{"id":1914,"date":"2015-11-12T18:30:43","date_gmt":"2015-11-12T18:30:43","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&#038;p=1914"},"modified":"2015-11-12T18:30:43","modified_gmt":"2015-11-12T18:30:43","slug":"graphing-parabolas-with-vertices-at-the-origin","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/graphing-parabolas-with-vertices-at-the-origin\/","title":{"raw":"Graphing Parabolas with Vertices at the Origin","rendered":"Graphing Parabolas with Vertices at the Origin"},"content":{"raw":"<p>In <a class=\"target-chapter\" href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/introduction-to-the-ellipse\/\" target=\"_blank\">The Ellipse<\/a>, we saw that an <strong>ellipse<\/strong> is formed when a plane cuts through a right circular cone. If the plane is parallel to the edge of the cone, an unbounded curve is formed. This curve is a <strong>parabola<\/strong>.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202317\/CNX_Precalc_Figure_10_03_0022.jpg\" alt=\"\" width=\"487\" height=\"425\" data-media-type=\"image\/jpg\"\/><b>Figure 2.<\/b> Parabola[\/caption]\n\nLike the ellipse and <strong>hyperbola<\/strong>, the parabola can also be defined by a set of points in the coordinate plane. A parabola is the set of all points [latex]\\left(x,y\\right)[\/latex] in a plane that are the same distance from a fixed line, called the <strong>directrix<\/strong>, and a fixed point (the <strong>focus<\/strong>) not on the directrix.\n\nWe previously learned about a parabola\u2019s vertex and axis of symmetry. Now we extend the discussion to include other key features of the parabola. See Figure 3. Notice that the axis of symmetry passes through the focus and vertex and is perpendicular to the directrix. The vertex is the midpoint between the directrix and the focus.\n\nThe line segment that passes through the focus and is parallel to the directrix is called the <strong>latus rectum<\/strong>. The endpoints of the latus rectum lie on the curve. By definition, the distance [latex]d[\/latex] from the focus to any point [latex]P[\/latex] on the parabola is equal to the distance from [latex]P[\/latex] to the directrix.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202318\/CNX_Precalc_Figure_10_03_003n2.jpg\" alt=\"\" width=\"487\" height=\"291\" data-media-type=\"image\/jpg\"\/><b>Figure 3.<\/b> Key features of the parabola[\/caption]\n\nTo work with parabolas in the <strong>coordinate plane<\/strong>, we consider two cases: those with a vertex at the origin and those with a <strong>vertex<\/strong> at a point other than the origin. We begin with the former.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202320\/CNX_Precalc_Figure_10_03_0182.jpg\" alt=\"\" width=\"487\" height=\"292\" data-media-type=\"image\/jpg\"\/><b>Figure 4<\/b>[\/caption]\n\nLet [latex]\\left(x,y\\right)[\/latex] be a point on the parabola with vertex [latex]\\left(0,0\\right)[\/latex], focus [latex]\\left(0,p\\right)[\/latex], and directrix [latex]y= -p[\/latex]\u00a0as shown in Figure 4. The distance [latex]d[\/latex] from point [latex]\\left(x,y\\right)[\/latex] to point [latex]\\left(x,-p\\right)[\/latex]\u00a0on the directrix is the difference of the <em>y<\/em>-values: [latex]d=y+p[\/latex]. The distance from the focus [latex]\\left(0,p\\right)[\/latex] to the point [latex]\\left(x,y\\right)[\/latex] is also equal to [latex]d[\/latex] and can be expressed using the <strong>distance formula<\/strong>.