of the vertices are [latex]\\left(0,\\pm a\\right)[\/latex]<\/li>\n \t<li>the coordinates of the co-vertices are [latex]\\left(\\pm b,0\\right)[\/latex]<\/li>\n \t<li>the coordinates of the foci are [latex]\\left(0,\\pm c\\right)[\/latex]<\/li>\n \t<li>the equations of the asymptotes are [latex]y=\\pm \\frac{a}{b}x[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n \t<li>Solve for the coordinates of the foci using the equation [latex]c=\\pm \\sqrt{{a}^{2}+{b}^{2}}[\/latex].<\/li>\n \t<li>Plot the vertices, co-vertices, foci, and asymptotes in the coordinate plane, and draw a smooth curve to form the hyperbola.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Graphing a Hyperbola Centered at (0, 0) Given an Equation in Standard Form<\/h3>\nGraph the hyperbola given by the equation [latex]\\dfrac{{y}^{2}}{64}-\\dfrac{{x}^{2}}{36}=1[\/latex]. Identify and label the vertices, co-vertices, foci, and asymptotes.\n\n[reveal-answer q=\"419240\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"419240\"]\n\nThe standard form that applies to the given equation is [latex]\\dfrac{{y}^{2}}{{a}^{2}}-\\dfrac{{x}^{2}}{{b}^{2}}=1[\/latex]. Thus, the transverse axis is on the <em>y<\/em>-axis\n\nThe coordinates of the vertices are [latex]\\left(0,\\pm a\\right)=\\left(0,\\pm \\sqrt{64}\\right)=\\left(0,\\pm 8\\right)[\/latex]\n\nThe coordinates of the co-vertices are [latex]\\left(\\pm b,0\\right)=\\left(\\pm \\sqrt{36},\\text{ }0\\right)=\\left(\\pm 6,0\\right)[\/latex]\n\nThe coordinates of the foci are [latex]\\left(0,\\pm c\\right)[\/latex], where [latex]c=\\pm \\sqrt{{a}^{2}+{b}^{2}}[\/latex]. Solving for [latex]c[\/latex] we have\n<p style=\"text-align: center;\">[latex]\\begin{align} c&amp;=\\pm \\sqrt{{a}^{2}+{b}^{2}} \\\\ &amp;=\\pm \\sqrt{64+36} \\\\ &amp;=\\pm \\sqrt{100} \\\\ &amp;=\\pm 10 \\end{align}[\/latex]<\/p>\nTherefore, the coordinates of the foci are [latex]\\left(0,\\pm 10\\right)[\/latex]\n\nThe equations of the asymptotes are\n<p style=\"text-align: center;\">[latex]\\begin{align} y=\\pm \\frac{a}{b}x \\\\ y=\\pm \\frac{8}{6}x \\\\ y=\\pm \\frac{4}{3}x \\end{align}[\/latex]<\/p>\nPlot and label the vertices and co-vertices, and then sketch the central rectangle. Sides of the rectangle are parallel to the axes and pass through the vertices and co-vertices. Sketch and extend the diagonals of the central rectangle to show the asymptotes. The central rectangle and asymptotes provide the framework needed to sketch an accurate graph of the hyperbola. Label the foci and asymptotes, and draw a smooth curve to form the hyperbola.\n\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03203430\/CNX_Precalc_Figure_10_02_0062.jpg\" width=\"731\" height=\"593\">\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nGraph the hyperbola given by the equation [latex]\\dfrac{{x}^{2}}{144}-\\dfrac{{y}^{2}}{81}=1[\/latex]. Identify and label the vertices, co-vertices, foci, and asymptotes.\n\n[reveal-answer q=\"168390\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"168390\"]\n\nvertices: [latex]\\left(\\pm 12,0\\right)[\/latex]; co-vertices: [latex]\\left(0,\\pm 9\\right)[\/latex]; foci: [latex]\\left(\\pm 15,0\\right)[\/latex]; asymptotes: [latex]y=\\pm \\frac{3}{4}x[\/latex];\n\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02170323\/CNX_Precalc_Figure_10_02_0072.jpg\"><img class=\"aligncenter size-full wp-image-3268\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02170323\/CNX_Precalc_Figure_10_02_0072.jpg\" alt=\"\" width=\"731\" height=\"361\"><\/a>\n\n[\/hidden-answer]\n\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=65294&amp;theme=oea&amp;iframe_resize_id=mom2[\/embed]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nUse an online graphing tool to plot the equation [latex]\\dfrac{x^2}{a^2} - \\dfrac{y^2}{b^2} = 1[\/latex]. Adjust the values you use for [latex]a,b[\/latex] to values between [latex] 