12.3 The Parabola
Figure 1 Katherine Johnson’s pioneering mathematical work in the area of parabolic and other orbital calculations played a significant role in the development of U.S space flight. (credit: NASA)
Katherine Johnson is the pioneering NASA mathematician who was integral to the successful and safe flight and return of many human missions as well as satellites. Prior to the work featured in the movie Hidden Figures, she had already made major contributions to the space program. She provided trajectory analysis for the Mercury mission, in which Alan Shepard became the first American to reach space, and she and engineer Ted Sopinski authored a monumental paper regarding placing an object in a precise orbital position and having it return safely to Earth. Many of the orbits she determined were made up of parabolas, and her ability to combine different types of math enabled an unprecedented level of precision. Johnson said, “You tell me when you want it and where you want it to land, and I’ll do it backwards and tell you when to take off.”
Johnson’s work on parabolic orbits and other complex mathematics resulted in successful orbits, Moon landings, and the development of the Space Shuttle program. Applications of parabolas are also critical to other areas of science. 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.
12.3.1 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 \(\left( {x,y} \right)\) 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 \(d\) from the focus to any point \(P\) on the parabola is equal to the distance from \(P\) 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 \(\left( {x,y} \right)\) be a point on the parabola with vertex \(\left( {0,0} \right),\) focus \(\left( {0,p} \right),\) and directrix \(y\operatorname{=\ -}p\) as shown in Figure 4. The distance \(d\) from point \(\left( {x,y} \right)\) 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 \(\left( {x,y} \right)\) is also equal to \(d\) and can be expressed using the distance formula.
\[\begin{array}{l} {d = \sqrt{{(x - 0)}^{2} + {(y - p)}^{2}}} \\ {\mspace{9mu}\mspace{9mu}\mspace{9mu} = \sqrt{x^{2} + {(y - p)}^{2}}} \end{array}\]
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 \(\left( {x,y} \right)\) to \(\left( {0,p} \right)\) equals the distance from \(\left( {x,y} \right)\) to \((x\operatorname{,\ -}p).\)
\[\sqrt{x^{2} + \left( {y - p} \right)^{2}} = y + p\]
We then square both sides of the equation, expand the squared terms, and simplify by combining like terms.
\[\begin{matrix} {x^{2} + {(y - p)}^{2} = {(y + p)}^{2}} \\ {x^{2} + y^{2} - 2py + p^{2} = y^{2} + 2py + p^{2}} \\ {x^{2} - 2py = 2py} \\ {\mspace{9mu}\text{~~~~~~~~~}x^{2} = 4py} \end{matrix}\]
The equations of parabolas with vertex \(\left( {0,0} \right)\) 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.
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 | \(y^{2} = 4px\) | \(\left( {p,\mspace{9mu} 0} \right)\) | \(x = - p\) | \(\left( {p,\mspace{9mu} \pm 2p} \right)\) |
| \(y\)-axis | \(x^{2} = 4py\) | \(\left( {0,\mspace{9mu} p} \right)\) | \(y = - p\) | \(\left( {\pm 2p,\mspace{9mu} p} \right)\) |
Table 1
Figure 5 (a) When \(p > 0\) and the axis of symmetry is the \(x\)-axis, the parabola opens right. (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. 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
Graph \(y^{2} = 24x.\) Identify and label the focus, directrix, and endpoints of the latus rectum.
Solution (click to reveal)
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 \(\left( {p,0} \right) = \left( {6,0} \right)\)
- 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: \(\left( {6, \pm 12} \right)\)
Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. Figure 7
Figure 7
Graph \(y^{2} = -16x.\) Identify and label the focus, directrix, and endpoints of the latus rectum.
Solution (click to reveal)
Focus: \(\left( {- 4,0} \right);\) Directrix: \(x = 4;\) Endpoints of the latus rectum: \(\left( {- 4, \pm 8} \right)\)
Graph \(x^{2} = -6y.\) Identify and label the focus, directrix, and endpoints of the latus rectum.