\n<\/p><div style=\"text-align: center;\">[latex]\\begin{array}{l}d=\\sqrt{{\\left(x - 0\\right)}^{2}+{\\left(y-p\\right)}^{2}}\\hfill \\\\ =\\sqrt{{x}^{2}+{\\left(y-p\\right)}^{2}}\\hfill \\end{array}[\/latex]<\/div>\nSet the two expressions for [latex]d[\/latex] equal to each other and solve for [latex]y[\/latex] to derive the equation of the parabola. We do this because the distance from [latex]\\left(x,y\\right)[\/latex] to [latex]\\left(0,p\\right)[\/latex] equals the distance from [latex]\\left(x,y\\right)[\/latex] to [latex]\\left(x, -p\\right)[\/latex].\n<div style=\"text-align: center;\">[latex]\\sqrt{{x}^{2}+{\\left(y-p\\right)}^{2}}=y+p[\/latex]<\/div>\nWe then square both sides of the equation, expand the squared terms, and simplify by combining like terms.\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}{x}^{2}+{\\left(y-p\\right)}^{2}={\\left(y+p\\right)}^{2}\\\\ {x}^{2}+{y}^{2}-2py+{p}^{2}={y}^{2}+2py+{p}^{2}\\\\ {x}^{2}-2py=2py\\\\ \\text{ }{x}^{2}=4py\\end{array}[\/latex]<\/div>\nThe equations of parabolas with vertex [latex]\\left(0,0\\right)[\/latex] are [latex]{y}^{2}=4px[\/latex] when the <em>x<\/em>-axis is the axis of symmetry and [latex]{x}^{2}=4py[\/latex] when the <em>y<\/em>-axis is the axis of symmetry. These standard forms are given below, along with their general graphs and key features.\n<div class=\"textbox\">\n<h3>A General Note: Standard Forms of Parabolas with Vertex (0, 0)<\/h3>\nThe table below\u00a0and Figure 5\u00a0summarize the standard features of parabolas with a vertex at the origin.\n<table id=\"Table_10_03_01\" summary=\"..\"><tbody><tr><td><strong>Axis of Symmetry<\/strong><\/td>\n<td><strong>Equation<\/strong><\/td>\n<td><strong>Focus<\/strong><\/td>\n<td><strong>Directrix<\/strong><\/td>\n<td><strong>Endpoints of Latus Rectum<\/strong><\/td>\n<\/tr><tr><td><em>x<\/em>-axis<\/td>\n<td>[latex]{y}^{2}=4px[\/latex]<\/td>\n<td>[latex]\\left(p,\\text{ }0\\right)[\/latex]<\/td>\n<td>[latex]x=-p[\/latex]<\/td>\n<td>[latex]\\left(p,\\text{ }\\pm 2p\\right)[\/latex]<\/td>\n<\/tr><tr><td><em>y<\/em>-axis<\/td>\n<td>[latex]{x}^{2}=4py[\/latex]<\/td>\n<td>[latex]\\left(0,\\text{ }p\\right)[\/latex]<\/td>\n<td>[latex]y=-p[\/latex]<\/td>\n<td>[latex]\\left(\\pm 2p,\\text{ }p\\right)[\/latex]<\/td>\n<\/tr><\/tbody><\/table>\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202321\/CNX_Precalc_Figure_10_03_004n2.jpg\" alt=\"\" width=\"975\" height=\"721\" data-media-type=\"image\/jpg\"\/><b>Figure 5.<\/b> (a) When [latex]p&gt;0[\/latex] and the axis of symmetry is the x-axis, the parabola opens right. (b) When [latex]p&lt;0[\/latex] and the axis of symmetry is the x-axis, the parabola opens left. (c) When [latex]p&lt;0[\/latex] and the axis of symmetry is the y-axis, the parabola opens up. (d) When [latex]\\text{ }p&lt;0\\text{ }[\/latex] and the axis of symmetry is the y-axis, the parabola opens down.[\/caption]<\/div>\nThe key features of a parabola are its vertex, axis of symmetry, focus, directrix, and latus rectum. When given a standard equation for a parabola centered at the origin, we can easily identify the key features to graph the parabola.\n\nA line is said to be tangent to a curve if it intersects the curve at exactly one point. If we sketch lines tangent to the parabola at the endpoints of the latus rectum, these lines intersect on the axis of symmetry, as shown in Figure 6.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202322\/CNX_Precalc_Figure_10_03_0052.jpg\" alt=\"\" width=\"487\" height=\"514\" data-media-type=\"image\/jpg\"\/><b>Figure 6<\/b>[\/caption]\n\n<div class=\"textbox\">\n<h3>How To: Given a standard form equation for a parabola centered at (0, 0), sketch the graph.<\/h3>\n<ul><li>Determine which of the standard forms applies to the given equation: [latex]{y}^{2}=4px[\/latex] or [latex]{x}^{2}=4py[\/latex].