1, 20[\/latex]. Your task in this exercise is to graph a hyperbola and then calculate and add the following features to the graph:\n<ul>\n \t<li>vertices<\/li>\n \t<li>co-vertices<\/li>\n \t<li>foci<\/li>\n \t<li>asymptotes<\/li>\n<\/ul>\n<\/div>\n<h2>Graphing Hyperbolas Not Centered at the Origin<\/h2>\nGraphing hyperbolas centered at a point [latex]\\left(h,k\\right)[\/latex] other than the origin is similar to graphing ellipses centered at a point other than the origin. We use the standard forms [latex]\\dfrac{{\\left(x-h\\right)}^{2}}{{a}^{2}}-\\dfrac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex] for horizontal hyperbolas, and [latex]\\dfrac{{\\left(y-k\\right)}^{2}}{{a}^{2}}-\\dfrac{{\\left(x-h\\right)}^{2}}{{b}^{2}}=1[\/latex] for vertical hyperbolas. From these standard form equations we can easily calculate and plot key features of the graph: the coordinates of its center, vertices, co-vertices, and foci; the equations of its asymptotes; and the positions of the transverse and conjugate axes.\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\nWriting the equation of a hyperbola not centered at the origin uses graph transformation techniques in the same way that writing the equation of an ellipse does. The center is represented by [latex]\\left(h, k\\right)[\/latex] as it was in the ellipse.\n\nLook for similarities and differences in the construction of ellipses, hyperbolas, and parabolas to help you build your intuition about these objects.\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a general form for a hyperbola centered at [latex]\\left(h,k\\right)[\/latex], sketch the graph.<\/h3>\n<ul>\n \t<li>Convert the general form to that standard form. Determine which of the standard forms applies to the given equation.<\/li>\n \t<li>Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the center, vertices, co-vertices, foci; and equations for the asymptotes.\n<ul>\n \t<li>If the equation is in the form [latex]\\dfrac{{\\left(x-h\\right)}^{2}}{{a}^{2}}-\\dfrac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex], then\n<ul>\n \t<li>the transverse axis is parallel to the <em>x<\/em>-axis<\/li>\n \t<li>the center is [latex]\\left(h,k\\right)[\/latex]<\/li>\n \t<li>the coordinates of the vertices are [latex]\\left(h\\pm a,k\\right)[\/latex]<\/li>\n \t<li>the coordinates of the co-vertices are [latex]\\left(h,k\\pm b\\right)[\/latex]<\/li>\n \t<li>the coordinates of the foci are [latex]\\left(h\\pm c,k\\right)[\/latex]<\/li>\n \t<li>the equations of the asymptotes are [latex]y=\\pm \\frac{b}{a}\\left(x-h\\right)+k[\/latex]<\/li>\n<\/ul>\n<\/li>\n \t<li>If the equation is in the form [latex]\\dfrac{{\\left(y-k\\right)}^{2}}{{a}^{2}}-\\dfrac{{\\left(x-h\\right)}^{2}}{{b}^{2}}=1[\/latex], then\n<ul>\n \t<li>the transverse axis is parallel to the <em>y<\/em>-axis<\/li>\n \t<li>the center is [latex]\\left(h,k\\right)[\/latex]<\/li>\n \t<li>the coordinates of the vertices are [latex]\\left(h,k\\pm a\\right)[\/latex]<\/li>\n \t<li>the coordinates of the co-vertices are [latex]\\left(h\\pm b,k\\right)[\/latex]<\/li>\n \t<li>the coordinates of the foci are [latex]\\left(h,k\\pm c\\right)[\/latex]<\/li>\n \t<li>the equations of the asymptotes are [latex]y=\\pm \\frac{a}{b}\\left(x-h\\right)+k[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n \t<li>Solve for the coordinates of the foci using the equation [latex]c=\\pm \\sqrt{{a}^{2}+{b}^{2}}[\/latex].<\/li>\n \t<li>Plot the center, vertices, co-vertices, foci, and asymptotes in the coordinate plane and draw a smooth curve to form the hyperbola.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3>recall Complete the square<\/h3>\nYou will need to complete the square twice when rewriting the general form of a hyperbola in the same way you did with the ellipse. Here's the form to help you.