{"id":250,"date":"2025-02-13T22:45:34","date_gmt":"2025-02-13T22:45:34","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/the-parabola\/"},"modified":"2025-10-22T23:15:14","modified_gmt":"2025-10-22T23:15:14","slug":"the-parabola","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/the-parabola\/","title":{"raw":"Parabolas: Learn It 1","rendered":"Parabolas: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Identify the vertex, focus, directrix, and endpoints of the latus rectum.<\/li>\r\n \t<li>Write equations of parabolas in standard form.<\/li>\r\n \t<li>Graph parabolas.<\/li>\r\n \t<li>Solve applied problems involving parabolas.<\/li>\r\n<\/ul>\r\n<\/section>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182915\/CNX_Precalc_Figure_10_03_001n2.jpg\" alt=\"Description in caption\" width=\"487\" height=\"325\" \/> The Olympic torch concludes its journey around the world when it is used to light the Olympic cauldron during the opening ceremony. (credit: Ken Hackman, U.S. Air Force)[\/caption]\r\n\r\nDid you know that the Olympic torch is lit several months before the start of the games? The ceremonial method for lighting the flame is the same as in ancient times. The ceremony takes place at the Temple of Hera in Olympia, Greece, and is rooted in Greek mythology, paying tribute to Prometheus, who stole fire from Zeus to give to all humans. One of eleven acting priestesses places the torch at the focus of a parabolic mirror, which focuses light rays from the sun to ignite the flame.\r\n\r\nParabolic mirrors (or reflectors) are able to capture energy and focus it to a single point. The advantages of this property are evidenced by the vast list of parabolic objects we use every day: satellite dishes, suspension bridges, telescopes, microphones, spotlights, and car headlights, to name a few. Parabolic reflectors are also used in alternative energy devices, such as solar cookers and water heaters, because they are inexpensive to manufacture and need little maintenance. In this section we will explore the parabola and its uses, including low-cost, energy-efficient solar designs.\r\n<h2>Graphing Parabolas with Vertices at the Origin<\/h2>\r\nWhen a plane cuts through a cone and is parallel to the edge of the cone, an unbounded curve is formed. This curve is a <strong>parabola<\/strong>.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182917\/CNX_Precalc_Figure_10_03_0022.jpg\" alt=\"\" width=\"487\" height=\"425\" \/> Parabola[\/caption]\r\n\r\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.\r\n\r\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. 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.\r\n\r\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.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182919\/CNX_Precalc_Figure_10_03_003n2.jpg\" alt=\"\" width=\"487\" height=\"291\" \/> Key features of the parabola[\/caption]\r\n\r\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.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182921\/CNX_Precalc_Figure_10_03_0182.jpg\" alt=\"\" width=\"487\" height=\"292\" \/>\r\n\r\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]. The distance [latex]d[\/latex] from point [latex]\\left(x,y\\right)[\/latex] to point [latex]\\left(x,-p\\right)[\/latex] on 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>.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}d&amp;=\\sqrt{{\\left(x - 0\\right)}^{2}+{\\left(y-p\\right)}^{2}} \\\\ &amp;=\\sqrt{{x}^{2}+{\\left(y-p\\right)}^{2}} \\end{align}[\/latex]<\/div>\r\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].\r\n<div style=\"text-align: center;\">[latex]\\sqrt{{x}^{2}+{\\left(y-p\\right)}^{2}}=y+p[\/latex]<\/div>\r\nWe then square both sides of the equation, expand the squared terms, and simplify by combining like terms.\r\n<div style=\"text-align: center;\">[latex]\\begin{gathered}{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\\\\ {x}^{2}=4py\\end{gathered}[\/latex]<\/div>\r\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.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>A General Note: Standard Forms of Parabolas with Vertex (0, 0)<\/h3>\r\nThe table and graph below summarize the standard features of parabolas with a vertex at the origin.