<\/li>\n\t<li>Use the standard form identified in Step 1 to determine the axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum.\n<ul><li>If the equation is in the form [latex]{y}^{2}=4px[\/latex], then\n<ul><li>the axis of symmetry is the <em>x<\/em>-axis, [latex]y=0[\/latex]<\/li>\n\t<li>set [latex]4p[\/latex] equal to the coefficient of <em>x <\/em>in the given equation to solve for [latex]p[\/latex]. If [latex]p&gt;0[\/latex], the parabola opens right. If [latex]p&lt;0[\/latex], the parabola opens left.<\/li>\n\t<li>use [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(p,0\\right)[\/latex]<\/li>\n\t<li>use [latex]p[\/latex] to find the equation of the directrix, [latex]x=-p[\/latex]<\/li>\n\t<li>use [latex]p[\/latex] to find the endpoints of the latus rectum, [latex]\\left(p,\\pm 2p\\right)[\/latex]. Alternately, substitute [latex]x=p[\/latex] into the original equation.<\/li>\n<\/ul><\/li>\n\t<li>If the equation is in the form [latex]{x}^{2}=4py[\/latex], then\n<ul><li>the axis of symmetry is the <em>y<\/em>-axis, [latex]x=0[\/latex]<\/li>\n\t<li>set [latex]4p[\/latex] equal to the coefficient of <em>y <\/em>in the given equation to solve for [latex]p[\/latex]. If [latex]p&gt;0[\/latex], the parabola opens up. If [latex]p&lt;0[\/latex], the parabola opens down.<\/li>\n\t<li>use [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(0,p\\right)[\/latex]<\/li>\n\t<li>use [latex]p[\/latex] to find equation of the directrix, [latex]y=-p[\/latex]<\/li>\n\t<li>use [latex]p[\/latex] to find the endpoints of the latus rectum, [latex]\\left(\\pm 2p,p\\right)[\/latex]<\/li>\n<\/ul><\/li>\n<\/ul><\/li>\n\t<li>Plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.<\/li>\n<\/ul><\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Graphing a Parabola with Vertex (0, 0) and the <em>x<\/em>-axis as the Axis of Symmetry<\/h3>\nGraph [latex]{y}^{2}=24x[\/latex]. Identify and label the <strong>focus<\/strong>, <strong>directrix<\/strong>, and endpoints of the <strong>latus rectum<\/strong>.\n\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\nThe standard form that applies to the given equation is [latex]{y}^{2}=4px[\/latex]. Thus, the axis of symmetry is the <em>x<\/em>-axis. It follows that:\n<div>\n<ul><li>[latex]24=4p[\/latex], so [latex]p=6[\/latex]. Since [latex]p&gt;0[\/latex], the parabola opens right\u00a0the coordinates of the focus are [latex]\\left(p,0\\right)=\\left(6,0\\right)[\/latex]<\/li>\n\t<li>the equation of the directrix is [latex]x=-p=-6[\/latex]<\/li>\n\t<li>the endpoints of the latus rectum have the same <em>x<\/em>-coordinate at the focus. To find the endpoints, substitute [latex]x=6[\/latex] into the original equation: [latex]\\left(6,\\pm 12\\right)[\/latex]<\/li>\n<\/ul><\/div>\nNext we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the <strong>parabola<\/strong>.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202323\/CNX_Precalc_Figure_10_03_0192.jpg\" alt=\"\" width=\"487\" height=\"376\" data-media-type=\"image\/jpg\"\/><b>Figure 7<\/b>[\/caption]\n\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 1<\/h3>\nGraph [latex]{y}^{2}=-16x[\/latex]. Identify and label the focus, directrix, and endpoints of the latus rectum.\n\n<a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-25\/\" target=\"_blank\">Solution<\/a>\n\n<\/div>\n<div\/>\n<div class=\"textbox shaded\">\n<h3>Example 2: Graphing a Parabola with Vertex (0, 0) and the <em>y<\/em>-axis as the Axis of Symmetry<\/h3>\nGraph [latex]{x}^{2}=-6y[\/latex]. Identify and label the <strong>focus<\/strong>, <strong>directrix<\/strong>, and endpoints of the <strong>latus rectum<\/strong>.