7.3: The Parabola
In this section, you will:
- Graph parabolas with vertices at the origin.
- Write equations of parabolas in standard form.
- Graph parabolas with vertices not at the origin.
- Solve applied problems involving parabolas.
Figure 1 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)
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 1 ), 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.
Graphing Parabolas with Vertices at the Origin
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 2 .
Figure 2 Parabola
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 (x,y)(x,y) 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 3 . Notice that the axis of symmetry passes through the focus and vertex and is perpendicular to the directrix. The vertex is the midpoint between the directrix and the focus.
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 dd from the focus to any point PP on the parabola is equal to the distance from PP to the directrix.
Figure 3 Key features of the parabola
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.
Figure 4
Let (x,y)(x,y) be a point on the parabola with vertex (0,0),(0,0), focus (0,p),(0,p), and directrix y= −py= −p as shown in Figure 4 . The distance dd from point (x,y)(x,y) to point (x,−p)(x,−p) on the directrix is the difference of the y -values: d=y+p.d=y+p. The distance from the focus (0,p)(0,p) to the point (x,y)(x,y) is also equal to dd and can be expressed using the distance formula.
d=(x−0)2+(y−p)2−−−−−−−−−−−−−−−√=x2+(y−p)2−−−−−−−−−−√d=(x−0)2+(y−p)2=x2+(y−p)2
Set the two expressions for dd equal to each other and solve for yy to derive the equation of the parabola. We do this because the distance from (x,y)(x,y) to (0,p)(0,p) equals the distance from (x,y)(x,y) to (x, −p).(x, −p).
x2+(y−p)2−−−−−−−−−−√=y+px2+(y−p)2=y+p
We then square both sides of the equation, expand the squared terms, and simplify by combining like terms.
x2+(y−p)2=(y+p)2x2+y2−2py+p2=y2+2py+p2x2−2py=2py x2=4pyx2+(y−p)2=(y+p)2x2+y2−2py+p2=y2+2py+p2x2−2py=2py x2=4py
The equations of parabolas with vertex (0,0)(0,0) are y2=4pxy2=4px when the x -axis is the axis of symmetry and x2=4pyx2=4py when the y -axis is the axis of symmetry. These standard forms are given below, along with their general graphs and key features.
Table 1 and Figure 5 summarize the standard features of parabolas with a vertex at the origin.
| Axis of Symmetry | Equation | Focus | Directrix | Endpoints of Latus Rectum |
| x -axis | y2=4pxy2=4px | (p,0)(p,0) | x=−px=−p | (p,±2p)(p,±2p) |
| y -axis | x2=4pyx2=4py | (0,p)(0,p) | y=−py=−p | (±2p,p)(±2p,p) |
Table 1
Figure 5 (a) When p>0p>0 and the axis of symmetry is the x -axis, the parabola opens right. (b) When p<0p<0 and the axis of symmetry is the x -axis, the parabola opens left. (c) When p>0p>0 and the axis of symmetry is the y -axis, the parabola opens up. (d) When p<0p<0 and the axis of symmetry is the y -axis, the parabola opens down.
The key features of a parabola are its vertex, axis of symmetry, focus, directrix, and latus rectum. See Figure 5 . 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 6 .
Figure 6
Given a standard form equation for a parabola centered at (0, 0), sketch the graph.
- Determine which of the standard forms applies to the given equation: y2=4pxy2=4px or x2=4py.x2=4py.
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Use the standard form identified in Step 1 to determine the axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum.
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If the equation is in the form y2=4px,y2=4px, then
- the axis of symmetry is the x -axis, y=0y=0
- set 4p4p equal to the coefficient of x in the given equation to solve for p.p. If p>0,p>0, the parabola opens right. If p<0,p<0, the parabola opens left.
- use p p to find the coordinates of the focus, (p,0)(p,0)
- use pp to find the equation of the directrix, x=−px=−p
- use pp to find the endpoints of the latus rectum, (p,±2p).(p,±2p). Alternately, substitute x=px=p into the original equation.
-
If the equation is in the form x2=4py,x2=4py, then
- the axis of symmetry is the y -axis, x=0x=0
- set 4p4p equal to the coefficient of y in the given equation to solve for p.p. If p>0,p>0, the parabola opens up. If p<0,p<0, the parabola opens down.
- use pp to find the coordinates of the focus, (0,p)(0,p)
- use pp to find equation of the directrix, y=−py=−p
- use pp to find the endpoints of the latus rectum, (±2p,p)(±2p,p)
-
If the equation is in the form y2=4px,y2=4px, then
- Plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.
EXAMPLE 1
Graphing a Parabola with Vertex (0, 0) and the x -axis as the Axis of Symmetry
Graph y2=24x.y2=24x. Identify and label the focus, directrix, and endpoints of the latus rectum.
- Answer
-
Graph y2=−16x.y2=−16x. Identify and label the focus, directrix, and endpoints of the latus rectum.
EXAMPLE 2
Graphing a Parabola with Vertex (0, 0) and the y -axis as the Axis of Symmetry
Graph x2=−6y.x2=−6y. Identify and label the focus, directrix, and endpoints of the latus rectum.
- Answer
-
Graph x2=8y.x2=8y. Identify and label the focus, directrix, and endpoints of the latus rectum.
Writing Equations of Parabolas in Standard Form
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.
-
Determine whether the axis of symmetry is the
x
- or
y
-axis.