\r\n<table id=\"Table_10_03_01\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td><strong>Axis of Symmetry<\/strong><\/td>\r\n<td><strong>Equation<\/strong><\/td>\r\n<td><strong>Focus<\/strong><\/td>\r\n<td><strong>Directrix<\/strong><\/td>\r\n<td><strong>Endpoints of Latus Rectum<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>x<\/em>-axis<\/td>\r\n<td>[latex]{y}^{2}=4px[\/latex]<\/td>\r\n<td>[latex]\\left(p,\\text{ }0\\right)[\/latex]<\/td>\r\n<td>[latex]x=-p[\/latex]<\/td>\r\n<td>[latex]\\left(p,\\text{ }\\pm 2p\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>y<\/em>-axis<\/td>\r\n<td>[latex]{x}^{2}=4py[\/latex]<\/td>\r\n<td>[latex]\\left(0,\\text{ }p\\right)[\/latex]<\/td>\r\n<td>[latex]y=-p[\/latex]<\/td>\r\n<td>[latex]\\left(\\pm 2p,\\text{ }p\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182924\/CNX_Precalc_Figure_10_03_004n2.jpg\" alt=\"\" width=\"975\" height=\"721\" \/> (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]Type your <em>Key Takeaway<\/em> text here<\/section>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.\r\n\r\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.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182926\/CNX_Precalc_Figure_10_03_0052.jpg\" alt=\"\" width=\"487\" height=\"514\" \/>\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a standard form equation for a parabola centered at (0, 0), sketch the graph.<\/strong>\r\n<ul>\r\n \t<li>Determine which of the standard forms applies to the given equation: [latex]{y}^{2}=4px[\/latex] or [latex]{x}^{2}=4py[\/latex].<\/li>\r\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.\r\n<ul>\r\n \t<li>If the equation is in the form [latex]{y}^{2}=4px[\/latex], then\r\n<ul>\r\n \t<li>the axis of symmetry is the <em>x<\/em>-axis, [latex]y=0[\/latex]<\/li>\r\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>\r\n \t<li>use [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(p,0\\right)[\/latex]<\/li>\r\n \t<li>use [latex]p[\/latex] to find the equation of the directrix, [latex]x=-p[\/latex]<\/li>\r\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>\r\n<\/ul>\r\n<\/li>\r\n \t<li>If the equation is in the form [latex]{x}^{2}=4py[\/latex], then\r\n<ul>\r\n \t<li>the axis of symmetry is the <em>y<\/em>-axis, [latex]x=0[\/latex]<\/li>\r\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>\r\n \t<li>use [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(0,p\\right)[\/latex]<\/li>\r\n \t<li>use [latex]p[\/latex] to find equation of the directrix, [latex]y=-p[\/latex]<\/li>\r\n \t<li>use [latex]p[\/latex] to find the endpoints of the latus rectum, [latex]\\left(\\pm 2p,p\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Graph [latex]{y}^{2}=24x[\/latex]. Identify and label the <strong>focus<\/strong>, <strong>directrix<\/strong>, and endpoints of the <strong>latus rectum<\/strong>.[reveal-answer q=\"325223\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"325223\"]\r\n\r\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:\r\n<div>\r\n<ul>\r\n \t<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>\r\n \t<li>the equation of the directrix is [latex]x=-p=-6[\/latex]<\/li>\r\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>\r\n<\/ul>\r\n<\/div>\r\nNext we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the <strong>parabola<\/strong>.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182929\/CNX_Precalc_Figure_10_03_0192.jpg\" alt=\"\" width=\"487\" height=\"376\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<div class=\"bcc-box bcc-success\">\r\n\r\nGraph [latex]{y}^{2}=-16x[\/latex]. Identify and label the focus, directrix, and endpoints of the latus rectum.\r\n\r\n[reveal-answer q=\"373133\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"373133\"]\r\n\r\nFocus: [latex]\\left(-4,0\\right)[\/latex]; Directrix: [latex]x=4[\/latex]; Endpoints of the latus rectum: [latex]\\left(-4,\\pm 8\\right)[\/latex]\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182957\/CNX_Precalc_Figure_10_03_0062.jpg\" alt=\"\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Graph [latex]{x}^{2}=-6y[\/latex]. Identify and label the <strong>focus<\/strong>, <strong>directrix<\/strong>, and endpoints of the <strong>latus rectum<\/strong>.[reveal-answer q=\"659906\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"659906\"]\r\n\r\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:\r\n<div>\r\n<ul>\r\n \t<li>[latex]-6=4p[\/latex], so [latex]p=-\\frac{3}{2}[\/latex]. Since [latex]p&lt;0[\/latex], the parabola opens down.