\n\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\nThe standard form that applies to the given equation is [latex]{x}^{2}=4py[\/latex]. Thus, the axis of symmetry is the <em>y<\/em>-axis. It follows that:\n<div>\n<ul><li>[latex]-6=4p[\/latex], so [latex]p=-\\frac{3}{2}[\/latex]. Since [latex]p&lt;0[\/latex], the parabola opens down.<\/li>\n\t<li>the coordinates of the focus are [latex]\\left(0,p\\right)=\\left(0,-\\frac{3}{2}\\right)[\/latex]<\/li>\n\t<li>the equation of the directrix is [latex]y=-p=\\frac{3}{2}[\/latex]<\/li>\n\t<li>the endpoints of the latus rectum can be found by substituting [latex]\\text{ }y=\\frac{3}{2}\\text{ }[\/latex] into the original equation, [latex]\\left(\\pm 3,-\\frac{3}{2}\\right)[\/latex]<\/li>\n<\/ul><\/div>\nNext we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the <strong>parabola<\/strong>.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202325\/CNX_Precalc_Figure_10_03_007n2.jpg\" alt=\"\" width=\"487\" height=\"327\" data-media-type=\"image\/jpg\"\/><b>Figure 8<\/b>[\/caption]\n\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 2<\/h3>\nGraph [latex]{x}^{2}=8y[\/latex]. Identify and label the focus, directrix, and endpoints of the latus rectum.\n\n<a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-25\/\" target=\"_blank\">Solution<\/a>\n\n<\/div>\n<h2>Writing Equations of Parabolas in Standard Form<\/h2>\nIn the previous examples, we used the standard form equation of a parabola to calculate the locations of its key features. We can also use the calculations in reverse to write an equation for a parabola when given its key features.\n<div class=\"textbox\">\n<h3>How To: Given its focus and directrix, write the equation for a parabola in standard form.<\/h3>\n<ul><li>Determine whether the axis of symmetry is the <em>x<\/em>- or <em>y<\/em>-axis.\n<ul><li>If the given coordinates of the focus have the form [latex]\\left(p,0\\right)[\/latex], then the axis of symmetry is the <em>x<\/em>-axis. Use the standard form [latex]{y}^{2}=4px[\/latex].<\/li>\n\t<li>If the given coordinates of the focus have the form [latex]\\left(0,p\\right)[\/latex], then the axis of symmetry is the <em>y<\/em>-axis. Use the standard form [latex]{x}^{2}=4py[\/latex].<\/li>\n<\/ul><\/li>\n\t<li>Multiply [latex]4p[\/latex].<\/li>\n\t<li>Substitute the value from Step 2 into the equation determined in Step 1.<\/li>\n<\/ul><\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Writing the Equation of a Parabola in Standard Form Given its Focus and Directrix<\/h3>\nWhat is the equation for the <strong>parabola<\/strong> with <strong>focus<\/strong> [latex]\\left(-\\frac{1}{2},0\\right)[\/latex] and <strong>directrix<\/strong> [latex]x=\\frac{1}{2}?[\/latex]\n\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\nThe focus has the form [latex]\\left(p,0\\right)[\/latex], so the equation will have the form [latex]{y}^{2}=4px[\/latex].\n<div>\n<div>Multiplying [latex]4p[\/latex], we have [latex]4p=4\\left(-\\frac{1}{2}\\right)=-2[\/latex].\u00a0Substituting for [latex]4p[\/latex], we have [latex]{y}^{2}=4px=-2x[\/latex].<\/div>\n<\/div>\nTherefore, the equation for the parabola is [latex]{y}^{2}=-2x[\/latex].\n\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 3<\/h3>\nWhat is the equation for the parabola with focus [latex]\\left(0,\\frac{7}{2}\\right)[\/latex] and directrix [latex]y=-\\frac{7}{2}?[\/latex]\n\n<a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-25\/\" target=\"_blank\">Solution<\/a>\n\n<\/div>","rendered":"<p>In <a class=\"target-chapter\" href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/introduction-to-the-ellipse\/\" target=\"_blank\">The Ellipse<\/a>, we saw that an <strong>ellipse<\/strong> is formed when a plane cuts through a right circular cone. If the plane is parallel to the edge of the cone, an unbounded curve is formed. This curve is a <strong>parabola<\/strong>.