{"id":5240,"date":"2018-05-10T21:09:01","date_gmt":"2018-05-10T21:09:01","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/coreq-collegealgebra\/?post_type=chapter&#038;p=5240"},"modified":"2018-05-21T19:35:53","modified_gmt":"2018-05-21T19:35:53","slug":"the-parabola","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/chapter\/the-parabola\/","title":{"raw":"The Parabola*","rendered":"The Parabola*"},"content":{"raw":"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.\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\/896\/2016\/11\/03204531\/CNX_Precalc_Figure_10_03_001n2.jpg\" alt=\"Description in caption\" width=\"487\" height=\"325\" data-media-type=\"image\/jpg\" \/> 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\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>Parabolas with Vertices at the Origin<\/h2>\r\nIn The Ellipse, 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>.\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\/896\/2016\/11\/03204534\/CNX_Precalc_Figure_10_03_0022.jpg\" alt=\"\" width=\"487\" height=\"425\" data-media-type=\"image\/jpg\" \/> 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,\u00a0<\/strong>also called the\u00a0<strong>focal diameter<\/strong>. The endpoints of the focal diameter\u00a0lie 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\/896\/2016\/11\/03204536\/CNX_Precalc_Figure_10_03_003n2.jpg\" alt=\"\" width=\"487\" height=\"291\" data-media-type=\"image\/jpg\" \/> 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 class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204538\/CNX_Precalc_Figure_10_03_0182.jpg\" width=\"487\" height=\"292\" data-media-type=\"image\/jpg\" \/>\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]\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>.\r\n<p 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]<\/p>\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[latex]\\sqrt{{x}^{2}+{\\left(y-p\\right)}^{2}}=y+p[\/latex]\r\n\r\nWe then square both sides of the equation, expand the squared terms, and simplify by combining like terms.\r\n<p 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]<\/p>\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<div class=\"textbox\">\r\n<h3>A General Note: Standard Forms of Parabolas with Vertex (0, 0)<\/h3>\r\nThe table below summarizes the standard features of parabolas with a vertex at the origin.\r\n<table 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\u00a0Focal Diameter<\/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\/896\/2016\/11\/03204540\/CNX_Precalc_Figure_10_03_004n2.jpg\" alt=\"\" width=\"975\" height=\"721\" data-media-type=\"image\/jpg\" \/> (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>\r\nThe key features of a parabola are its vertex, axis of symmetry, focus, directrix, and focal diameter. 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 focal diameter, these lines intersect on the axis of symmetry.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204543\/CNX_Precalc_Figure_10_03_0052.jpg\" width=\"487\" height=\"514\" data-media-type=\"image\/jpg\" \/>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a standard form equation for a parabola centered at (0, 0), sketch the graph.<\/h3>\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 focal diameter.\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 focal diameter, [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 focal diameter, [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 focal diameter, and draw a smooth curve to form the parabola.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Graphing a Parabola with Vertex (0, 0) and the <em>x<\/em>-axis as the Axis of Symmetry<\/h3>\r\nGraph [latex]{y}^{2}=24x[\/latex]. Identify and label the <strong>focus<\/strong>, <strong>directrix<\/strong>, and endpoints of the <strong>focal diameter<\/strong>.\r\n\r\n[reveal-answer q=\"885427\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"885427\"]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:\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 focal diameter\u00a0have 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\nNext we plot the focus, directrix, and focal diameter, and draw a smooth curve to form the <strong>parabola<\/strong>.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204545\/CNX_Precalc_Figure_10_03_0192.jpg\" width=\"487\" height=\"376\" data-media-type=\"image\/jpg\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGraph [latex]{y}^{2}=-16x[\/latex]. Identify and label the focus, directrix, and endpoints of the focal diameter.\r\n\r\n[reveal-answer q=\"277220\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"277220\"]\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<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02180919\/leftopen.jpg\"><img class=\"aligncenter size-full wp-image-3276\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02180919\/leftopen.jpg\" alt=\"\" width=\"487\" height=\"366\" \/><\/a>\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=23511&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"550\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Graphing a Parabola with Vertex (0, 0) and the <em>y<\/em>-axis as the Axis of Symmetry<\/h3>\r\nGraph [latex]{x}^{2}=-6y[\/latex]. Identify and label the <strong>focus<\/strong>, <strong>directrix<\/strong>, and endpoints of the <strong>focal diameter<\/strong>.\r\n\r\n[reveal-answer q=\"951555\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"951555\"]\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<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 focal diameter 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\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 class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204547\/CNX_Precalc_Figure_10_03_007n2.jpg\" width=\"487\" height=\"327\" data-media-type=\"image\/jpg\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGraph [latex]{x}^{2}=8y[\/latex]. Identify and label the focus, directrix, and endpoints of the focal diameter.\r\n\r\n[reveal-answer q=\"824847\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"824847\"]Focus: [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\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02181513\/CNX_Precalc_Figure_10_03_0082.jpg\"><img class=\"aligncenter size-full wp-image-3277\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02181513\/CNX_Precalc_Figure_10_03_0082.jpg\" alt=\"\" width=\"487\" height=\"365\" \/><\/a>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nIn the graph below you will find the plot of a parabola whose axis of symmetry is the x-axis. The free variable p is allowed to slide between [latex]-10,10[\/latex]. Your task in this exercise is to add\u00a0the focus, directrix, and endpoints of the focal diameter in terms of the free variable, p.