15.3: The Parabola
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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 \( \PageIndex{ 1 } \)), 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
We previously 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 \( \PageIndex{ 2 } \).
Parabola
Like the ellipse and hyperbola, the parabola can also be defined by a set of points in the coordinate plane.
When studying 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 \( \PageIndex{ 3 } \). 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.
Key features of the parabola
The endpoints of the latus rectum lie on the curve. By definition, the distance \( d \) from the focus to any point \( P \) on the parabola is equal to the distance from \( P \) 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 \( (x,y) \) be a point on the parabola with vertex \( (0,0) \), focus \( (0,p) \), and directrix \( y = -p \) as shown in Figure \( \PageIndex{ 4 } \). The distance \( d \) from point \( (x,y) \) to point \( (x, -p) \) on the directrix is the difference of the \( y \)-values: \( d = y + p \). The distance from the focus \( (0,p) \) to the point \( (x,y) \) is also equal to \( d \) and can be expressed using the Distance Formula.\[ d = \sqrt{(x - 0)^2 + (y - p)^2} = \sqrt{x^2 + (y - p)^2}. \nonumber \]Set the two expressions for \( d \) equal to each other and solve for \( y \) to derive the equation of the parabola. We do this because the distance from \( (x,y) \) to \( (0,p) \) equals the distance from \( (x,y) \) to \( (x,-p \).\[ \sqrt{x^2 + (y - p)^2} = y + p \nonumber \]We then square both sides of the equation, expand the squared terms, and simplify by combining like terms.\[ \begin{array}{rrclcl} & x^2 + (y - p)^2 & = & (y + p)^2 & \quad & \left( \text{squaring both sides} \right) \\[6pt] \implies & x^2 + y^2 - 2py + p^2 & = & y^2 + 2 py + p^2 & \quad & \left( \text{distributing} \right) \\[6pt] \implies & x^2 - 2py & = & 2 py & \quad & \left( \text{subtracting }y^2\text{ and }p^2\text{ from both sides} \right) \\[6pt] \implies & x^2 & = & 4 py & \quad & \left( \text{adding }2py\text{ to both sides} \right) \\[6pt] \end{array} \nonumber \]The equations of parabolas with vertex \( (0,0) \) are \( y^2 = 4px \) when the \( x \)-axis is the axis of symmetry and \( x^2 = 4py \) when the \( y \)-axis is the axis of symmetry. These standard forms are given below, along with their general graphs and key features.
The table below and the following figure summarize the standard features of parabolas with a vertex at the origin.
| Axis of Symmetry | Equation | Focus | Directrix | Endpoints of Latus Rectum |
|---|---|---|---|---|
| \( x \)-axis | \( y^2 = 4px \) | \( (p,0) \) | \( x = -p \) | \( (p, \pm 2p) \) |
| \( y \)-axis | \( x^2 = 4py \) | \( (0,p) \) | \( y = -p \) | \( (\pm 2p, p) \) |
(b) When \( p < 0 \) and the axis of symmetry is the \( x \)-axis, the parabola opens left.
(c) When \( p > 0 \) and the axis of symmetry is the \( y \)-axis, the parabola opens up.
(d) When \( p < 0 \) 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. 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 \( \PageIndex{ 5 } \).
Graph \( y^2 = 24x \). Identify and label the focus, directrix, and endpoints of the latus rectum.
- Solution
-
The standard form that applies to the given equation is \( y^2 = 4px \). Thus, the axis of symmetry is the \( x \)-axis. It follows that:
- \( 24 = 4p \), so \( p=6 \). Since \( p>0 \), the parabola opens right
- the coordinates of the focus are \( (p,0)=(6,0) \)
- the equation of the directrix is \( x=−p=−6 \)
- the endpoints of the latus rectum have the same \( x \)-coordinate at the focus. To find the endpoints, substitute \( x=6 \) into the original equation: \( (6,\pm 12) \)
Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.
Figure \( \PageIndex{ 6 } \)
Graph \(y^2 = −16x\). Identify and label the focus, directrix, and endpoints of the latus rectum.
Graph \(x^2=−6y\). Identify and label the focus, directrix, and endpoints of the latus rectum.
- Solution
-
The standard form that applies to the given equation is \( x^2 = 4py \). Thus, the axis of symmetry is the \( y \)-axis. It follows that:
- \( -6 = 4p \), so \( p=-\frac{3}{2} \). Since \( p < 0 \), the parabola opens down
- the coordinates of the focus are \( (0,p)=\left( 0,-\frac{3}{2} \right) \)
- the equation of the directrix is \( y=−p=\frac{3}{2} \)
- the endpoints of the latus rectum can be found by substituting \( y=\frac{3}{2} \) into the original equation: \( \left( \pm 3, -\frac{3}{2} \right) \)
Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.
Figure \( \PageIndex{ 7 } \)
Graph \(x^2 = 8y\). 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.
What is the equation for the parabola with focus \( \left(−\frac{1}{2},0\right) \) and directrix \( x=\frac{1}{2} \)?
- Solution
-
The focus has the form \( (p,0) \), so the equation will have the form \( y^2=4px \). Multiplying \( 4p \), we have \( 4p=4\left( -\frac{1}{2} \right)=−2 \). Substituting for \( 4p \), we have \( y^2=4px=−2x \). Therefore, the equation for the parabola is \( y^2=−2x \).
What is the equation for the parabola with focus \( \left( 0,\frac{7}{2} \right) \) and directrix \( y=−\frac{7}{2} \)?