Solution (click to reveal)
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 \(\left( {0,p} \right) = \left( {0, - \frac{3}{2}} \right)\)
- 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 8
Graph \(x^{2} = 8y.\) Identify and label the focus, directrix, and endpoints of the latus rectum.
Solution (click to reveal)
Focus: \(\left( {0,2} \right);\) Directrix: \(y = -2;\) Endpoints of the latus rectum: \(\left( {\pm 4,2} \right).\)
12.3.2 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.
12.3.3 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 \(\left( {h,k} \right).\) This translation results in the standard form of the equation we saw previously with \(x\) replaced by \(\left( {x - h} \right)\) and \(y\) replaced by \(\left( {y - k} \right).\)
To graph parabolas with a vertex \(\left( {h,k} \right)\) other than the origin, we use the standard form \(\left( {y - k} \right)^{2} = 4p\left( {x - h} \right)\) for parabolas that have an axis of symmetry parallel to the \(x\)-axis, and \(\left( {x - h} \right)^{2} = 4p\left( {y - k} \right)\) 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 \(\left( {h,k} \right).\)
| Axis of Symmetry | Equation | Focus | Directrix | Endpoints of Latus Rectum |
| \(y = k\) | \(\left( {y - k} \right)^{2} = 4p\left( {x - h} \right)\) | \(\left( {h + p,\mspace{9mu} k} \right)\) | \(x = h - p\) | \(\left( {h + p,\mspace{9mu} k \pm 2p} \right)\) |
| \(x = h\) | \(\left( {x - h} \right)^{2} = 4p\left( {y - k} \right)\) | \(\left( {h,\mspace{9mu} k + p} \right)\) | \(y = k - p\) | \(\left( {h \pm 2p,\mspace{9mu} k + p} \right)\) |
Table 2
Figure 9 (a) When \(p > 0,\) the parabola opens right. (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 \(\left( {y - 1} \right)^{2} = -16\left( {x + 3} \right).\) Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.
Solution (click to reveal)
The standard form that applies to the given equation is \(\left( {y - k} \right)^{2} = 4p\left( {x - h} \right).\) Thus, the axis of symmetry is parallel to the \(x\)-axis. It follows that:
- the vertex is \(\left( {h,k} \right) = \left( {- 3,1} \right)\)
- 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 \(\left( {h + p,k} \right) = \left( {-3 + (-4),1} \right) = \left( {-7,1} \right)\)
- the equation of the directrix is \(x = h - p = -3 - (-4) = 1\)
- the endpoints of the latus rectum are \(\left( {h + p,k \pm 2p} \right) = \left( {-3 + (-4),1 \pm 2(-4)} \right),\) or \(\left( {-7,-7} \right)\) and \(\left( {-7,9} \right)\)
Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. See Figure 10.
Figure 10
Graph \(\left( {y + 1} \right)^{2} = 4\left( {x - 8} \right).\) Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.
Solution (click to reveal)
Vertex: \(\left( {8, - 1} \right);\) Axis of symmetry: \(y = -1;\) Focus: \(\left( {9, - 1} \right);\) Directrix: \(x = 7;\) Endpoints of the latus rectum: \(\left( {9, - 3} \right)\) and \(\left( {9,1} \right).\)
Graph \(x^{2} - 8x - 28y - 208 = 0.\) Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.
Solution (click to reveal)
Start by writing the equation of the parabola in standard form. The standard form that applies to the given equation is \(\left( {x - h} \right)^{2} = 4p\left( {y - k} \right).\) 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.