\r\n\r\nFor example, to add the focus, you would define a point, [latex](p,0)[\/latex] .\r\n\r\nhttps:\/\/www.desmos.com\/calculator\/wunbnybenw\r\n[reveal-answer q=\"864413\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"864413\"]\r\n\r\nhttps:\/\/www.desmos.com\/calculator\/b3buagwzwl\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>Writing Equations of Parabolas in Standard Form<\/h2>\r\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.\r\n<div class=\"textbox\">\r\n<h3>How To: Given its focus and directrix, write the equation for a parabola in standard form.<\/h3>\r\n<ul>\r\n \t<li>Determine whether the axis of symmetry is the <em>x<\/em>- or <em>y<\/em>-axis.\r\n<ul>\r\n \t<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>\r\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>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Multiply [latex]4p[\/latex].<\/li>\r\n \t<li>Substitute the value from Step 2 into the equation determined in Step 1.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Writing the Equation of a Parabola in Standard Form Given its Focus and Directrix<\/h3>\r\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]\r\n\r\n[reveal-answer q=\"259208\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"259208\"]\r\n\r\nThe focus has the form [latex]\\left(p,0\\right)[\/latex], so the equation will have the form [latex]{y}^{2}=4px[\/latex].\r\n\r\nMultiplying [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].\r\n\r\nTherefore, the equation for the parabola is [latex]{y}^{2}=-2x[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\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]?\r\n\r\n[reveal-answer q=\"886076\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"886076\"][latex]{x}^{2}=14y[\/latex][\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29711&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Parabolas with Vertices Not at the Origin<\/h2>\r\nLike other graphs we\u2019ve worked with, the graph of a parabola can be translated. If a parabola is translated [latex]h[\/latex] units horizontally and [latex]k[\/latex] units vertically, the vertex will be [latex]\\left(h,k\\right)[\/latex]. This translation results in the standard form of the equation we saw previously with [latex]x[\/latex] replaced by [latex]\\left(x-h\\right)[\/latex] and [latex]y[\/latex] replaced by [latex]\\left(y-k\\right)[\/latex].\r\n\r\nTo graph parabolas with a vertex [latex]\\left(h,k\\right)[\/latex] other than the origin, we use the standard form [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex] for parabolas that have an axis of symmetry parallel to the <em>x<\/em>-axis, and [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex] for parabolas that have an axis of symmetry parallel to the <em>y<\/em>-axis. These standard forms are given below, along with their general graphs and key features.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Standard Forms of Parabolas with Vertex (<em>h<\/em>, <em>k<\/em>)<\/h3>\r\nThe table summarizes the standard features of parabolas with a vertex at a point [latex]\\left(h,k\\right)[\/latex].\r\n<table 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 focal diameter<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]y=k[\/latex]<\/td>\r\n<td>[latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(h+p,\\text{ }k\\right)[\/latex]<\/td>\r\n<td>[latex]x=h-p[\/latex]<\/td>\r\n<td>[latex]\\left(h+p,\\text{ }k\\pm 2p\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]x=h[\/latex]<\/td>\r\n<td>[latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(h,\\text{ }k+p\\right)[\/latex]<\/td>\r\n<td>[latex]y=k-p[\/latex]<\/td>\r\n<td>[latex]\\left(h\\pm 2p,\\text{ }k+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\/896\/2016\/11\/03204550\/CNX_Precalc_Figure_10_03_0092.jpg\" alt=\"\" width=\"975\" height=\"901\" data-media-type=\"image\/jpg\" \/> (a) When [latex]p&gt;0[\/latex], the parabola opens right. (b) When [latex]p&lt;0[\/latex], the parabola opens left. (c) When [latex]p&gt;0[\/latex], the parabola opens up. (d) When [latex]p&lt;0[\/latex], the parabola opens down.[\/caption]<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a standard form equation for a parabola centered at (<em>h<\/em>, <em>k<\/em>), sketch the graph.<\/h3>\r\n<ol>\r\n \t<li>Determine which of the standard forms applies to the given equation: [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex] or [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex].<\/li>\r\n \t<li>Use the standard form identified in Step 1 to determine the vertex, axis of symmetry, focus, equation of the directrix, and endpoints of the focal diameter.\r\n<ol>\r\n \t<li>If the equation is in the form [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex], then:\r\n<ul>\r\n \t<li>use the given equation to identify [latex]h[\/latex] and [latex]k[\/latex] for the vertex, [latex]\\left(h,k\\right)[\/latex]<\/li>\r\n \t<li>use the value of [latex]k[\/latex] to determine the axis of symmetry, [latex]y=k[\/latex]<\/li>\r\n \t<li>set [latex]4p[\/latex] equal to the coefficient of [latex]\\left(x-h\\right)[\/latex] 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]h,k[\/latex], and [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(h+p,\\text{ }k\\right)[\/latex]<\/li>\r\n \t<li>use [latex]h[\/latex] and [latex]p[\/latex] to find the equation of the directrix, [latex]x=h-p[\/latex]<\/li>\r\n \t<li>use [latex]h,k[\/latex], and [latex]p[\/latex] to find the endpoints of the focal diameter, [latex]\\left(h+p,k\\pm 2p\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>If the equation is in the form [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex], then:\r\n<ul>\r\n \t<li>use the given equation to identify [latex]h[\/latex] and [latex]k[\/latex] for the vertex, [latex]\\left(h,k\\right)[\/latex]<\/li>\r\n \t<li>use the value of [latex]h[\/latex] to determine the axis of symmetry, [latex]x=h[\/latex]<\/li>\r\n \t<li>set [latex]4p[\/latex] equal to the coefficient of [latex]\\left(y-k\\right)[\/latex] 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]h,k[\/latex], and [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(h,\\text{ }k+p\\right)[\/latex]<\/li>\r\n \t<li>use [latex]k[\/latex] and [latex]p[\/latex] to find the equation of the directrix, [latex]y=k-p[\/latex]<\/li>\r\n \t<li>use [latex]h,k[\/latex], and [latex]p[\/latex] to find the endpoints of the focal diameter, [latex]\\left(h\\pm 2p,\\text{ }k+p\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Graphing a Parabola with Vertex (<em>h<\/em>, <em>k<\/em>) and Axis of Symmetry Parallel to the <em>x<\/em>-axis<\/h3>\r\nGraph [latex]{\\left(y - 1\\right)}^{2}=-16\\left(x+3\\right)[\/latex]. Identify and label the <strong>vertex<\/strong>, <strong>axis of symmetry<\/strong>, <strong>focus<\/strong>, <strong>directrix<\/strong>, and endpoints of the <strong>focal diameter<\/strong>.