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202317\/CNX_Precalc_Figure_10_03_0022.jpg\" alt=\"\" width=\"487\" height=\"425\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2.<\/b> Parabola<\/p>\n<\/div>\n<p>Like the ellipse and <strong>hyperbola<\/strong>, the parabola can also be defined by a set of points in the coordinate plane. A parabola is the set of all points [latex]\\left(x,y\\right)[\/latex] in a plane that are the same distance from a fixed line, called the <strong>directrix<\/strong>, and a fixed point (the <strong>focus<\/strong>) not on the directrix.<\/p>\n<p>We previously learned about a parabola\u2019s vertex and axis of symmetry. Now we extend the discussion to include other key features of the parabola. See Figure 3. Notice that the axis of symmetry passes through the focus and vertex and is perpendicular to the directrix. The vertex is the midpoint between the directrix and the focus.<\/p>\n<p>The line segment that passes through the focus and is parallel to the directrix is called the <strong>latus rectum<\/strong>. The endpoints of the latus rectum lie on the curve. By definition, the distance [latex]d[\/latex] from the focus to any point [latex]P[\/latex] on the parabola is equal to the distance from [latex]P[\/latex] to the directrix.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202318\/CNX_Precalc_Figure_10_03_003n2.jpg\" alt=\"\" width=\"487\" height=\"291\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3.<\/b> Key features of the parabola<\/p>\n<\/div>\n<p>To work with parabolas in the <strong>coordinate plane<\/strong>, we consider two cases: those with a vertex at the origin and those with a <strong>vertex<\/strong> at a point other than the origin. We begin with the former.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202320\/CNX_Precalc_Figure_10_03_0182.jpg\" alt=\"\" width=\"487\" height=\"292\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<p>Let [latex]\\left(x,y\\right)[\/latex] be a point on the parabola with vertex [latex]\\left(0,0\\right)[\/latex], focus [latex]\\left(0,p\\right)[\/latex], and directrix [latex]y= -p[\/latex]\u00a0as shown in Figure 4. The distance [latex]d[\/latex] from point [latex]\\left(x,y\\right)[\/latex] to point [latex]\\left(x,-p\\right)[\/latex]\u00a0on the directrix is the difference of the <em>y<\/em>-values: [latex]d=y+p[\/latex]. The distance from the focus [latex]\\left(0,p\\right)[\/latex] to the point [latex]\\left(x,y\\right)[\/latex] is also equal to [latex]d[\/latex] and can be expressed using the <strong>distance formula<\/strong>.\n<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}d=\\sqrt{{\\left(x - 0\\right)}^{2}+{\\left(y-p\\right)}^{2}}\\hfill \\\\ =\\sqrt{{x}^{2}+{\\left(y-p\\right)}^{2}}\\hfill \\end{array}[\/latex]<\/div>\n<p>Set the two expressions for [latex]d[\/latex] equal to each other and solve for [latex]y[\/latex] to derive the equation of the parabola. We do this because the distance from [latex]\\left(x,y\\right)[\/latex] to [latex]\\left(0,p\\right)[\/latex] equals the distance from [latex]\\left(x,y\\right)[\/latex] to [latex]\\left(x, -p\\right)[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\sqrt{{x}^{2}+{\\left(y-p\\right)}^{2}}=y+p[\/latex]<\/div>\n<p>We then square both sides of the equation, expand the squared terms, and simplify by combining like terms.