\n<p style=\"text-align: center;\">Completing the Square<\/p>\nGiven an expression of the form [latex]a\\left(x^2+bx\\right)[\/latex], add [latex]\\left(\\dfrac{b}{2}\\right)^2[\/latex] inside the parentheses, then subtract [latex]a\\left(\\dfrac{b}{2}\\right)^2[\/latex] to counteract the change you made.\n\nIf completing the square on one side of an equation, you may either subtract the value of&nbsp;[latex]a\\left(\\dfrac{b}{2}\\right)^2[\/latex] from that side, or add it to the other to maintain equality.\n\nThen factor the perfect square trinomial you created inside the original parentheses.\n\nExample\n\n[latex]\\qquad a\\left(x^2+bx\\right)[\/latex]\n\n[latex]=a\\left(x^2+bx+ \\left(\\dfrac{b}{2}\\right)^2\\right)-a\\left(\\dfrac{b}{2}\\right)^2[\/latex]\n\n[latex]=a\\left(x+ \\dfrac{b}{2}\\right)^2-a\\left(\\dfrac{b}{2}\\right)^2[\/latex]\n\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Graphing a Hyperbola Centered at (<em>h<\/em>, <em>k<\/em>) Given an Equation in General Form<\/h3>\nGraph the hyperbola given by the equation [latex]9{x}^{2}-4{y}^{2}-36x - 40y - 388=0[\/latex]. Identify and label the center, vertices, co-vertices, foci, and asymptotes.\n\n[reveal-answer q=\"227013\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"227013\"]\n\nStart by expressing the equation in standard form. Group terms that contain the same variable, and move the constant to the opposite side of the equation.\n<p style=\"text-align: center;\">[latex]\\left(9{x}^{2}-36x\\right)-\\left(4{y}^{2}+40y\\right)=388[\/latex]<\/p>\nFactor the leading coefficient of each expression.\n<p style=\"text-align: center;\">[latex]9\\left({x}^{2}-4x\\right)-4\\left({y}^{2}+10y\\right)=388[\/latex]<\/p>\nComplete the square twice. Remember to balance the equation by adding the same constants to each side.\n<p style=\"text-align: center;\">[latex]9\\left({x}^{2}-4x+4\\right)-4\\left({y}^{2}+10y+25\\right)=388+9\\cdot4 - 4\\cdot25[\/latex]<\/p>\nRewrite as perfect squares.\n<p style=\"text-align: center;\">[latex]9{\\left(x - 2\\right)}^{2}-4{\\left(y+5\\right)}^{2}=324[\/latex]<\/p>\nDivide both sides by the constant term to place the equation in standard form.\n<p style=\"text-align: center;\">[latex]\\dfrac{{\\left(x - 2\\right)}^{2}}{36}-\\dfrac{{\\left(y+5\\right)}^{2}}{81}=1[\/latex]<\/p>\nThe standard form that applies to the given equation is [latex]\\dfrac{{\\left(x-h\\right)}^{2}}{{a}^{2}}-\\dfrac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex], where [latex]{a}^{2}=36[\/latex] and [latex]{b}^{2}=81[\/latex], or [latex]a=6[\/latex] and [latex]b=9[\/latex]. Thus, the transverse axis is parallel to the <em>x<\/em>-axis. It follows that:\n<p style=\"text-align: center;\">the center of the ellipse is [latex]\\left(h,k\\right)=\\left(2,-5\\right)[\/latex]<\/p>\n\n<ul>\n \t<li>the coordinates of the vertices are [latex]\\left(h\\pm a,k\\right)=\\left(2\\pm 6,-5\\right)[\/latex], or [latex]\\left(-4,-5\\right)[\/latex] and [latex]\\left(8,-5\\right)[\/latex]<\/li>\n \t<li>the coordinates of the co-vertices are [latex]\\left(h,k\\pm b\\right)=\\left(2,-5\\pm 9\\right)[\/latex], or [latex]\\left(2,-14\\right)[\/latex] and [latex]\\left(2,4\\right)[\/latex]<\/li>\n \t<li>the coordinates of the foci are [latex]\\left(h\\pm c,k\\right)[\/latex], where [latex]c=\\pm \\sqrt{{a}^{2}+{b}^{2}}[\/latex]. Solving for [latex]c[\/latex], we have<\/li>\n<\/ul>\n<p style=\"text-align: center;\">[latex]c=\\pm \\sqrt{36+81}=\\pm \\sqrt{117}=\\pm 3\\sqrt{13}[\/latex]<\/p>\nTherefore, the coordinates of the foci are [latex]\\left(2 - 3\\sqrt{13},-5\\right)[\/latex] and [latex]\\left(2+3\\sqrt{13},-5\\right)[\/latex].\n\nThe equations of the asymptotes are [latex]y=\\pm \\frac{b}{a}\\left(x-h\\right)+k=\\pm \\frac{3}{2}\\left(x - 2\\right)-5[\/latex].\n\nNext, we plot and label the center, vertices, co-vertices, foci, and asymptotes and draw smooth curves to form the hyperbola.