\r\n\r\n[reveal-answer q=\"943688\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"943688\"]\r\n\r\nThe standard form that applies to the given equation is [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex]. Thus, the axis of symmetry is parallel to the <em>x<\/em>-axis. It follows that:\r\n<ul>\r\n \t<li>the vertex is [latex]\\left(h,k\\right)=\\left(-3,1\\right)[\/latex]<\/li>\r\n \t<li>the axis of symmetry is [latex]y=k=1[\/latex]<\/li>\r\n \t<li>[latex]-16=4p[\/latex], so [latex]p=-4[\/latex]. Since [latex]p&lt;0[\/latex], the parabola opens left.<\/li>\r\n \t<li>the coordinates of the focus are [latex]\\left(h+p,k\\right)=\\left(-3+\\left(-4\\right),1\\right)=\\left(-7,1\\right)[\/latex]<\/li>\r\n \t<li>the equation of the directrix is [latex]x=h-p=-3-\\left(-4\\right)=1[\/latex]<\/li>\r\n \t<li>the endpoints of the focal diameter are [latex]\\left(h+p,k\\pm 2p\\right)=\\left(-3+\\left(-4\\right),1\\pm 2\\left(-4\\right)\\right)[\/latex], or [latex]\\left(-7,-7\\right)[\/latex] and [latex]\\left(-7,9\\right)[\/latex]<\/li>\r\n<\/ul>\r\nNext we plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204553\/CNX_Precalc_Figure_10_03_0102.jpg\" width=\"487\" height=\"480\" data-media-type=\"image\/jpg\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGraph [latex]{\\left(y+1\\right)}^{2}=4\\left(x - 8\\right)[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter.\r\n\r\n[reveal-answer q=\"626564\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"626564\"]\r\n\r\nVertex: [latex]\\left(8,-1\\right)[\/latex]; Axis of symmetry: [latex]y=-1[\/latex]; Focus: [latex]\\left(9,-1\\right)[\/latex]; Directrix: [latex]x=7[\/latex]; Endpoints of the latus rectum: [latex]\\left(9,-3\\right)[\/latex] and [latex]\\left(9,1\\right)[\/latex].\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02182612\/CNX_Precalc_Figure_10_03_0112.jpg\"><img class=\"aligncenter size-full wp-image-3280\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02182612\/CNX_Precalc_Figure_10_03_0112.jpg\" alt=\"\" width=\"487\" height=\"522\" \/><\/a>\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=86148&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"550\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Graphing a Parabola from an Equation Given in General Form<\/h3>\r\nGraph [latex]{x}^{2}-8x - 28y - 208=0[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter.\r\n\r\n[reveal-answer q=\"335853\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"335853\"]\r\n\r\nStart by writing the equation of the <strong>parabola<\/strong> in standard form. The standard form that applies to the given equation is [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex]. Thus, the axis of symmetry is parallel to the <em>y<\/em>-axis. To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable [latex]x[\/latex] in order to complete the square.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}-8x - 28y - 208=0\\hfill \\\\ \\text{ }{x}^{2}-8x=28y+208\\hfill \\\\ \\text{ }{x}^{2}-8x+16=28y+208+16\\hfill \\\\ \\text{ }{\\left(x - 4\\right)}^{2}=28y+224\\hfill \\\\ \\text{ }{\\left(x - 4\\right)}^{2}=28\\left(y+8\\right)\\hfill \\\\ \\text{ }{\\left(x - 4\\right)}^{2}=4\\cdot 7\\cdot \\left(y+8\\right)\\hfill \\end{array}[\/latex]<\/p>\r\nIt follows that:\r\n<ul>\r\n \t<li>the vertex is [latex]\\left(h,k\\right)=\\left(4,-8\\right)[\/latex]<\/li>\r\n \t<li>the axis of symmetry is [latex]x=h=4[\/latex]<\/li>\r\n \t<li>since [latex]p=7,p&gt;0[\/latex] and so the parabola opens up<\/li>\r\n \t<li>the coordinates of the focus are [latex]\\left(h,k+p\\right)=\\left(4,-8+7\\right)=\\left(4,-1\\right)[\/latex]<\/li>\r\n \t<li>the equation of the directrix is [latex]y=k-p=-8 - 7=-15[\/latex]<\/li>\r\n \t<li>the endpoints of the focal diameter are [latex]\\left(h\\pm 2p,k+p\\right)=\\left(4\\pm 2\\left(7\\right),-8+7\\right)[\/latex], or [latex]\\left(-10,-1\\right)[\/latex] and [latex]\\left(18,-1\\right)[\/latex]<\/li>\r\n<\/ul>\r\nNext we plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204555\/CNX_Precalc_Figure_10_03_0122.jpg\" width=\"487\" height=\"258\" data-media-type=\"image\/jpg\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGraph [latex]{\\left(x+2\\right)}^{2}=-20\\left(y - 3\\right)[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter.\r\n\r\n[reveal-answer q=\"665189\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"665189\"]Vertex: [latex]\\left(-2,3\\right)[\/latex]; Axis of symmetry: [latex]x=-2[\/latex]; Focus: [latex]\\left(-2,-2\\right)[\/latex]; Directrix: [latex]y=8[\/latex]; Endpoints of the latus rectum: [latex]\\left(-12,-2\\right)[\/latex] and [latex]\\left(8,-2\\right)[\/latex].\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02182942\/CNX_Precalc_Figure_10_03_0132.jpg\"><img class=\"aligncenter size-full wp-image-3281\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02182942\/CNX_Precalc_Figure_10_03_0132.jpg\" alt=\"\" width=\"731\" height=\"438\" \/><\/a>\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=86124&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"950\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nIn the graph below you will find a parabola whose vertex is [latex](h,k)[\/latex]. The equation used to generate the parabola is [latex](y-k)^2 = 4p(x-h)[\/latex]. Your task in this exercise is to plot the\u00a0vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter in terms of the free variables [latex]y,k,p[\/latex].\r\n\r\n&nbsp;\r\n\r\nFor example, to plot the vertex, you would create a point [latex](h,k)[\/latex].\r\n\r\nhttps:\/\/www.desmos.com\/calculator\/8bj8n72isi\r\n[reveal-answer q=\"759110\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"759110\"]\r\n\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\nhttps:\/\/www.desmos.com\/calculator\/gvpu51ti7n\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>Solving Applied Problems Involving Parabolas<\/h2>\r\nAs we mentioned at the beginning of the section, parabolas are used to design many objects we use every day, such as telescopes, suspension bridges, microphones, and radar equipment. Parabolic mirrors, such as the one used to light the Olympic torch, have a very unique reflecting property. When rays of light parallel to the parabola\u2019s <strong>axis of symmetry<\/strong> are directed toward any surface of the mirror, the light is reflected directly to the focus.\u00a0This is why the Olympic torch is ignited when it is held at the focus of the parabolic mirror.