<\/li>\r\n \t<li>the coordinates of the focus are [latex]\\left(0,p\\right)=\\left(0,-\\frac{3}{2}\\right)[\/latex]<\/li>\r\n \t<li>the equation of the directrix is [latex]y=-p=\\frac{3}{2}[\/latex]<\/li>\r\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>\r\n<\/ul>\r\n<\/div>\r\nNext we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the <strong>parabola<\/strong>.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182931\/CNX_Precalc_Figure_10_03_007n2.jpg\" alt=\"\" width=\"487\" height=\"327\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<div class=\"bcc-box bcc-success\">\r\n\r\nGraph [latex]{x}^{2}=8y[\/latex]. Identify and label the focus, directrix, and endpoints of the latus rectum.\r\n\r\n[reveal-answer q=\"588760\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"588760\"]\r\n\r\nFocus: [latex]\\left(0,2\\right)[\/latex]; Directrix: [latex]y=-2[\/latex]; Endpoints of the latus rectum: [latex]\\left(\\pm 4,2\\right)[\/latex].\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182959\/CNX_Precalc_Figure_10_03_0082.jpg\" alt=\"\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]174039[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Identify the vertex, focus, directrix, and endpoints of the latus rectum.<\/li>\n<li>Write equations of parabolas in standard form.<\/li>\n<li>Graph parabolas.<\/li>\n<li>Solve applied problems involving parabolas.<\/li>\n<\/ul>\n<\/section>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182915\/CNX_Precalc_Figure_10_03_001n2.jpg\" alt=\"Description in caption\" width=\"487\" height=\"325\" \/><figcaption class=\"wp-caption-text\">The Olympic torch concludes its journey around the world when it is used to light the Olympic cauldron during the opening ceremony. (credit: Ken Hackman, U.S. Air Force)<\/figcaption><\/figure>\n<p>Did you know that the Olympic torch is lit several months before the start of the games? The ceremonial method for lighting the flame is the same as in ancient times. The ceremony takes place at the Temple of Hera in Olympia, Greece, and is rooted in Greek mythology, paying tribute to Prometheus, who stole fire from Zeus to give to all humans. One of eleven acting priestesses places the torch at the focus of a parabolic mirror, which focuses light rays from the sun to ignite the flame.<\/p>\n<p>Parabolic mirrors (or reflectors) are able to capture energy and focus it to a single point. The advantages of this property are evidenced by the vast list of parabolic objects we use every day: satellite dishes, suspension bridges, telescopes, microphones, spotlights, and car headlights, to name a few. Parabolic reflectors are also used in alternative energy devices, such as solar cookers and water heaters, because they are inexpensive to manufacture and need little maintenance. In this section we will explore the parabola and its uses, including low-cost, energy-efficient solar designs.<\/p>\n<h2>Graphing Parabolas with Vertices at the Origin<\/h2>\n<p>When a plane cuts through a cone and is parallel to the edge of the cone, an unbounded curve is formed. This curve is a <strong>parabola<\/strong>.<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182917\/CNX_Precalc_Figure_10_03_0022.jpg\" alt=\"\" width=\"487\" height=\"425\" \/><figcaption class=\"wp-caption-text\">Parabola<\/figcaption><\/figure>\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. 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<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182919\/CNX_Precalc_Figure_10_03_003n2.jpg\" alt=\"\" width=\"487\" height=\"291\" \/><figcaption class=\"wp-caption-text\">Key features of the parabola<\/figcaption><\/figure>\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<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182921\/CNX_Precalc_Figure_10_03_0182.jpg\" alt=\"\" width=\"487\" height=\"292\" \/><\/p>\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]. The distance [latex]d[\/latex] from point [latex]\\left(x,y\\right)[\/latex] to point [latex]\\left(x,-p\\right)[\/latex] on 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>.