\[\begin{array}{l} {x^{2} - 8x - 28y - 208 = 0} \\ {\mspace{9mu}\text{~~~~~~~~~~~~~~~~~~~~}x^{2} - 8x = 28y + 208} \\ {\mspace{9mu}\text{~~~~~~~~~~~}x^{2} - 8x + 16 = 28y + 208 + 16} \\ {\mspace{9mu}\text{~~~~~~~~~~~~~~~~~~~}{(x - 4)}^{2} = 28y + 224} \\ {\mspace{9mu}\text{~~~~~~~~~~~~~~~~~~~}{(x - 4)}^{2} = 28(y + 8)} \\ {\mspace{9mu}\text{~~~~~~~~~~~~~~~~~~~}{(x - 4)}^{2} = 4 \cdot 7 \cdot (y + 8)} \end{array}\]
It follows that:
- the vertex is \(\left( {h,k} \right) = \left( {4,-8} \right)\)
- 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 \(\left( {h,k + p} \right) = \left( {4,-8 + 7} \right) = \left( {4,-1} \right)\)
- the equation of the directrix is \(y = k - p = -8 - 7 = -15\)
- the endpoints of the latus rectum are \(\left( {h \pm 2p,k + p} \right) = \left( {4 \pm 2(7),-8 + 7} \right),\) or \(\left( {-10,-1} \right)\) and \(\left( {18,-1} \right)\)
Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. See Figure 11.
Figure 11
Graph \(\left( {x + 2} \right)^{2} = -20\left( {y - 3} \right).\) Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.
Solution (click to reveal)
Vertex: \(\left( {- 2,3} \right);\) Axis of symmetry: \(x = -2;\) Focus: \(\left( {- 2, - 2} \right);\) Directrix: \(y = 8;\) Endpoints of the latus rectum: \(\left( {- 12, - 2} \right)\) and \(\left( {8, - 2} \right).\)
12.3.4 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.
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
Solution (click to reveal)
- ⓐ 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 \(p = 1.7.\)
\[\begin{array}{ll} {x^{2} = 4py} & {\begin{array}{llll} & & & \end{array}\text{Standard~form~of~upward-facing~parabola~with~vertex~(0,0)}} \\ {x^{2} = 4(1.7)y} & {\begin{array}{llll} & & & \end{array}\text{Substitute~1}\text{.7~for~}p.} \\ {x^{2} = 6.8y} & {\begin{array}{llll} & & & \end{array}\text{Multiply}.} \end{array}\]
- ⓑ 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.
\[\begin{array}{ll} {\mspace{9mu}\text{~~~~~~~}x^{2} = 6.8y} & {\text{Equation~found~in~part~(a)}.} \\ {{(2.25)}^{2} = 6.8y} & {\text{Substitute~2}\text{.25~for~}x.} \\ {\mspace{9mu}\text{~~~~~~~~~}y \approx 0.74~} & {\text{Solve~for~}y.} \end{array}\]
The dish is about 0.74 inches deep.
Section Exercises
Verbal
1. Define a parabola in terms of its focus and directrix.
Solution (click to reveal)
A parabola is the set of points in the plane that lie equidistant from a fixed point, the focus, and a fixed line, the directrix.
2. If the equation of a parabola is written in standard form and \(p\) is positive and the directrix is a vertical line, then what can we conclude about its graph?
3. If the equation of a parabola is written in standard form and \(p\) is negative and the directrix is a horizontal line, then what can we conclude about its graph?
Solution (click to reveal)
The graph will open down.
4. What is the effect on the graph of a parabola if its equation in standard form has increasing values of \(p\text{?}\)
5. As the graph of a parabola becomes wider, what will happen to the distance between the focus and directrix?
Solution (click to reveal)
The distance between the focus and directrix will increase.
Algebraic
For the following exercises, determine whether the given equation is a parabola. If so, rewrite the equation in standard form.
6. \(y^{2} = 4 - x^{2}\)
7. \(y = 4x^{2}\)
Solution (click to reveal)
yes \(x^{2} = 4\left( \frac{1}{16} \right)y\)
8. \(3x^{2} - 6y^{2} = 12\)
9. \(\left( {y - 3} \right)^{2} = 8\left( {x - 2} \right)\)
Solution (click to reveal)
yes \(\left( {y - 3} \right)^{2} = 4(2)\left( {x - 2} \right)\)
10. \(y^{2} + 12x - 6y - 51 = 0\)
For the following exercises, rewrite the given equation in standard form, and then determine the vertex \((V),\) focus \((F),\) and directrix \((d)\) of the parabola.