\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\/896\/2016\/11\/03204557\/CNX_Precalc_Figure_10_03_0142.jpg\" width=\"487\" height=\"362\" data-media-type=\"image\/jpg\" \/> Reflecting property of parabolas[\/caption]\r\n\r\nParabolic mirrors have the ability to focus the sun\u2019s energy to a single point, raising the temperature hundreds of degrees in a matter of seconds. Thus, parabolic mirrors are featured in many low-cost, energy efficient solar products, such as solar cookers, solar heaters, and even travel-sized fire starters.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving Applied Problems Involving Parabolas<\/h3>\r\nA cross-section of a design for a travel-sized solar fire starter. The sun\u2019s rays reflect off the parabolic mirror toward an object attached to the igniter. Because the igniter is located at the focus of the parabola, the reflected rays cause the object to burn in just seconds.\r\n<ol>\r\n \t<li>Find the equation of the parabola that models the fire starter. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane.<\/li>\r\n \t<li>Use the equation found in part (a) to find the depth of the fire starter.<\/li>\r\n<\/ol>\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204559\/CNX_Precalc_Figure_10_03_0162.jpg\" alt=\"\" width=\"487\" height=\"217\" data-media-type=\"image\/jpg\" \/> Cross-section of a travel-sized solar fire starter[\/caption]\r\n\r\n[reveal-answer q=\"424264\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"424264\"]\r\n<ol>\r\n \t<li>The vertex of the dish is the origin of the coordinate plane, so the parabola will take the standard form [latex]{x}^{2}=4py[\/latex], where [latex]p&gt;0[\/latex]. The igniter, which is the focus, is 1.7 inches above the vertex of the dish. Thus we have [latex]p=1.7[\/latex].\r\n[latex]\\begin{array}{ll}{x}^{2}=4py\\hfill &amp; \\begin{array}{cccc}&amp; &amp; &amp; \\end{array}\\text{Standard form of upward-facing parabola with vertex (0,0)}\\hfill \\\\ {x}^{2}=4\\left(1.7\\right)y\\hfill &amp; \\begin{array}{cccc}&amp; &amp; &amp; \\end{array}\\text{Substitute 1}\\text{.7 for }p.\\hfill \\\\ {x}^{2}=6.8y\\hfill &amp; \\begin{array}{cccc}&amp; &amp; &amp; \\end{array}\\text{Multiply}.\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>The dish extends [latex]\\frac{4.5}{2}=2.25[\/latex] inches on either side of the origin. We can substitute 2.25 for [latex]x[\/latex] in the equation from part (a) to find the depth of the dish.\r\n[latex]\\begin{array}{ll}\\text{ }{x}^{2}=6.8y\\hfill &amp; \\text{Equation found in part (a)}.\\hfill \\\\ {\\left(2.25\\right)}^{2}=6.8y\\hfill &amp; \\text{Substitute 2}\\text{.25 for }x.\\hfill \\\\ \\text{ }y\\approx 0.74 \\hfill &amp; \\text{Solve for }y.\\hfill \\end{array}[\/latex]\r\nThe dish is about 0.74 inches deep.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nBalcony-sized solar cookers have been designed for families living in India. The top of a dish has a diameter of 1600 mm. The sun\u2019s rays reflect off the parabolic mirror toward the \"cooker,\" which is placed 320 mm from the base.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Find an equation that models a cross-section of the solar cooker. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane, and that the parabola opens to the right (i.e., has the <em>x<\/em>-axis as its axis of symmetry).<\/li>\r\n \t<li>Use the equation found in part (a) to find the depth of the cooker.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"535409\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"535409\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]{y}^{2}=1280x[\/latex]<\/li>\r\n \t<li>The depth of the cooker is 500 mm<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=87077&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"650\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>Key Equations<\/h2>\r\n<table summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td>Parabola, vertex at origin, axis of symmetry on <em>x<\/em>-axis<\/td>\r\n<td>[latex]{y}^{2}=4px[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Parabola, vertex at origin, axis of symmetry on <em>y<\/em>-axis<\/td>\r\n<td>[latex]{x}^{2}=4py[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Parabola, vertex at [latex]\\left(h,k\\right)[\/latex], axis of symmetry on <em>x<\/em>-axis<\/td>\r\n<td>[latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Parabola, vertex at [latex]\\left(h,k\\right)[\/latex], axis of symmetry on <em>y<\/em>-axis<\/td>\r\n<td>[latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>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 directrix, and a fixed point (the focus) not on the directrix.<\/li>\r\n \t<li>The standard form of a parabola with vertex [latex]\\left(0,0\\right)[\/latex] and the <em>x<\/em>-axis as its axis of symmetry can be used to graph the parabola. 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>The standard form of a parabola with vertex [latex]\\left(0,0\\right)[\/latex] and the <em>y<\/em>-axis as its axis of symmetry can be used to graph the parabola. 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>When given the focus and directrix of a parabola, we can write its equation in standard form.<\/li>\r\n \t<li>The standard form of a parabola with vertex [latex]\\left(h,k\\right)[\/latex] and axis of symmetry parallel to the <em>x<\/em>-axis can be used to graph the parabola. 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>The standard form of a parabola with vertex [latex]\\left(h,k\\right)[\/latex] and axis of symmetry parallel to the <em>y<\/em>-axis can be used to graph the parabola. 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>Real-world situations can be modeled using the standard equations of parabolas. For instance, given the diameter and focus of a cross-section of a parabolic reflector, we can find an equation that models its sides.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<strong>directrix<\/strong> a line perpendicular to the axis of symmetry of a parabola; a line such that the ratio of the distance between the points on the conic and the focus to the distance to the directrix is constant\r\n\r\n<strong>focus (of a parabola)<\/strong> a fixed point in the interior of a parabola that lies on the axis of symmetry\r\n\r\n<strong>focal diameter (latus rectum)<\/strong> the line segment that passes through the focus of a parabola parallel to the directrix, with endpoints on the parabola\r\n\r\n<strong>parabola<\/strong> 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 directrix, and a fixed point (the focus) not on the directrix","rendered":"<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<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204531\/CNX_Precalc_Figure_10_03_001n2.jpg\" alt=\"Description in caption\" width=\"487\" height=\"325\" data-media-type=\"image\/jpg\" \/><\/p>\n<p 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)<\/p>\n<\/div>\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>Parabolas with Vertices at the Origin<\/h2>\n<p>In The Ellipse, 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\/wp-content\/uploads\/sites\/896\/2016\/11\/03204534\/CNX_Precalc_Figure_10_03_0022.jpg\" alt=\"\" width=\"487\" height=\"425\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">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. 