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 \( h \) units horizontally and \( k \) units vertically, the vertex will be \( (h,k) \). This translation results in the standard form of the equation we saw previously with \( x \) replaced by \( (x−h) \) and \( y \) replaced by \( (y−k) \).
To graph parabolas with a vertex \( (h,k) \) other than the origin, we use the standard form \( (y−k)^2=4p(x−h) \) for parabolas that have an axis of symmetry parallel to the \( x \)-axis, and \((x−h)^2=4p(y−k)\) 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.
The table below and the following figure summarize the standard features of parabolas with a vertex at the origin.
| Axis of Symmetry | Equation | Focus | Directrix | Endpoints of Latus Rectum |
|---|---|---|---|---|
| \( y = k \) | \( (y - k)^2 = 4p(x - h) \) | \( (h + p,k) \) | \( x = h-p \) | \( (h+p, k \pm 2p) \) |
| \( x = h \) | \( (x - h)^2 = 4p(y - k) \) | \( (h,k+p) \) | \( y = k-p \) | \( (h \pm 2p, k + p) \) |
(b) When \( p < 0 \), the parabola opens left.
(c) When \( p > 0 \), the parabola opens up.
(d) When \( p < 0 \), the parabola opens down.
Graph \( (y−1)^2=−16(x+3) \). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.
- Solution
-
The standard form that applies to the given equation is \( (y−k)^2=4p(x−h) \). Thus, the axis of symmetry is parallel to the\( x \)-axis. It follows that:
- the vertex is \( (h,k)=(−3,1) \)
- the axis of symmetry is \( y=k=1 \)
- \( −16=4p \), so \( p=−4 \). Since \( p<0 \), the parabola opens left.
- the coordinates of the focus are \( (h+p,k)=(−3+(−4),1)=(−7,1) \)
- the equation of the directrix is \( x=h−p=−3−(−4)=1 \)
- the endpoints of the latus rectum are \( (h+p,k \pm 2p)=(−3+(−4),1 \pm 2(−4)) \), or \( (−7,−7) \) and \( (−7,9) \)
Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. See Figure \( \PageIndex{ 8 } \).
Figure \( \PageIndex{ 8 } \)
Graph \( (y+1)^2=4(x−8) \). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.
Graph \( x^2−8x−28y−208=0 \). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.
- Solution
-
Start by writing the equation of the parabola in standard form. The standard form that applies to the given equation is \( (x−h)^2=4p(y−k) \). Thus, the axis of symmetry is parallel to the \( y \)-axis. To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable \( x \) in order to complete the square.\[ \begin{array}{rrcl} & x^2−8x−28y−208 & = & 0 \\[6pt] \implies & x^2−8x & = & 28y+208 \\[6pt] \implies & x2−8x+16 & = & 28y+208+16 \\[6pt] \implies & (x−4)^2 & = & 28y+224 \\[6pt] \implies & (x−4)^2 & = & 28(y+8) \\[6pt] \implies & (x−4)^2 & = & 4 \cdot 7 (y+8) \\[6pt] \end{array} \nonumber \]It follows that:
- the vertex is \( (h,k)=(4,−8) \)
- the axis of symmetry is \( x=h=4 \)
- since \( p=7 \), \( p>0 \) and so the parabola opens up
- the coordinates of the focus are \( (h,k+p)=(4,−8+7)=(4,−1) \)
- the equation of the directrix is \( y=k−p=−8−7=−15 \)
- the endpoints of the latus rectum are \( (h \pm 2p,k+p)=(4 \pm 2(7),−8+7) \), or \( (−10,−1) \) and \( (18,−1) \)
Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. See Figure \( \PageIndex{ 9 } \).
Figure \( \PageIndex{ 9 } \)
Graph \( (x+2)^2=−20(y−3) \). 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 \( \PageIndex{ 12 } \). This is why the Olympic torch is ignited when it is held at the focus of the parabolic mirror.
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.
A cross-section of a design for a travel-sized solar fire starter is shown in Figure \( \PageIndex{ 11 } \). 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 (a) to find the depth of the fire starter.
Cross-section of a travel-sized solar fire starter
- Solution
-
- The vertex of the dish is the origin of the coordinate plane, so the parabola will take the standard form \( x^2=4py \), where \( p>0 \). The igniter, which is the focus, is 1.7 inches above the vertex of the dish. Thus we have \( p=1.7 \).\[ \begin{array}{rrclcl} & x^2 & = & 4py & \quad & \left( \text{Standard form of upward-facing parabola with vertex }(0,0) \right) \\[6pt] \implies & x^2 & = & 4(1.7)y & \quad & \left( \text{substituting} \right) \\[6pt] \implies & x^2 & = & 6.8y & \quad & \left( \text{multiplying} \right) \\[6pt] \end{array} \nonumber \]
- The dish extends \(\frac{4.5}{2}=2.25\) inches on either side of the origin. We can substitute 2.25 for \( x \) in the equation from part (a) to find the depth of the dish.\[ \begin{array}{rrclcl} & x^2 & = & 6.8y & & \\[6pt] \implies & (2.25)^2 & = & 6.8y & \quad & \left( \text{substituting} \right) \\[6pt] \implies & 0.74 & \approx & y & \quad & \left( \text{solving for }y \right) \\[6pt] \end{array} \nonumber \]The dish is about 0.74 inches deep.
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 (a) to find the depth of the cooker.