11. \(x = 8y^{2}\)
Solution (click to reveal)
\(y^{2} = \frac{1}{8}x,V:(0,0);F:\left( {\frac{1}{32},0} \right);d:x = - \frac{1}{32}\)
12. \(y = \frac{1}{4}x^{2}\)
13. \(y = -4x^{2}\)
Solution (click to reveal)
\(x^{2} = - \frac{1}{4}y,V:\left( {0,0} \right);F:\left( {0, - \frac{1}{16}} \right);d:y = \frac{1}{16}\)
14. \(x = \frac{1}{8}y^{2}\)
15. \(x = 36y^{2}\)
Solution (click to reveal)
\(y^{2} = \frac{1}{36}x,V:\left( {0,0} \right);F:\left( {\frac{1}{144},0} \right);d:x = - \frac{1}{144}\)
16. \(x = \frac{1}{36}y^{2}\)
17. \(\left( {x - 1} \right)^{2} = 4\left( {y - 1} \right)\)
Solution (click to reveal)
\(\left( {x - 1} \right)^{2} = 4\left( {y - 1} \right),V:\left( {1,1} \right);F:\left( {1,2} \right);d:y = 0\)
18. \(\left( {y - 2} \right)^{2} = \frac{4}{5}\left( {x + 4} \right)\)
19. \(\left( {y - 4} \right)^{2} = 2\left( {x + 3} \right)\)
Solution (click to reveal)
\(\left( {y - 4} \right)^{2} = 2\left( {x + 3} \right),V:\left( {- 3,4} \right);F:\left( {- \frac{5}{2},4} \right);d:x = - \frac{7}{2}\)
20. \(\left( {x + 1} \right)^{2} = 2\left( {y + 4} \right)\)
21. \(\left( {x + 4} \right)^{2} = 24\left( {y + 1} \right)\)
Solution (click to reveal)
\(\left( {x + 4} \right)^{2} = 24\left( {y + 1} \right),V:\left( {- 4, - 1} \right);F:\left( {- 4,5} \right);d:y = -7\)
22. \(\left( {y + 4} \right)^{2} = 16\left( {x + 4} \right)\)
23. \(y^{2} + 12x - 6y + 21 = 0\)
Solution (click to reveal)
\(\left( {y - 3} \right)^{2} = -12\left( {x + 1} \right),V:\left( {- 1,3} \right);F:\left( {- 4,3} \right);d:x = 2\)
24. \(x^{2} - 4x - 24y + 28 = 0\)
25. \(5x^{2} - 50x - 4y + 113 = 0\)
Solution (click to reveal)
\(\left( {x - 5} \right)^{2} = \frac{4}{5}\left( {y + 3} \right),V:\left( {5, - 3} \right);F:\left( {5, - \frac{14}{5}} \right);d:y = - \frac{16}{5}\)
26. \(y^{2} - 24x + 4y - 68 = 0\)
27. \(x^{2} - 4x + 2y - 6 = 0\)
Solution (click to reveal)
\(\left( {x - 2} \right)^{2} = -2\left( {y - 5} \right),V:\left( {2,5} \right);F:\left( {2,\frac{9}{2}} \right);d:y = \frac{11}{2}\)
28. \(y^{2} - 6y + 12x - 3 = 0\)
29. \(3y^{2} - 4x - 6y + 23 = 0\)
Solution (click to reveal)
\(\left( {y - 1} \right)^{2} = \frac{4}{3}\left( {x - 5} \right),V:\left( {5,1} \right);F:\left( {\frac{16}{3},1} \right);d:x = \frac{14}{3}\)
30. \(x^{2} + 4x + 8y - 4 = 0\)
Graphical
For the following exercises, graph the parabola, labeling the focus and the directrix.