\n\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03203432\/CNX_Precalc_Figure_10_02_0082.jpg\" width=\"487\" height=\"443\">\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nGraph the hyperbola given by the standard form of an equation [latex]\\dfrac{{\\left(y+4\\right)}^{2}}{100}-\\dfrac{{\\left(x - 3\\right)}^{2}}{64}=1[\/latex]. Identify and label the center, vertices, co-vertices, foci, and asymptotes.\n\n[reveal-answer q=\"563114\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"563114\"]\n\ncenter: [latex]\\left(3,-4\\right)[\/latex]; vertices: [latex]\\left(3,-14\\right)[\/latex] and [latex]\\left(3,6\\right)[\/latex]; co-vertices: [latex]\\left(-5,-4\\right)[\/latex]; and [latex]\\left(11,-4\\right)[\/latex]; foci: [latex]\\left(3,-4 - 2\\sqrt{41}\\right)[\/latex] and [latex]\\left(3,-4+2\\sqrt{41}\\right)[\/latex]; asymptotes: [latex]y=\\pm \\frac{5}{4}\\left(x - 3\\right)-4[\/latex]\n\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02170636\/CNX_Precalc_Figure_10_02_0092.jpg\"><img class=\"aligncenter size-full wp-image-3269\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02170636\/CNX_Precalc_Figure_10_02_0092.jpg\" alt=\"\" width=\"487\" height=\"514\"><\/a>\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nUsing an online graphing calculator, plot a hyperbola not centered at the origin.\n\nThe equation used to generate the graph is [latex]\\dfrac{(x-h)^2}{a^2} - \\dfrac{(y-k)^2}{b^2} = 1[\/latex].\n\nAdjust the values you use for [latex]a,b[\/latex] to values between [latex] 1, 20[\/latex], and the variables [latex]h,k[\/latex] to numbers between [latex]10,10[\/latex]. Your task in this exercise is to graph the hyperbola and then calculate and add the following features to the graph:\n<ul>\n \t<li>center<\/li>\n \t<li>vertices<\/li>\n \t<li>co-vertices<\/li>\n \t<li>foci<\/li>\n \t<li>asymptotes<\/li>\n<\/ul>\n<\/div>\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Graph a hyperbola centered at the origin.<\/li>\n<li>Graph a hyperbola not centered at the origin.<\/li>\n<\/ul>\n<\/div>\n<p>When we have an equation in standard form for a hyperbola centered at the origin, we can interpret its parts to identify the key features of its graph: the center, vertices, co-vertices, asymptotes, foci, and lengths and positions of the transverse and conjugate axes. To graph hyperbolas centered at the origin, we use the standard form [latex]\\dfrac{{x}^{2}}{{a}^{2}}-\\dfrac{{y}^{2}}{{b}^{2}}=1[\/latex] for horizontal hyperbolas and the standard form [latex]\\dfrac{{y}^{2}}{{a}^{2}}-\\dfrac{{x}^{2}}{{b}^{2}}=1[\/latex] for vertical hyperbolas.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a standard form equation for a hyperbola centered at [latex]\\left(0,0\\right)[\/latex], sketch the graph.<\/h3>\n<ul>\n<li>Determine which of the standard forms applies to the given equation.<\/li>\n<li>Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the vertices, co-vertices, and foci; and the equations for the asymptotes.\n<ul>\n<li>If the equation is in the form [latex]\\dfrac{{x}^{2}}{{a}^{2}}-\\dfrac{{y}^{2}}{{b}^{2}}=1[\/latex], then\n<ul>\n<li>the transverse axis is on the <em>x<\/em>-axis<\/li>\n<li>the coordinates of the vertices are [latex]\\left(\\pm a,0\\right)[\/latex]<\/li>\n<li>the coordinates of the co-vertices are [latex]\\left(0,\\pm b\\right)[\/latex]<\/li>\n<li>the coordinates of the foci are [latex]\\left(\\pm c,0\\right)[\/latex]<\/li>\n<li>the equations of the asymptotes are [latex]y=\\pm \\frac{b}{a}x[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li>If the equation is in the form [latex]\\dfrac{{y}^{2}}{{a}^{2}}-\\dfrac{{x}^{2}}{{b}^{2}}=1[\/latex], then\n<ul>\n<li>the transverse axis is on the <em>y<\/em>-axis<\/li>\n<li>the coordinates of the vertices are [latex]\\left(0,\\pm a\\right)[\/latex]<\/li>\n<li>the coordinates of the co-vertices are [latex]\\left(\\pm b,0\\right)[\/latex]<\/li>\n<li>the coordinates of the foci are [latex]\\left(0,\\pm c\\right)[\/latex]<\/li>\n<li>the equations of the asymptotes are [latex]y=\\pm \\frac{a}{b}x[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<li>Solve for the coordinates of the foci using the equation [latex]c=\\pm \\sqrt{{a}^{2}+{b}^{2}}[\/latex].