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,\u00a0<\/strong>also called the\u00a0<strong>focal diameter<\/strong>. The endpoints of the focal diameter\u00a0lie 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\/wp-content\/uploads\/sites\/896\/2016\/11\/03204536\/CNX_Precalc_Figure_10_03_003n2.jpg\" alt=\"\" width=\"487\" height=\"291\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">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<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\/03204538\/CNX_Precalc_Figure_10_03_0182.jpg\" width=\"487\" height=\"292\" data-media-type=\"image\/jpg\" alt=\"image\" \/><\/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]\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>.<\/p>\n<p 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]<\/p>\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].<br \/>\n[latex]\\sqrt{{x}^{2}+{\\left(y-p\\right)}^{2}}=y+p[\/latex]<\/p>\n<p>We then square both sides of the equation, expand the squared terms, and simplify by combining like terms.<\/p>\n<p 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]<\/p>\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 summarizes the standard features of parabolas with a vertex at the origin.<\/p>\n<table 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\u00a0Focal Diameter<\/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\/wp-content\/uploads\/sites\/896\/2016\/11\/03204540\/CNX_Precalc_Figure_10_03_004n2.jpg\" alt=\"\" width=\"975\" height=\"721\" data-media-type=\"image\/jpg\" \/><\/p>\n<p 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.<\/p>\n<\/div>\n<\/div>\n<p>The key features of a parabola are its vertex, axis of symmetry, focus, directrix, and focal diameter. 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 focal diameter, these lines intersect on the axis of symmetry.<\/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\/03204543\/CNX_Precalc_Figure_10_03_0052.jpg\" width=\"487\" height=\"514\" data-media-type=\"image\/jpg\" alt=\"image\" \/><\/p>\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 focal diameter.\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 focal diameter, [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 focal diameter, [latex]\\left(\\pm 2p,p\\right)[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<li>Plot the focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: 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>focal diameter<\/strong>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q885427\">Solution<\/span><\/p>\n<div id=\"q885427\" class=\"hidden-answer\" style=\"display: none\">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<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 focal diameter\u00a0have 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<p>Next we plot the focus, directrix, and focal diameter, and draw a smooth curve to form the <strong>parabola<\/strong>.<\/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\/03204545\/CNX_Precalc_Figure_10_03_0192.jpg\" width=\"487\" height=\"376\" data-media-type=\"image\/jpg\" alt=\"image\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Graph [latex]{y}^{2}=-16x[\/latex]. Identify and label the focus, directrix, and endpoints of the focal diameter.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q277220\">Solution<\/span><\/p>\n<div id=\"q277220\" 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><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02180919\/leftopen.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3276\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02180919\/leftopen.jpg\" alt=\"\" width=\"487\" height=\"366\" \/><\/a><\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=23511&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"550\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: 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>focal diameter<\/strong>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q951555\">Solution<\/span><\/p>\n<div id=\"q951555\" 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<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 focal diameter 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<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\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204547\/CNX_Precalc_Figure_10_03_007n2.jpg\" width=\"487\" height=\"327\" data-media-type=\"image\/jpg\" alt=\"image\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Graph [latex]{x}^{2}=8y[\/latex]. Identify and label the focus, directrix, and endpoints of the focal diameter.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q824847\">Solution<\/span><\/p>\n<div id=\"q824847\" class=\"hidden-answer\" style=\"display: none\">Focus: [latex]\\left(0,2\\right)[\/latex]; Directrix: [latex]y=-2[\/latex]; Endpoints of the latus rectum: [latex]\\left(\\pm 4,2\\right)[\/latex].<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02181513\/CNX_Precalc_Figure_10_03_0082.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3277\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02181513\/CNX_Precalc_Figure_10_03_0082.jpg\" alt=\"\" width=\"487\" height=\"365\" \/><\/a><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>In the graph below you will find the plot of a parabola whose axis of symmetry is the x-axis. The free variable p is allowed to slide between [latex]-10,10[\/latex]. Your task in this exercise is to add\u00a0the focus, directrix, and endpoints of the focal diameter in terms of the free variable, p.<\/p>\n<p>For example, to add the focus, you would define a point, [latex](p,0)[\/latex] .<\/p>\n<p>https:\/\/www.desmos.com\/calculator\/wunbnybenw<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q864413\">Show Answer<\/span><\/p>\n<div id=\"q864413\" class=\"hidden-answer\" style=\"display: none\">\n<p>https:\/\/www.desmos.com\/calculator\/b3buagwzwl<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\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 exercises\">\n<h3>Example: 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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q259208\">Solution<\/span><\/p>\n<div id=\"q259208\" class=\"hidden-answer\" style=\"display: none\">\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<p>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].