Parabolas were previously presented in function form. This book will introduce the conic section applications of parabolas. The connection to conic sections will not change the characteristics of parabolas. However some may seem new. A parabola is the set of points in a plane that are the same distance from a given point and a given line in that plane. The given point is called the focus, and the line is called the directrix. The midpoint of the perpendicular segment from the focus to the directrix is called the vertex of the parabola. The line that passes through the vertex and focus is called the axis of symmetry.
Page 2ClareGladwinRESD
In this section, you will:
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 (see (Figure)), which focuses light rays from the sun to ignite the flame. 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.
In The Ellipse, we saw that an ellipse 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 parabola. See (Figure). Like the ellipse and hyperbola, 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 directrix, and a fixed point (the focus) not on the directrix. In Quadratic Functions, we learned about a parabola’s vertex and axis of symmetry. Now we extend the discussion to include other key features of the parabola. See (Figure). 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. The line segment that passes through the focus and is parallel to the directrix is called the latus rectum. 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. To work with parabolas in the coordinate plane, we consider two cases: those with a vertex at the origin and those with a vertex at a point other than the origin. We begin with the former. 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] [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] 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] [latex]\sqrt{{x}^{2}+{\left(y-p\right)}^{2}}=y+p[/latex] We then square both sides of the equation, expand the squared terms, and simplify by combining like terms. [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] The equations of parabolas with vertex[latex]\,\left(0,0\right)\,[/latex]are[latex]\,{y}^{2}=4px\,[/latex]when the x-axis is the axis of symmetry and[latex]\,{x}^{2}=4py\,[/latex]when the y-axis is the axis of symmetry. These standard forms are given below, along with their general graphs and key features.
(Figure) and (Figure) summarize the standard features of parabolas with a vertex at the origin.
The key features of a parabola are its vertex, axis of symmetry, focus, directrix, and latus rectum. See (Figure). When given a standard equation for a parabola centered at the origin, we can easily identify the key features to graph the parabola. A line is said to be tangent to a curve if it intersects the curve at exactly one point. If we sketch lines tangent to the parabola at the endpoints of the latus rectum, these lines intersect on the axis of symmetry, as shown in (Figure).
Given a standard form equation for a parabola centered at (0, 0), sketch the graph.
Graph[latex]\,{y}^{2}=24x.\,[/latex]Identify and label the focus, directrix, and endpoints of the latus rectum.
Graph[latex]\,{y}^{2}=-16x.\,[/latex]Identify and label the focus, directrix, and endpoints of the latus rectum.
Graph[latex]\,{x}^{2}=-6y.\,[/latex]Identify and label the focus, directrix, and endpoints of the latus rectum.
Graph[latex]\,{x}^{2}=8y.\,[/latex]Identify and label the focus, directrix, and endpoints of the latus rectum. 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.
Given its focus and directrix, write the equation for a parabola in standard form.
What is the equation for the parabola with focus[latex]\,\left(-\frac{1}{2},0\right)\,[/latex]and directrix[latex]\,x=\frac{1}{2}?[/latex]
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]
Like other graphs we’ve 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] 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 x-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 y-axis. These standard forms are given below, along with their general graphs and key features.
(Figure) and (Figure) summarize the standard features of parabolas with a vertex at a point[latex]\,\left(h,k\right).[/latex]
Given a standard form equation for a parabola centered at (h, k), sketch the graph.
Graph[latex]\,{\left(y-1\right)}^{2}=-16\left(x+3\right).\,[/latex]Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.
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 latus rectum.
Graph[latex]\,{x}^{2}-8x-28y-208=0.\,[/latex]Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.
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 latus rectum. 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’s axis of symmetry are directed toward any surface of the mirror, the light is reflected directly to the focus. See (Figure). This is why the Olympic torch is ignited when it is held at the focus of the parabolic mirror. Parabolic mirrors have the ability to focus the sun’s 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.
A cross-section of a design for a travel-sized solar fire starter is shown in (Figure). The sun’s 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.