<\/li>\n<li>Plot the vertices, co-vertices, foci, and asymptotes in the coordinate plane, and draw a smooth curve to form the hyperbola.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Graphing a Hyperbola Centered at (0, 0) Given an Equation in Standard Form<\/h3>\n<p>Graph the hyperbola given by the equation [latex]\\dfrac{{y}^{2}}{64}-\\dfrac{{x}^{2}}{36}=1[\/latex]. Identify and label the vertices, co-vertices, foci, and asymptotes.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q419240\">Show Solution<\/span><\/p>\n<div id=\"q419240\" class=\"hidden-answer\" style=\"display: none\">\n<p>The standard form that applies to the given equation is [latex]\\dfrac{{y}^{2}}{{a}^{2}}-\\dfrac{{x}^{2}}{{b}^{2}}=1[\/latex]. Thus, the transverse axis is on the <em>y<\/em>-axis<\/p>\n<p>The coordinates of the vertices are [latex]\\left(0,\\pm a\\right)=\\left(0,\\pm \\sqrt{64}\\right)=\\left(0,\\pm 8\\right)[\/latex]<\/p>\n<p>The coordinates of the co-vertices are [latex]\\left(\\pm b,0\\right)=\\left(\\pm \\sqrt{36},\\text{ }0\\right)=\\left(\\pm 6,0\\right)[\/latex]<\/p>\n<p>The coordinates of the foci are [latex]\\left(0,\\pm c\\right)[\/latex], where [latex]c=\\pm \\sqrt{{a}^{2}+{b}^{2}}[\/latex]. Solving for [latex]c[\/latex] we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} c&=\\pm \\sqrt{{a}^{2}+{b}^{2}} \\\\ &=\\pm \\sqrt{64+36} \\\\ &=\\pm \\sqrt{100} \\\\ &=\\pm 10 \\end{align}[\/latex]<\/p>\n<p>Therefore, the coordinates of the foci are [latex]\\left(0,\\pm 10\\right)[\/latex]<\/p>\n<p>The equations of the asymptotes are<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} y=\\pm \\frac{a}{b}x \\\\ y=\\pm \\frac{8}{6}x \\\\ y=\\pm \\frac{4}{3}x \\end{align}[\/latex]<\/p>\n<p>Plot and label the vertices and co-vertices, and then sketch the central rectangle. Sides of the rectangle are parallel to the axes and pass through the vertices and co-vertices. Sketch and extend the diagonals of the central rectangle to show the asymptotes. The central rectangle and asymptotes provide the framework needed to sketch an accurate graph of the hyperbola. Label the foci and asymptotes, and draw a smooth curve to form the hyperbola.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03203430\/CNX_Precalc_Figure_10_02_0062.jpg\" width=\"731\" height=\"593\" alt=\"image\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Graph the hyperbola given by the equation [latex]\\dfrac{{x}^{2}}{144}-\\dfrac{{y}^{2}}{81}=1[\/latex]. Identify and label the vertices, co-vertices, foci, and asymptotes.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q168390\">Show Solution<\/span><\/p>\n<div id=\"q168390\" class=\"hidden-answer\" style=\"display: none\">\n<p>vertices: [latex]\\left(\\pm 12,0\\right)[\/latex]; co-vertices: [latex]\\left(0,\\pm 9\\right)[\/latex]; foci: [latex]\\left(\\pm 15,0\\right)[\/latex]; asymptotes: [latex]y=\\pm \\frac{3}{4}x[\/latex];<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02170323\/CNX_Precalc_Figure_10_02_0072.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3268\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02170323\/CNX_Precalc_Figure_10_02_0072.jpg\" alt=\"\" width=\"731\" height=\"361\" \/><\/a><\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm65294\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=65294&#38;theme=oea&#38;iframe_resize_id=ohm65294&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use an online graphing tool to plot the equation [latex]\\dfrac{x^2}{a^2} - \\dfrac{y^2}{b^2} = 1[\/latex]. Adjust the values you use for [latex]a,b[\/latex] to values between [latex]1, 