<\/p>\n<p>Therefore, the equation for the parabola is [latex]{y}^{2}=-2x[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q886076\">Solution<\/span><\/p>\n<div id=\"q886076\" class=\"hidden-answer\" style=\"display: none\">[latex]{x}^{2}=14y[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29711&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<h2>Parabolas with Vertices Not at the Origin<\/h2>\n<p>Like other graphs we\u2019ve worked with, the graph of a parabola can be translated. If a parabola is translated [latex]h[\/latex] units horizontally and [latex]k[\/latex] units vertically, the vertex will be [latex]\\left(h,k\\right)[\/latex]. This translation results in the standard form of the equation we saw previously with [latex]x[\/latex] replaced by [latex]\\left(x-h\\right)[\/latex] and [latex]y[\/latex] replaced by [latex]\\left(y-k\\right)[\/latex].<\/p>\n<p>To graph parabolas with a vertex [latex]\\left(h,k\\right)[\/latex] other than the origin, we use the standard form [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex] for parabolas that have an axis of symmetry parallel to the <em>x<\/em>-axis, and [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex] for parabolas that have an axis of symmetry parallel to the <em>y<\/em>-axis. 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 (<em>h<\/em>, <em>k<\/em>)<\/h3>\n<p>The table summarizes the standard features of parabolas with a vertex at a point [latex]\\left(h,k\\right)[\/latex].<\/p>\n<table 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 focal diameter<\/strong><\/td>\n<\/tr>\n<tr>\n<td>[latex]y=k[\/latex]<\/td>\n<td>[latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex]<\/td>\n<td>[latex]\\left(h+p,\\text{ }k\\right)[\/latex]<\/td>\n<td>[latex]x=h-p[\/latex]<\/td>\n<td>[latex]\\left(h+p,\\text{ }k\\pm 2p\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]x=h[\/latex]<\/td>\n<td>[latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex]<\/td>\n<td>[latex]\\left(h,\\text{ }k+p\\right)[\/latex]<\/td>\n<td>[latex]y=k-p[\/latex]<\/td>\n<td>[latex]\\left(h\\pm 2p,\\text{ }k+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\/wp-content\/uploads\/sites\/896\/2016\/11\/03204550\/CNX_Precalc_Figure_10_03_0092.jpg\" alt=\"\" width=\"975\" height=\"901\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">(a) When [latex]p&gt;0[\/latex], the parabola opens right. (b) When [latex]p&lt;0[\/latex], the parabola opens left. (c) When [latex]p&gt;0[\/latex], the parabola opens up. (d) When [latex]p&lt;0[\/latex], the parabola opens down.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a standard form equation for a parabola centered at (<em>h<\/em>, <em>k<\/em>), sketch the graph.<\/h3>\n<ol>\n<li>Determine which of the standard forms applies to the given equation: [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex] or [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex].<\/li>\n<li>Use the standard form identified in Step 1 to determine the vertex, axis of symmetry, focus, equation of the directrix, and endpoints of the focal diameter.\n<ol>\n<li>If the equation is in the form [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex], then:\n<ul>\n<li>use the given equation to identify [latex]h[\/latex] and [latex]k[\/latex] for the vertex, [latex]\\left(h,k\\right)[\/latex]<\/li>\n<li>use the value of [latex]k[\/latex] to determine the axis of symmetry, [latex]y=k[\/latex]<\/li>\n<li>set [latex]4p[\/latex] equal to the coefficient of [latex]\\left(x-h\\right)[\/latex] 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]h,k[\/latex], and [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(h+p,\\text{ }k\\right)[\/latex]<\/li>\n<li>use [latex]h[\/latex] and [latex]p[\/latex] to find the equation of the directrix, [latex]x=h-p[\/latex]<\/li>\n<li>use [latex]h,k[\/latex], and [latex]p[\/latex] to find the endpoints of the focal diameter, [latex]\\left(h+p,k\\pm 2p\\right)[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li>If the equation is in the form [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex], then:\n<ul>\n<li>use the given equation to identify [latex]h[\/latex] and [latex]k[\/latex] for the vertex, [latex]\\left(h,k\\right)[\/latex]<\/li>\n<li>use the value of [latex]h[\/latex] to determine the axis of symmetry, [latex]x=h[\/latex]<\/li>\n<li>set [latex]4p[\/latex] equal to the coefficient of [latex]\\left(y-k\\right)[\/latex] 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]h,k[\/latex], and [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(h,\\text{ }k+p\\right)[\/latex]<\/li>\n<li>use [latex]k[\/latex] and [latex]p[\/latex] to find the equation of the directrix, [latex]y=k-p[\/latex]<\/li>\n<li>use [latex]h,k[\/latex], and [latex]p[\/latex] to find the endpoints of the focal diameter, [latex]\\left(h\\pm 2p,\\text{ }k+p\\right)[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/li>\n<li>Plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Graphing a Parabola with Vertex (<em>h<\/em>, <em>k<\/em>) and Axis of Symmetry Parallel to the <em>x<\/em>-axis<\/h3>\n<p>Graph [latex]{\\left(y - 1\\right)}^{2}=-16\\left(x+3\\right)[\/latex]. Identify and label the <strong>vertex<\/strong>, <strong>axis of symmetry<\/strong>, <strong>focus<\/strong>, <strong>directrix<\/strong>, and endpoints of the <strong>focal diameter<\/strong>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q943688\">Solution<\/span><\/p>\n<div id=\"q943688\" class=\"hidden-answer\" style=\"display: none\">\n<p>The standard form that applies to the given equation is [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex]. Thus, the axis of symmetry is parallel to the <em>x<\/em>-axis. It follows that:<\/p>\n<ul>\n<li>the vertex is [latex]\\left(h,k\\right)=\\left(-3,1\\right)[\/latex]<\/li>\n<li>the axis of symmetry is [latex]y=k=1[\/latex]<\/li>\n<li>[latex]-16=4p[\/latex], so [latex]p=-4[\/latex]. Since [latex]p<0[\/latex], the parabola opens left.<\/li>\n<li>the coordinates of the focus are [latex]\\left(h+p,k\\right)=\\left(-3+\\left(-4\\right),1\\right)=\\left(-7,1\\right)[\/latex]<\/li>\n<li>the equation of the directrix is [latex]x=h-p=-3-\\left(-4\\right)=1[\/latex]<\/li>\n<li>the endpoints of the focal diameter are [latex]\\left(h+p,k\\pm 2p\\right)=\\left(-3+\\left(-4\\right),1\\pm 2\\left(-4\\right)\\right)[\/latex], or [latex]\\left(-7,-7\\right)[\/latex] and [latex]\\left(-7,9\\right)[\/latex]<\/li>\n<\/ul>\n<p>Next we plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.<\/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\/03204553\/CNX_Precalc_Figure_10_03_0102.jpg\" width=\"487\" height=\"480\" data-media-type=\"image\/jpg\" alt=\"image\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Graph [latex]{\\left(y+1\\right)}^{2}=4\\left(x - 8\\right)[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q626564\">Solution<\/span><\/p>\n<div id=\"q626564\" class=\"hidden-answer\" style=\"display: none\">\n<p>Vertex: [latex]\\left(8,-1\\right)[\/latex]; Axis of symmetry: [latex]y=-1[\/latex]; Focus: [latex]\\left(9,-1\\right)[\/latex]; Directrix: [latex]x=7[\/latex]; Endpoints of the latus rectum: [latex]\\left(9,-3\\right)[\/latex] and [latex]\\left(9,1\\right)[\/latex].