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’s rays reflect off the parabolic mirror toward the “cooker,” which is placed 320 mm from the base.
Key Concepts
Define a parabola in terms of its focus and directrix.
If the equation of a parabola is written in standard form and[latex]\,p\,[/latex]is positive and the directrix is a vertical line, then what can we conclude about its graph?
If the equation of a parabola is written in standard form and[latex]\,p\,[/latex]is negative and the directrix is a horizontal line, then what can we conclude about its graph?
What is the effect on the graph of a parabola if its equation in standard form has increasing values of [latex]\,p\text{?}[/latex]
As the graph of a parabola becomes wider, what will happen to the distance between the focus and directrix?
For the following exercises, determine whether the given equation is a parabola. If so, rewrite the equation in standard form.
[latex]{y}^{2}=4-{x}^{2}[/latex]
[latex]y=4{x}^{2}[/latex]
[latex]3{x}^{2}-6{y}^{2}=12[/latex]
[latex]{\left(y-3\right)}^{2}=8\left(x-2\right)[/latex]
[latex]{y}^{2}+12x-6y-51=0[/latex] For the following exercises, rewrite the given equation in standard form, and then determine the vertex[latex]\,\left(V\right),[/latex] focus[latex]\,\left(F\right),[/latex] and directrix[latex]\text{ }\left(d\right)\text{ }[/latex]of the parabola.
[latex]x=8{y}^{2}[/latex]
[latex]y=\frac{1}{4}{x}^{2}[/latex]
[latex]y=-4{x}^{2}[/latex]
[latex]x=\frac{1}{8}{y}^{2}[/latex]
[latex]x=36{y}^{2}[/latex]
[latex]x=\frac{1}{36}{y}^{2}[/latex]
[latex]{\left(x-1\right)}^{2}=4\left(y-1\right)[/latex]
[latex]{\left(y-2\right)}^{2}=\frac{4}{5}\left(x+4\right)[/latex]
[latex]{\left(y-4\right)}^{2}=2\left(x+3\right)[/latex]
[latex]{\left(x+1\right)}^{2}=2\left(y+4\right)[/latex]
[latex]{\left(x+4\right)}^{2}=24\left(y+1\right)[/latex]
[latex]{\left(y+4\right)}^{2}=16\left(x+4\right)[/latex]
[latex]{y}^{2}+12x-6y+21=0[/latex]
[latex]{x}^{2}-4x-24y+28=0[/latex]
[latex]5{x}^{2}-50x-4y+113=0[/latex]
[latex]{y}^{2}-24x+4y-68=0[/latex]
[latex]{x}^{2}-4x+2y-6=0[/latex]
[latex]{y}^{2}-6y+12x-3=0[/latex]
[latex]3{y}^{2}-4x-6y+23=0[/latex]
[latex]{x}^{2}+4x+8y-4=0[/latex]
For the following exercises, graph the parabola, labeling the focus and the directrix.
[latex]x=\frac{1}{8}{y}^{2}[/latex]
[latex]y=36{x}^{2}[/latex]
[latex]y=\frac{1}{36}{x}^{2}[/latex]
[latex]y=-9{x}^{2}[/latex]
[latex]{\left(y-2\right)}^{2}=-\frac{4}{3}\left(x+2\right)[/latex]
[latex]-5{\left(x+5\right)}^{2}=4\left(y+5\right)[/latex]
[latex]-6{\left(y+5\right)}^{2}=4\left(x-4\right)[/latex]
[latex]{y}^{2}-6y-8x+1=0[/latex]
[latex]{x}^{2}+8x+4y+20=0[/latex]
[latex]3{x}^{2}+30x-4y+95=0[/latex]
[latex]{y}^{2}-8x+10y+9=0[/latex]
[latex]{x}^{2}+4x+2y+2=0[/latex]
[latex]{y}^{2}+2y-12x+61=0[/latex]
[latex]-2{x}^{2}+8x-4y-24=0[/latex] For the following exercises, find the equation of the parabola given information about its graph.