20[\/latex]. Your task in this exercise is to graph a hyperbola and then calculate and add the following features to the graph:<\/p>\n<ul>\n<li>vertices<\/li>\n<li>co-vertices<\/li>\n<li>foci<\/li>\n<li>asymptotes<\/li>\n<\/ul>\n<\/div>\n<h2>Graphing Hyperbolas Not Centered at the Origin<\/h2>\n<p>Graphing hyperbolas centered at a point [latex]\\left(h,k\\right)[\/latex] other than the origin is similar to graphing ellipses centered at a point other than the origin. We use the standard forms [latex]\\dfrac{{\\left(x-h\\right)}^{2}}{{a}^{2}}-\\dfrac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex] for horizontal hyperbolas, and [latex]\\dfrac{{\\left(y-k\\right)}^{2}}{{a}^{2}}-\\dfrac{{\\left(x-h\\right)}^{2}}{{b}^{2}}=1[\/latex] for vertical hyperbolas. From these standard form equations we can easily calculate and plot key features of the graph: the coordinates of its center, vertices, co-vertices, and foci; the equations of its asymptotes; and the positions of the transverse and conjugate axes.<\/p>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>Writing the equation of a hyperbola not centered at the origin uses graph transformation techniques in the same way that writing the equation of an ellipse does. The center is represented by [latex]\\left(h, k\\right)[\/latex] as it was in the ellipse.<\/p>\n<p>Look for similarities and differences in the construction of ellipses, hyperbolas, and parabolas to help you build your intuition about these objects.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a general form for a hyperbola centered at [latex]\\left(h,k\\right)[\/latex], sketch the graph.<\/h3>\n<ul>\n<li>Convert the general form to that standard form. Determine which of the standard forms applies to the given equation.<\/li>\n<li>Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the center, vertices, co-vertices, foci; and equations for the asymptotes.\n<ul>\n<li>If the equation is in the form [latex]\\dfrac{{\\left(x-h\\right)}^{2}}{{a}^{2}}-\\dfrac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex], then\n<ul>\n<li>the transverse axis is parallel to the <em>x<\/em>-axis<\/li>\n<li>the center is [latex]\\left(h,k\\right)[\/latex]<\/li>\n<li>the coordinates of the vertices are [latex]\\left(h\\pm a,k\\right)[\/latex]<\/li>\n<li>the coordinates of the co-vertices are [latex]\\left(h,k\\pm b\\right)[\/latex]<\/li>\n<li>the coordinates of the foci are [latex]\\left(h\\pm c,k\\right)[\/latex]<\/li>\n<li>the equations of the asymptotes are [latex]y=\\pm \\frac{b}{a}\\left(x-h\\right)+k[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li>If the equation is in the form [latex]\\dfrac{{\\left(y-k\\right)}^{2}}{{a}^{2}}-\\dfrac{{\\left(x-h\\right)}^{2}}{{b}^{2}}=1[\/latex], then\n<ul>\n<li>the transverse axis is parallel to the <em>y<\/em>-axis<\/li>\n<li>the center is [latex]\\left(h,k\\right)[\/latex]<\/li>\n<li>the coordinates of the vertices are [latex]\\left(h,k\\pm a\\right)[\/latex]<\/li>\n<li>the coordinates of the co-vertices are [latex]\\left(h\\pm b,k\\right)[\/latex]<\/li>\n<li>the coordinates of the foci are [latex]\\left(h,k\\pm c\\right)[\/latex]<\/li>\n<li>the equations of the asymptotes are [latex]y=\\pm \\frac{a}{b}\\left(x-h\\right)+k[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<li>Solve for the coordinates of the foci using the equation [latex]c=\\pm \\sqrt{{a}^{2}+{b}^{2}}[\/latex].<\/li>\n<li>Plot the center, vertices, co-vertices, foci, and asymptotes in the coordinate plane and draw a smooth curve to form the hyperbola.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3>recall Complete the square<\/h3>\n<p>You will need to complete the square twice when rewriting the general form of a hyperbola in the same way you did with the ellipse. Here&#8217;s the form to help you.