<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02182612\/CNX_Precalc_Figure_10_03_0112.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3280\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02182612\/CNX_Precalc_Figure_10_03_0112.jpg\" alt=\"\" width=\"487\" height=\"522\" \/><\/a><\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=86148&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"550\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Graphing a Parabola from an Equation Given in General Form<\/h3>\n<p>Graph [latex]{x}^{2}-8x - 28y - 208=0[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q335853\">Solution<\/span><\/p>\n<div id=\"q335853\" class=\"hidden-answer\" style=\"display: none\">\n<p>Start by writing the equation of the <strong>parabola<\/strong> in standard form. The standard form that applies to the given equation is [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex]. Thus, the axis of symmetry is parallel to the <em>y<\/em>-axis. To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable [latex]x[\/latex] in order to complete the square.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}-8x - 28y - 208=0\\hfill \\\\ \\text{ }{x}^{2}-8x=28y+208\\hfill \\\\ \\text{ }{x}^{2}-8x+16=28y+208+16\\hfill \\\\ \\text{ }{\\left(x - 4\\right)}^{2}=28y+224\\hfill \\\\ \\text{ }{\\left(x - 4\\right)}^{2}=28\\left(y+8\\right)\\hfill \\\\ \\text{ }{\\left(x - 4\\right)}^{2}=4\\cdot 7\\cdot \\left(y+8\\right)\\hfill \\end{array}[\/latex]<\/p>\n<p>It follows that:<\/p>\n<ul>\n<li>the vertex is [latex]\\left(h,k\\right)=\\left(4,-8\\right)[\/latex]<\/li>\n<li>the axis of symmetry is [latex]x=h=4[\/latex]<\/li>\n<li>since [latex]p=7,p>0[\/latex] and so the parabola opens up<\/li>\n<li>the coordinates of the focus are [latex]\\left(h,k+p\\right)=\\left(4,-8+7\\right)=\\left(4,-1\\right)[\/latex]<\/li>\n<li>the equation of the directrix is [latex]y=k-p=-8 - 7=-15[\/latex]<\/li>\n<li>the endpoints of the focal diameter are [latex]\\left(h\\pm 2p,k+p\\right)=\\left(4\\pm 2\\left(7\\right),-8+7\\right)[\/latex], or [latex]\\left(-10,-1\\right)[\/latex] and [latex]\\left(18,-1\\right)[\/latex]<\/li>\n<\/ul>\n<p>Next we plot the vertex, axis of symmetry, focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.<\/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\/03204555\/CNX_Precalc_Figure_10_03_0122.jpg\" width=\"487\" height=\"258\" data-media-type=\"image\/jpg\" alt=\"image\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Graph [latex]{\\left(x+2\\right)}^{2}=-20\\left(y - 3\\right)[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q665189\">Solution<\/span><\/p>\n<div id=\"q665189\" class=\"hidden-answer\" style=\"display: none\">Vertex: [latex]\\left(-2,3\\right)[\/latex]; Axis of symmetry: [latex]x=-2[\/latex]; Focus: [latex]\\left(-2,-2\\right)[\/latex]; Directrix: [latex]y=8[\/latex]; Endpoints of the latus rectum: [latex]\\left(-12,-2\\right)[\/latex] and [latex]\\left(8,-2\\right)[\/latex].<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02182942\/CNX_Precalc_Figure_10_03_0132.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3281\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/02182942\/CNX_Precalc_Figure_10_03_0132.jpg\" alt=\"\" width=\"731\" height=\"438\" \/><\/a><\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=86124&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"950\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>In the graph below you will find a parabola whose vertex is [latex](h,k)[\/latex]. The equation used to generate the parabola is [latex](y-k)^2 = 4p(x-h)[\/latex]. Your task in this exercise is to plot the\u00a0vertex, axis of symmetry, focus, directrix, and endpoints of the focal diameter in terms of the free variables [latex]y,k,p[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p>For example, to plot the vertex, you would create a point [latex](h,k)[\/latex].<\/p>\n<p>https:\/\/www.desmos.com\/calculator\/8bj8n72isi<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q759110\">Show Answer<\/span><\/p>\n<div id=\"q759110\" class=\"hidden-answer\" style=\"display: none\">\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<p>https:\/\/www.desmos.com\/calculator\/gvpu51ti7n<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Solving Applied Problems Involving Parabolas<\/h2>\n<p>As we mentioned at the beginning of the section, parabolas are used to design many objects we use every day, such as telescopes, suspension bridges, microphones, and radar equipment. Parabolic mirrors, such as the one used to light the Olympic torch, have a very unique reflecting property. When rays of light parallel to the parabola\u2019s <strong>axis of symmetry<\/strong> are directed toward any surface of the mirror, the light is reflected directly to the focus.\u00a0This is why the Olympic torch is ignited when it is held at the focus of the parabolic mirror.<\/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\/wp-content\/uploads\/sites\/896\/2016\/11\/03204557\/CNX_Precalc_Figure_10_03_0142.jpg\" width=\"487\" height=\"362\" data-media-type=\"image\/jpg\" alt=\"image\" \/><\/p>\n<p class=\"wp-caption-text\">Reflecting property of parabolas<\/p>\n<\/div>\n<p>Parabolic mirrors have the ability to focus the sun\u2019s energy to a single point, raising the temperature hundreds of degrees in a matter of seconds. Thus, parabolic mirrors are featured in many low-cost, energy efficient solar products, such as solar cookers, solar heaters, and even travel-sized fire starters.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solving Applied Problems Involving Parabolas<\/h3>\n<p>A cross-section of a design for a travel-sized solar fire starter. The sun\u2019s rays reflect off the parabolic mirror toward an object attached to the igniter. Because the igniter is located at the focus of the parabola, the reflected rays cause the object to burn in just seconds.<\/p>\n<ol>\n<li>Find the equation of the parabola that models the fire starter. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane.<\/li>\n<li>Use the equation found in part (a) to find the depth of the fire starter.<\/li>\n<\/ol>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03204559\/CNX_Precalc_Figure_10_03_0162.jpg\" alt=\"\" width=\"487\" height=\"217\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">Cross-section of a travel-sized solar fire starter<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q424264\">Solution<\/span><\/p>\n<div id=\"q424264\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>The vertex of the dish is the origin of the coordinate plane, so the parabola will take the standard form [latex]{x}^{2}=4py[\/latex], where [latex]p>0[\/latex]. The igniter, which is the focus, is 1.7 inches above the vertex of the dish. Thus we have [latex]p=1.7[\/latex].