Vertex is[latex]\,\left(0,0\right);[/latex]directrix is[latex]\,y=4,[/latex] focus is[latex]\,\left(0,-4\right).[/latex]
Vertex is[latex]\,\left(0,0\right);\,[/latex]directrix is[latex]\,x=4,[/latex] focus is[latex]\,\left(-4,0\right).[/latex]
Vertex is[latex]\,\left(2,2\right);\,[/latex] directrix is[latex]\,x=2-\sqrt{2},[/latex] focus is[latex]\,\left(2+\sqrt{2},2\right).[/latex]
Vertex is[latex]\,\left(-2,3\right);\,[/latex]directrix is[latex]\,x=-\frac{7}{2},[/latex] focus is[latex]\,\left(-\frac{1}{2},3\right).[/latex]
Vertex is[latex]\,\left(\sqrt{2},-\sqrt{3}\right);[/latex] directrix is[latex]\,x=2\sqrt{2},[/latex] focus is[latex]\,\left(0,-\sqrt{3}\right).[/latex]
Vertex is[latex]\,\left(1,2\right);\,[/latex]directrix is[latex]\,y=\frac{11}{3},[/latex] focus is[latex]\,\left(1,\frac{1}{3}\right).[/latex] For the following exercises, determine the equation for the parabola from its graph.
For the following exercises, the vertex and endpoints of the latus rectum of a parabola are given. Find the equation.
[latex]V\left(0,0\right),\text{Endpoints }\left(2,1\right),\left(-2,1\right)[/latex]
[latex]V\left(0,0\right),\text{Endpoints }\left(-2,4\right),\left(-2,-4\right)[/latex]
[latex]V\left(1,2\right),\text{Endpoints }\left(-5,5\right),\left(7,5\right)[/latex]
[latex]V\left(-3,-1\right),\text{Endpoints }\left(0,5\right),\left(0,-7\right)[/latex]
[latex]V\left(4,-3\right),\text{Endpoints }\left(5,-\frac{7}{2}\right),\left(3,-\frac{7}{2}\right)[/latex]
The mirror in an automobile headlight has a parabolic cross-section with the light bulb at the focus. On a schematic, the equation of the parabola is given as[latex]\,{x}^{2}=4y.\,[/latex]At what coordinates should you place the light bulb?
If we want to construct the mirror from the previous exercise such that the focus is located at[latex]\,\left(0,0.25\right),[/latex] what should the equation of the parabola be?
A satellite dish is shaped like a paraboloid of revolution. This means that it can be formed by rotating a parabola around its axis of symmetry. The receiver is to be located at the focus. If the dish is 12 feet across at its opening and 4 feet deep at its center, where should the receiver be placed?
Consider the satellite dish from the previous exercise. If the dish is 8 feet across at the opening and 2 feet deep, where should we place the receiver?
A searchlight is shaped like a paraboloid of revolution. A light source is located 1 foot from the base along the axis of symmetry. If the opening of the searchlight is 3 feet across, find the depth.
If the searchlight from the previous exercise has the light source located 6 inches from the base along the axis of symmetry and the opening is 4 feet, find the depth.
An arch is in the shape of a parabola. It has a span of 100 feet and a maximum height of 20 feet. Find the equation of the parabola, and determine the height of the arch 40 feet from the center.
If the arch from the previous exercise has a span of 160 feet and a maximum height of 40 feet, find the equation of the parabola, and determine the distance from the center at which the height is 20 feet.
An object is projected so as to follow a parabolic path given by[latex]\,y=-{x}^{2}+96x,[/latex]where[latex]\,x\,[/latex]is the horizontal distance traveled in feet and[latex]\,y\,[/latex]is the height. Determine the maximum height the object reaches.
For the object from the previous exercise, assume the path followed is given by[latex]\,y=-0.5{x}^{2}+80x.\,[/latex] directrix 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 focus (of a parabola) a fixed point in the interior of a parabola that lies on the axis of symmetry latus rectum the line segment that passes through the focus of a parabola parallel to the directrix, with endpoints on the parabola parabola 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 |