<\/p>\n<p style=\"text-align: center;\">Completing the Square<\/p>\n<p>Given an expression of the form [latex]a\\left(x^2+bx\\right)[\/latex], add [latex]\\left(\\dfrac{b}{2}\\right)^2[\/latex] inside the parentheses, then subtract [latex]a\\left(\\dfrac{b}{2}\\right)^2[\/latex] to counteract the change you made.<\/p>\n<p>If completing the square on one side of an equation, you may either subtract the value of&nbsp;[latex]a\\left(\\dfrac{b}{2}\\right)^2[\/latex] from that side, or add it to the other to maintain equality.<\/p>\n<p>Then factor the perfect square trinomial you created inside the original parentheses.<\/p>\n<p>Example<\/p>\n<p>[latex]\\qquad a\\left(x^2+bx\\right)[\/latex]<\/p>\n<p>[latex]=a\\left(x^2+bx+ \\left(\\dfrac{b}{2}\\right)^2\\right)-a\\left(\\dfrac{b}{2}\\right)^2[\/latex]<\/p>\n<p>[latex]=a\\left(x+ \\dfrac{b}{2}\\right)^2-a\\left(\\dfrac{b}{2}\\right)^2[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Graphing a Hyperbola Centered at (<em>h<\/em>, <em>k<\/em>) Given an Equation in General Form<\/h3>\n<p>Graph the hyperbola given by the equation [latex]9{x}^{2}-4{y}^{2}-36x - 40y - 388=0[\/latex]. Identify and label the center, vertices, co-vertices, foci, and asymptotes.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q227013\">Show Solution<\/span><\/p>\n<div id=\"q227013\" class=\"hidden-answer\" style=\"display: none\">\n<p>Start by expressing the equation in standard form. Group terms that contain the same variable, and move the constant to the opposite side of the equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(9{x}^{2}-36x\\right)-\\left(4{y}^{2}+40y\\right)=388[\/latex]<\/p>\n<p>Factor the leading coefficient of each expression.<\/p>\n<p style=\"text-align: center;\">[latex]9\\left({x}^{2}-4x\\right)-4\\left({y}^{2}+10y\\right)=388[\/latex]<\/p>\n<p>Complete the square twice. Remember to balance the equation by adding the same constants to each side.<\/p>\n<p style=\"text-align: center;\">[latex]9\\left({x}^{2}-4x+4\\right)-4\\left({y}^{2}+10y+25\\right)=388+9\\cdot4 - 4\\cdot25[\/latex]<\/p>\n<p>Rewrite as perfect squares.<\/p>\n<p style=\"text-align: center;\">[latex]9{\\left(x - 2\\right)}^{2}-4{\\left(y+5\\right)}^{2}=324[\/latex]<\/p>\n<p>Divide both sides by the constant term to place the equation in standard form.<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{{\\left(x - 2\\right)}^{2}}{36}-\\dfrac{{\\left(y+5\\right)}^{2}}{81}=1[\/latex]<\/p>\n<p>The standard form that applies to the given equation is [latex]\\dfrac{{\\left(x-h\\right)}^{2}}{{a}^{2}}-\\dfrac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex], where [latex]{a}^{2}=36[\/latex] and [latex]{b}^{2}=81[\/latex], or [latex]a=6[\/latex] and [latex]b=9[\/latex]. Thus, the transverse axis is parallel to the <em>x<\/em>-axis. It follows that:<\/p>\n<p style=\"text-align: center;\">the center of the ellipse is [latex]\\left(h,k\\right)=\\left(2,-5\\right)[\/latex]<\/p>\n<ul>\n<li>the coordinates of the vertices are [latex]\\left(h\\pm a,k\\right)=\\left(2\\pm 6,-5\\right)[\/latex], or [latex]\\left(-4,-5\\right)[\/latex] and [latex]\\left(8,-5\\right)[\/latex]<\/li>\n<li>the coordinates of the co-vertices are [latex]\\left(h,k\\pm b\\right)=\\left(2,-5\\pm 9\\right)[\/latex], or [latex]\\left(2,-14\\right)[\/latex] and [latex]\\left(2,4\\right)[\/latex]<\/li>\n<li>the coordinates of the foci are [latex]\\left(h\\pm c,k\\right)[\/latex], where [latex]c=\\pm \\sqrt{{a}^{2}+{b}^{2}}[\/latex]. Solving for [latex]c[\/latex], we have<\/li>\n<\/ul>\n<p style=\"text-align: center;\">[latex]c=\\pm \\sqrt{36+81}=\\pm \\sqrt{117}=\\pm 3\\sqrt{13}[\/latex]<\/p>\n<p>Therefore, the coordinates of the foci are [latex]\\left(2 - 3\\sqrt{13},-5\\right)[\/latex] and [latex]\\left(2+3\\sqrt{13},-5\\right)[\/latex].<\/p>\n<p>The equations of the asymptotes are [latex]y=\\pm \\frac{b}{a}\\left(x-h\\right)+k=\\pm \\frac{3}{2}\\left(x - 2\\right)-5[\/latex].