<br \/>\n[latex]\\begin{array}{ll}{x}^{2}=4py\\hfill & \\begin{array}{cccc}& & & \\end{array}\\text{Standard form of upward-facing parabola with vertex (0,0)}\\hfill \\\\ {x}^{2}=4\\left(1.7\\right)y\\hfill & \\begin{array}{cccc}& & & \\end{array}\\text{Substitute 1}\\text{.7 for }p.\\hfill \\\\ {x}^{2}=6.8y\\hfill & \\begin{array}{cccc}& & & \\end{array}\\text{Multiply}.\\hfill \\end{array}[\/latex]<\/li>\n<li>The dish extends [latex]\\frac{4.5}{2}=2.25[\/latex] inches on either side of the origin. We can substitute 2.25 for [latex]x[\/latex] in the equation from part (a) to find the depth of the dish.<br \/>\n[latex]\\begin{array}{ll}\\text{ }{x}^{2}=6.8y\\hfill & \\text{Equation found in part (a)}.\\hfill \\\\ {\\left(2.25\\right)}^{2}=6.8y\\hfill & \\text{Substitute 2}\\text{.25 for }x.\\hfill \\\\ \\text{ }y\\approx 0.74 \\hfill & \\text{Solve for }y.\\hfill \\end{array}[\/latex]<br \/>\nThe dish is about 0.74 inches deep.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Balcony-sized solar cookers have been designed for families living in India. The top of a dish has a diameter of 1600 mm. The sun\u2019s rays reflect off the parabolic mirror toward the &#8220;cooker,&#8221; which is placed 320 mm from the base.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Find an equation that models a cross-section of the solar cooker. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane, and that the parabola opens to the right (i.e., has the <em>x<\/em>-axis as its axis of symmetry).<\/li>\n<li>Use the equation found in part (a) to find the depth of the cooker.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q535409\">Solution<\/span><\/p>\n<div id=\"q535409\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]{y}^{2}=1280x[\/latex]<\/li>\n<li>The depth of the cooker is 500 mm<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=87077&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"650\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Key Equations<\/h2>\n<table summary=\"..\">\n<tbody>\n<tr>\n<td>Parabola, vertex at origin, axis of symmetry on <em>x<\/em>-axis<\/td>\n<td>[latex]{y}^{2}=4px[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Parabola, vertex at origin, axis of symmetry on <em>y<\/em>-axis<\/td>\n<td>[latex]{x}^{2}=4py[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Parabola, vertex at [latex]\\left(h,k\\right)[\/latex], axis of symmetry on <em>x<\/em>-axis<\/td>\n<td>[latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Parabola, vertex at [latex]\\left(h,k\\right)[\/latex], axis of symmetry on <em>y<\/em>-axis<\/td>\n<td>[latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>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 directrix, and a fixed point (the focus) not on the directrix.<\/li>\n<li>The standard form of a parabola with vertex [latex]\\left(0,0\\right)[\/latex] and the <em>x<\/em>-axis as its axis of symmetry can be used to graph the parabola. If [latex]p>0[\/latex], the parabola opens right. If [latex]p<0[\/latex], the parabola opens left.<\/li>\n<li>The standard form of a parabola with vertex [latex]\\left(0,0\\right)[\/latex] and the <em>y<\/em>-axis as its axis of symmetry can be used to graph the parabola. If [latex]p>0[\/latex], the parabola opens up. If [latex]p<0[\/latex], the parabola opens down.<\/li>\n<li>When given the focus and directrix of a parabola, we can write its equation in standard form.<\/li>\n<li>The standard form of a parabola with vertex [latex]\\left(h,k\\right)[\/latex] and axis of symmetry parallel to the <em>x<\/em>-axis can be used to graph the parabola. If [latex]p>0[\/latex], the parabola opens right. If [latex]p<0[\/latex], the parabola opens left.<\/li>\n<li>The standard form of a parabola with vertex [latex]\\left(h,k\\right)[\/latex] and axis of symmetry parallel to the <em>y<\/em>-axis can be used to graph the parabola. If [latex]p>0[\/latex], the parabola opens up. If [latex]p<0[\/latex], the parabola opens down.<\/li>\n<li>Real-world situations can be modeled using the standard equations of parabolas. For instance, given the diameter and focus of a cross-section of a parabolic reflector, we can find an equation that models its sides.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<p><strong>directrix<\/strong> a line perpendicular to the axis of symmetry of a parabola; a line such that the ratio of the distance between the points on the conic and the focus to the distance to the directrix is constant<\/p>\n<p><strong>focus (of a parabola)<\/strong> a fixed point in the interior of a parabola that lies on the axis of symmetry<\/p>\n<p><strong>focal diameter (latus rectum)<\/strong> the line segment that passes through the focus of a parabola parallel to the directrix, with endpoints on the parabola<\/p>\n<p><strong>parabola<\/strong> 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 directrix, and a fixed point (the focus) not on the directrix<\/p>\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-5240\">\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 Parabolas Interactive. <strong>Authored by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.desmos.com\/calculator\/wunbnybenw\">https:\/\/www.desmos.com\/calculator\/wunbnybenw<\/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 Parabolas - With Solutions Interactive. <strong>Authored by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.desmos.com\/calculator\/b3buagwzwl\">https:\/\/www.desmos.com\/calculator\/b3buagwzwl<\/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 Parabolas With Vertex (H,K) Interactive. <strong>Authored by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.desmos.com\/calculator\/8bj8n72isi\">https:\/\/www.desmos.com\/calculator\/8bj8n72isi<\/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 23511, 86148, 86124, 87082. <strong>Authored by<\/strong>: Roy Shahbazian. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 23711. <strong>Authored by<\/strong>: Caren, McClure. <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":160,"menu_order":4,"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 23511, 86148, 86124, 87082\",\"author\":\"Roy Shahbazian\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Question ID 23711\",\"author\":\"Caren, McClure\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"original\",\"description\":\"Graphing Parabolas Interactive\",\"author\":\"Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/www.desmos.com\/calculator\/wunbnybenw\",\"project\":\"\",\"license\":\"pd\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Graphing Parabolas - 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