- If the given coordinates of the focus have the form (p,0),(p,0), then the axis of symmetry is the x -axis. Use the standard form y2=4px.y2=4px.
- If the given coordinates of the focus have the form (0,p),(0,p), then the axis of symmetry is the y -axis. Use the standard form x2=4py.x2=4py.
- Multiply 4p.4p.
- Substitute the value from Step 2 into the equation determined in Step 1.
EXAMPLE 3
Writing the Equation of a Parabola in Standard Form Given its Focus and Directrix
What is the equation for the parabola with focus (−12,0)(−12,0) and directrix x=12?x=12?
- Answer
-
What is the equation for the parabola with focus (0,72)(0,72) and directrix y=−72?y=−72?
Graphing Parabolas with Vertices Not at the Origin
Like other graphs we’ve worked with, the graph of a parabola can be translated. If a parabola is translated hh units horizontally and kk units vertically, the vertex will be (h,k).(h,k). This translation results in the standard form of the equation we saw previously with xx replaced by (x−h)(x−h) and yy replaced by (y−k).(y−k).
To graph parabolas with a vertex (h,k)(h,k) other than the origin, we use the standard form (y−k)2=4p(x−h)(y−k)2=4p(x−h) for parabolas that have an axis of symmetry parallel to the x -axis, and (x−h)2=4p(y−k)(x−h)2=4p(y−k) 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.
Table 2 and Figure 9 summarize the standard features of parabolas with a vertex at a point (h,k).(h,k).
| Axis of Symmetry | Equation | Focus | Directrix | Endpoints of Latus Rectum |
| y=ky=k | (y−k)2=4p(x−h)(y−k)2=4p(x−h) | (h+p,k)(h+p,k) | x=h−px=h−p | (h+p,k±2p)(h+p,k±2p) |
| x=hx=h | (x−h)2=4p(y−k)(x−h)2=4p(y−k) | (h,k+p)(h,k+p) | y=k−py=k−p | (h±2p,k+p)(h±2p,k+p) |
Table 2
Figure 9 (a) When p>0,p>0, the parabola opens right. (b) When p<0,p<0, the parabola opens left. (c) When p>0,p>0, the parabola opens up. (d) When p<0,p<0, the parabola opens down.
Given a standard form equation for a parabola centered at ( h , k ), sketch the graph.
- Determine which of the standard forms applies to the given equation: (y−k)2=4p(x−h)(y−k)2=4p(x−h) or (x−h)2=4p(y−k).(x−h)2=4p(y−k).
-
Use the standard form identified in Step 1 to determine the vertex, axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum.
-
If the equation is in the form (y−k)2=4p(x−h),(y−k)2=4p(x−h), then:
- use the given equation to identify h h and kk for the vertex, (h,k)(h,k)
- use the value of kk to determine the axis of symmetry, y=ky=k
- set 4p4p equal to the coefficient of (x−h)(x−h) in the given equation to solve for p.p. If p>0,p>0, the parabola opens right. If p<0,p<0, the parabola opens left.
- use h,k,h,k, and pp to find the coordinates of the focus, (h+p,k)(h+p,k)
- use hh and pp to find the equation of the directrix, x=h−px=h−p
- use h,k,h,k, and pp to find the endpoints of the latus rectum, (h+p,k±2p)(h+p,k±2p)
-
If the equation is in the form (x−h)2=4p(y−k),(x−h)2=4p(y−k), then:
- use the given equation to identify hh and kk for the vertex, (h,k)(h,k)
- use the value of hh to determine the axis of symmetry, x=hx=h
- set 4p4p equal to the coefficient of (y−k)(y−k) in the given equation to solve for p.p. If p>0,p>0, the parabola opens up. If p<0,p<0, the parabola opens down.
- use h,k,h,k, and pp to find the coordinates of the focus, (h,k+p)(h,k+p)
- use kk and pp to find the equation of the directrix, y=k−py=k−p
- use h,k,h, k, and p p to find the endpoints of the latus rectum, (h±2p,k+p)(h±2p,k+p)
-
If the equation is in the form (y−k)2=4p(x−h),(y−k)2=4p(x−h), then:
- Plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.
EXAMPLE 4
Graphing a Parabola with Vertex ( h , k ) and Axis of Symmetry Parallel to the x -axis
Graph (y−1)2=−16(x+3).(y−1)2=−16(x+3). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.
- Answer
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Graph (y+1)2=4(x−8).(y+1)2=4(x−8). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.
EXAMPLE 5
Graphing a Parabola from an Equation Given in General Form
Graph x2−8x−28y−208=0.x2−8x−28y−208=0. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.
- Answer
-
Graph (x+2)2=−20(y−3).(x+2)2=−20(y−3). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.
Solving Applied Problems Involving Parabolas
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 12 . This is why the Olympic torch is ignited when it is held at the focus of the parabolic mirror.
Figure 12 Reflecting property of parabolas
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.
EXAMPLE 6
Solving Applied Problems Involving Parabolas
A cross-section of a design for a travel-sized solar fire starter is shown in Figure 13 . 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.
- ⓐ 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.
- ⓑ Use the equation found in part ⓐ to find the depth of the fire starter.
Figure 13 Cross-section of a travel-sized solar fire starter
- Answer
-
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.
ⓐ 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 x -axis as its axis of symmetry).
ⓑ Use the equation found in part ⓐ to find the depth of the cooker.
Access these online resources for additional instruction and practice with parabolas.