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}{x}^{2}+{\\left(y-p\\right)}^{2}={\\left(y+p\\right)}^{2}\\\\ {x}^{2}+{y}^{2}-2py+{p}^{2}={y}^{2}+2py+{p}^{2}\\\\ {x}^{2}-2py=2py\\\\ \\text{ }{x}^{2}=4py\\end{array}[\/latex]<\/div>\n<p>The equations of parabolas with vertex [latex]\\left(0,0\\right)[\/latex] are [latex]{y}^{2}=4px[\/latex] when the <em>x<\/em>-axis is the axis of symmetry and [latex]{x}^{2}=4py[\/latex] when the <em>y<\/em>-axis is the axis of symmetry. These standard forms are given below, along with their general graphs and key features.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Standard Forms of Parabolas with Vertex (0, 0)<\/h3>\n<p>The table below\u00a0and Figure 5\u00a0summarize the standard features of parabolas with a vertex at the origin.<\/p>\n<table id=\"Table_10_03_01\" summary=\"..\">\n<tbody>\n<tr>\n<td><strong>Axis of Symmetry<\/strong><\/td>\n<td><strong>Equation<\/strong><\/td>\n<td><strong>Focus<\/strong><\/td>\n<td><strong>Directrix<\/strong><\/td>\n<td><strong>Endpoints of Latus Rectum<\/strong><\/td>\n<\/tr>\n<tr>\n<td><em>x<\/em>-axis<\/td>\n<td>[latex]{y}^{2}=4px[\/latex]<\/td>\n<td>[latex]\\left(p,\\text{ }0\\right)[\/latex]<\/td>\n<td>[latex]x=-p[\/latex]<\/td>\n<td>[latex]\\left(p,\\text{ }\\pm 2p\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><em>y<\/em>-axis<\/td>\n<td>[latex]{x}^{2}=4py[\/latex]<\/td>\n<td>[latex]\\left(0,\\text{ }p\\right)[\/latex]<\/td>\n<td>[latex]y=-p[\/latex]<\/td>\n<td>[latex]\\left(\\pm 2p,\\text{ }p\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202321\/CNX_Precalc_Figure_10_03_004n2.jpg\" alt=\"\" width=\"975\" height=\"721\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5.<\/b> (a) When [latex]p&gt;0[\/latex] and the axis of symmetry is the x-axis, the parabola opens right. (b) When [latex]p&lt;0[\/latex] and the axis of symmetry is the x-axis, the parabola opens left. (c) When [latex]p&lt;0[\/latex] and the axis of symmetry is the y-axis, the parabola opens up. (d) When [latex]\\text{ }p&lt;0\\text{ }[\/latex] and the axis of symmetry is the y-axis, the parabola opens down.<\/p>\n<\/div>\n<\/div>\n<p>The key features of a parabola are its vertex, axis of symmetry, focus, directrix, and latus rectum. When given a standard equation for a parabola centered at the origin, we can easily identify the key features to graph the parabola.<\/p>\n<p>A line is said to be tangent to a curve if it intersects the curve at exactly one point. If we sketch lines tangent to the parabola at the endpoints of the latus rectum, these lines intersect on the axis of symmetry, as shown in Figure 6.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202322\/CNX_Precalc_Figure_10_03_0052.jpg\" alt=\"\" width=\"487\" height=\"514\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a standard form equation for a parabola centered at (0, 0), sketch the graph.<\/h3>\n<ul>\n<li>Determine which of the standard forms applies to the given equation: [latex]{y}^{2}=4px[\/latex] or [latex]{x}^{2}=4py[\/latex].<\/li>\n<li>Use the standard form identified in Step 1 to determine the axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum.\n<ul>\n<li>If the equation is in the form [latex]{y}^{2}=4px[\/latex], then\n<ul>\n<li>the axis of symmetry is the <em>x<\/em>-axis, [latex]y=0[\/latex]<\/li>\n<li>set [latex]4p[\/latex] equal to the coefficient of <em>x <\/em>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<li>use [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(p,0\\right)[\/latex]<\/li>\n<li>use [latex]p[\/latex] to find the equation of the directrix, [latex]x=-p[\/latex]<\/li>\n<li>use [latex]p[\/latex] to find the endpoints of the latus rectum, [latex]\\left(p,\\pm 2p\\right)[\/latex]. Alternately, substitute [latex]x=p[\/latex] into the original equation.