<\/p>\n<p>Next, we plot and label the center, vertices, co-vertices, foci, and asymptotes and draw smooth curves to form the hyperbola.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03203432\/CNX_Precalc_Figure_10_02_0082.jpg\" width=\"487\" height=\"443\" alt=\"image\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Graph the hyperbola given by the standard form of an equation [latex]\\dfrac{{\\left(y+4\\right)}^{2}}{100}-\\dfrac{{\\left(x - 3\\right)}^{2}}{64}=1[\/latex]. Identify and label the center, vertices, co-vertices, foci, and asymptotes.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q563114\">Show Solution<\/span><\/p>\n<div id=\"q563114\" class=\"hidden-answer\" style=\"display: none\">\n<p>center: [latex]\\left(3,-4\\right)[\/latex]; vertices: [latex]\\left(3,-14\\right)[\/latex] and [latex]\\left(3,6\\right)[\/latex]; co-vertices: [latex]\\left(-5,-4\\right)[\/latex]; and [latex]\\left(11,-4\\right)[\/latex]; foci: [latex]\\left(3,-4 - 2\\sqrt{41}\\right)[\/latex] and [latex]\\left(3,-4+2\\sqrt{41}\\right)[\/latex]; asymptotes: [latex]y=\\pm \\frac{5}{4}\\left(x - 3\\right)-4[\/latex]<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02170636\/CNX_Precalc_Figure_10_02_0092.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3269\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02170636\/CNX_Precalc_Figure_10_02_0092.jpg\" alt=\"\" width=\"487\" height=\"514\" \/><\/a><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Using an online graphing calculator, plot a hyperbola not centered at the origin.<\/p>\n<p>The equation used to generate the graph is [latex]\\dfrac{(x-h)^2}{a^2} - \\dfrac{(y-k)^2}{b^2} = 1[\/latex].<\/p>\n<p>Adjust the values you use for [latex]a,b[\/latex] to values between [latex]1, 20[\/latex], and the variables [latex]h,k[\/latex] to numbers between [latex]10,10[\/latex]. Your task in this exercise is to graph the hyperbola and then calculate and add the following features to the graph:<\/p>\n<ul>\n<li>center<\/li>\n<li>vertices<\/li>\n<li>co-vertices<\/li>\n<li>foci<\/li>\n<li>asymptotes<\/li>\n<\/ul>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-424\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Graphing Hyperbolae Interactive. <strong>Authored by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.desmos.com\/calculator\/du6j3vvisl\">https:\/\/www.desmos.com\/calculator\/du6j3vvisl<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><li>Graphing Hyperbolae - With Solutions Interactive. <strong>Authored by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.desmos.com\/calculator\/enstg1igwo\">https:\/\/www.desmos.com\/calculator\/enstg1igwo<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><li>Graphing Hyperbolae Not Centered at the Origin Interactive. <strong>Authored by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.desmos.com\/calculator\/r307oiry1a\">https:\/\/www.desmos.com\/calculator\/r307oiry1a<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><li>Graphing Hyperbolae Not Centered at the Origin - Solutions Interactive. <strong>Authored by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.desmos.com\/calculator\/jilkpkpse1\">https:\/\/www.desmos.com\/calculator\/jilkpkpse1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 65294. <strong>Authored by<\/strong>: Roberts,Janet. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY+GPL<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":11,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Question ID 65294\",\"author\":\"Roberts,Janet\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY+GPL\"},{\"type\":\"original\",\"description\":\"Graphing Hyperbolae Interactive\",\"author\":\"Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/www.desmos.com\/calculator\/du6j3vvisl\",\"project\":\"\",\"license\":\"pd\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Graphing Hyperbolae - 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