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}d&=\\sqrt{{\\left(x - 0\\right)}^{2}+{\\left(y-p\\right)}^{2}} \\\\ &=\\sqrt{{x}^{2}+{\\left(y-p\\right)}^{2}} \\end{align}[\/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{gathered}{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\\\\ {x}^{2}=4py\\end{gathered}[\/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<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>A General Note: Standard Forms of Parabolas with Vertex (0, 0)<\/h3>\n<p>The table and graph below summarize 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<figure style=\"width: 975px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182924\/CNX_Precalc_Figure_10_03_004n2.jpg\" alt=\"\" width=\"975\" height=\"721\" \/><figcaption class=\"wp-caption-text\">(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.<\/figcaption><\/figure>\n<p>Type your <em>Key Takeaway<\/em> text here<\/section>\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.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182926\/CNX_Precalc_Figure_10_03_0052.jpg\" alt=\"\" width=\"487\" height=\"514\" \/><\/p>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a standard form equation for a parabola centered at (0, 0), sketch the graph.<\/strong><\/p>\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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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 class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q325223\">Show Solution<\/button><\/p>\n<div id=\"q325223\" class=\"hidden-answer\" style=\"display: none\">\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<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182929\/CNX_Precalc_Figure_10_03_0192.jpg\" alt=\"\" width=\"487\" height=\"376\" \/><\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<div class=\"bcc-box bcc-success\">\n<p>Graph [latex]{y}^{2}=-16x[\/latex]. Identify and label the focus, directrix, and endpoints of the latus rectum.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q373133\">Show Solution<\/button><\/p>\n<div id=\"q373133\" class=\"hidden-answer\" style=\"display: none\">\n<p>Focus: [latex]\\left(-4,0\\right)[\/latex]; Directrix: [latex]x=4[\/latex]; Endpoints of the latus rectum: [latex]\\left(-4,\\pm 8\\right)[\/latex]<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182957\/CNX_Precalc_Figure_10_03_0062.jpg\" alt=\"\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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 class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q659906\">Show Solution<\/button><\/p>\n<div id=\"q659906\" class=\"hidden-answer\" style=\"display: none\">\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<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182931\/CNX_Precalc_Figure_10_03_007n2.jpg\" alt=\"\" width=\"487\" height=\"327\" \/><\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<div class=\"bcc-box bcc-success\">\n<p>Graph [latex]{x}^{2}=8y[\/latex]. Identify and label the focus, directrix, and endpoints of the latus rectum.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q588760\">Show Solution<\/button><\/p>\n<div id=\"q588760\" class=\"hidden-answer\" style=\"display: none\">\n<p>Focus: [latex]\\left(0,2\\right)[\/latex]; Directrix: [latex]y=-2[\/latex]; Endpoints of the latus rectum: [latex]\\left(\\pm 4,2\\right)[\/latex].<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27182959\/CNX_Precalc_Figure_10_03_0082.jpg\" alt=\"\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm174039\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174039&theme=lumen&iframe_resize_id=ohm174039&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":6,"menu_order":17,"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\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":522,"module-header":"learn_it","content_attributions":[{"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":""}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/250"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":8,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/250\/revisions"}],"predecessor-version":[{"id":4838,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/250\/revisions\/4838"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/522"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/250\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=250"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=250"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=250"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=250"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}