<\/li>\n<\/ul>\n<\/li>\n<li>If the equation is in the form [latex]{x}^{2}=4py[\/latex], then\n<ul>\n<li>the axis of symmetry is the <em>y<\/em>-axis, [latex]x=0[\/latex]<\/li>\n<li>set [latex]4p[\/latex] equal to the coefficient of <em>y <\/em>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<li>use [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(0,p\\right)[\/latex]<\/li>\n<li>use [latex]p[\/latex] to find equation of the directrix, [latex]y=-p[\/latex]<\/li>\n<li>use [latex]p[\/latex] to find the endpoints of the latus rectum, [latex]\\left(\\pm 2p,p\\right)[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<li>Plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Graphing a Parabola with Vertex (0, 0) and the <em>x<\/em>-axis as the Axis of Symmetry<\/h3>\n<p>Graph [latex]{y}^{2}=24x[\/latex]. Identify and label the <strong>focus<\/strong>, <strong>directrix<\/strong>, and endpoints of the <strong>latus rectum<\/strong>.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\n<p>The standard form that applies to the given equation is [latex]{y}^{2}=4px[\/latex]. Thus, the axis of symmetry is the <em>x<\/em>-axis. It follows that:<\/p>\n<div>\n<ul>\n<li>[latex]24=4p[\/latex], so [latex]p=6[\/latex]. Since [latex]p>0[\/latex], the parabola opens right\u00a0the coordinates of the focus are [latex]\\left(p,0\\right)=\\left(6,0\\right)[\/latex]<\/li>\n<li>the equation of the directrix is [latex]x=-p=-6[\/latex]<\/li>\n<li>the endpoints of the latus rectum have the same <em>x<\/em>-coordinate at the focus. To find the endpoints, substitute [latex]x=6[\/latex] into the original equation: [latex]\\left(6,\\pm 12\\right)[\/latex]<\/li>\n<\/ul>\n<\/div>\n<p>Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the <strong>parabola<\/strong>.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202323\/CNX_Precalc_Figure_10_03_0192.jpg\" alt=\"\" width=\"487\" height=\"376\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 7<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 1<\/h3>\n<p>Graph [latex]{y}^{2}=-16x[\/latex]. Identify and label the focus, directrix, and endpoints of the latus rectum.<\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-25\/\" target=\"_blank\">Solution<\/a><\/p>\n<\/div>\n<div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Graphing a Parabola with Vertex (0, 0) and the <em>y<\/em>-axis as the Axis of Symmetry<\/h3>\n<p>Graph [latex]{x}^{2}=-6y[\/latex]. Identify and label the <strong>focus<\/strong>, <strong>directrix<\/strong>, and endpoints of the <strong>latus rectum<\/strong>.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\n<p>The standard form that applies to the given equation is [latex]{x}^{2}=4py[\/latex]. Thus, the axis of symmetry is the <em>y<\/em>-axis. It follows that:<\/p>\n<div>\n<ul>\n<li>[latex]-6=4p[\/latex], so [latex]p=-\\frac{3}{2}[\/latex]. Since [latex]p<0[\/latex], the parabola opens down.<\/li>\n<li>the coordinates of the focus are [latex]\\left(0,p\\right)=\\left(0,-\\frac{3}{2}\\right)[\/latex]<\/li>\n<li>the equation of the directrix is [latex]y=-p=\\frac{3}{2}[\/latex]<\/li>\n<li>the endpoints of the latus rectum can be found by substituting [latex]\\text{ }y=\\frac{3}{2}\\text{ }[\/latex] into the original equation, [latex]\\left(\\pm 3,-\\frac{3}{2}\\right)[\/latex]<\/li>\n<\/ul>\n<\/div>\n<p>Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the <strong>parabola<\/strong>.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202325\/CNX_Precalc_Figure_10_03_007n2.jpg\" alt=\"\" width=\"487\" height=\"327\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 8<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 2<\/h3>\n<p>Graph [latex]{x}^{2}=8y[\/latex]. Identify and label the focus, directrix, and endpoints of the latus rectum.<\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-25\/\" target=\"_blank\">Solution<\/a><\/p>\n<\/div>\n<h2>Writing Equations of Parabolas in Standard Form<\/h2>\n<p>In the previous examples, we used the standard form equation of a parabola to calculate the locations of its key features. We can also use the calculations in reverse to write an equation for a parabola when given its key features.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given its focus and directrix, write the equation for a parabola in standard form.<\/h3>\n<ul>\n<li>Determine whether the axis of symmetry is the <em>x<\/em>&#8211; or <em>y<\/em>-axis.\n<ul>\n<li>If the given coordinates of the focus have the form [latex]\\left(p,0\\right)[\/latex], then the axis of symmetry is the <em>x<\/em>-axis. Use the standard form [latex]{y}^{2}=4px[\/latex].<\/li>\n<li>If the given coordinates of the focus have the form [latex]\\left(0,p\\right)[\/latex], then the axis of symmetry is the <em>y<\/em>-axis. Use the standard form [latex]{x}^{2}=4py[\/latex].<\/li>\n<\/ul>\n<\/li>\n<li>Multiply [latex]4p[\/latex].<\/li>\n<li>Substitute the value from Step 2 into the equation determined in Step 1.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Writing the Equation of a Parabola in Standard Form Given its Focus and Directrix<\/h3>\n<p>What is the equation for the <strong>parabola<\/strong> with <strong>focus<\/strong> [latex]\\left(-\\frac{1}{2},0\\right)[\/latex] and <strong>directrix<\/strong> [latex]x=\\frac{1}{2}?[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\n<p>The focus has the form [latex]\\left(p,0\\right)[\/latex], so the equation will have the form [latex]{y}^{2}=4px[\/latex].<\/p>\n<div>\n<div>Multiplying [latex]4p[\/latex], we have [latex]4p=4\\left(-\\frac{1}{2}\\right)=-2[\/latex].\u00a0Substituting for [latex]4p[\/latex], we have [latex]{y}^{2}=4px=-2x[\/latex].<\/div>\n<\/div>\n<p>Therefore, the equation for the parabola is [latex]{y}^{2}=-2x[\/latex].<\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 3<\/h3>\n<p>What is the equation for the parabola with focus [latex]\\left(0,\\frac{7}{2}\\right)[\/latex] and directrix [latex]y=-\\frac{7}{2}?[\/latex]<\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-25\/\" target=\"_blank\">Solution<\/a><\/p>\n<\/div>\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-1914\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><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":276,"menu_order":2,"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\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1914","chapter","type-chapter","status-publish","hentry"],"part":1904,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1914","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/users\/276"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1914\/revisions"}],"predecessor-version":[{"id":2205,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1914\/revisions\/2205"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/1904"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1914\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/media?parent=1914"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1914"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/contributor?post=1914"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/license?post=1914"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}