31. \(x = \frac{1}{8}y^{2}\)
Solution (click to reveal)
32. \(y = 36x^{2}\)
33. \(y = \frac{1}{36}x^{2}\)
Solution (click to reveal)
34. \(y = -9x^{2}\)
35. \(\left( {y - 2} \right)^{2} = - \frac{4}{3}\left( {x + 2} \right)\)
Solution (click to reveal)
36. \(-5\left( {x + 5} \right)^{2} = 4\left( {y + 5} \right)\)
37. \(-6\left( {y + 5} \right)^{2} = 4\left( {x - 4} \right)\)
Solution (click to reveal)
38. \(y^{2} - 6y - 8x + 1 = 0\)
39. \(x^{2} + 8x + 4y + 20 = 0\)
Solution (click to reveal)
40. \(3x^{2} + 30x - 4y + 95 = 0\)
41. \(y^{2} - 8x + 10y + 9 = 0\)
Solution (click to reveal)
42. \(x^{2} + 4x + 2y + 2 = 0\)
43. \(y^{2} + 2y - 12x + 61 = 0\)
Solution (click to reveal)
44. \(- 2x^{2} + 8x - 4y - 24 = 0\)
For the following exercises, find the equation of the parabola given information about its graph.
45. Vertex is \(\left( {0,0} \right);\) directrix is \(y = 4,\) focus is \(\left( {0,-4} \right).\)
Solution (click to reveal)
\(x^{2} = -16y\)
46. Vertex is \(\left( {0,0} \right);\) directrix is \(x = 4,\) focus is \(\left( {-4,0} \right).\)
47. Vertex is \(\left( {2,2} \right);\) directrix is \(x = 2 - \sqrt{2},\) focus is \(\left( {2 + \sqrt{2},2} \right).\)
Solution (click to reveal)
\(\left( {y - 2} \right)^{2} = 4\sqrt{2}\left( {x - 2} \right)\)
48. Vertex is \(\left( {-2,3} \right);\) directrix is \(x = - \frac{7}{2},\) focus is \(\left( {- \frac{1}{2},3} \right).\)
49. Vertex is \(\left( {\sqrt{2}, - \sqrt{3}} \right);\) directrix is \(x = 2\sqrt{2},\) focus is \(\left( {0, - \sqrt{3}} \right).\)
Solution (click to reveal)
\(\left( {y + \sqrt{3}} \right)^{2} = -4\sqrt{2}\left( {x - \sqrt{2}} \right)\)
50. Vertex is \(\left( {1,2} \right);\) directrix is \(y = \frac{11}{3},\) focus is \(\left( {1,\frac{1}{3}} \right).\)
For the following exercises, determine the equation for the parabola from its graph.
51.
Solution (click to reveal)
\(x^{2} = y\)
52.
53.
Solution (click to reveal)
\(\left( {y - 2} \right)^{2} = \frac{1}{4}\left( {x + 2} \right)\)
54.
55.
Solution (click to reveal)
\(\left( {y - \sqrt{3}} \right)^{2} = 4\sqrt{5}\left( {x + \sqrt{2}} \right)\)
Extensions
For the following exercises, the vertex and endpoints of the latus rectum of a parabola are given. Find the equation.
56. \(V\left( {0,0} \right)\), Endpoints \(\left( {2,1} \right)\), \(\left( {-2,1} \right)\)
57. \(V\left( {0,0} \right)\), Endpoints \(\left( {-2,4} \right)\), \(\left( {-2,-4} \right)\)
Solution (click to reveal)
\(y^{2} = -8x\)
58. \(V\left( {1,2} \right)\), Endpoints \(\left( {-5,5} \right)\), \(\left( {7,5} \right)\)
59. \(V\left( {-3,-1} \right)\), Endpoints \(\left( {0,5} \right)\), \(\left( {0,-7} \right)\)
Solution (click to reveal)
\(\left( {y + 1} \right)^{2} = 12\left( {x + 3} \right)\)
60. \(V\left( {4,-3} \right)\), Endpoints \(\left( {5, - \frac{7}{2}} \right)\), \(\left( {3, - \frac{7}{2}} \right)\)
Real-World Applications
61. 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 \(x^{2} = 4y.\) At what coordinates should you place the light bulb?
Solution (click to reveal)
\(\left( {0,1} \right)\)
62. If we want to construct the mirror from the previous exercise such that the focus is located at \(\left( {0,0.25} \right),\) what should the equation of the parabola be?
63. 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?
Solution (click to reveal)
At the point 2.25 feet above the vertex.
64. 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?
65. The reflector in 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.
Solution (click to reveal)
0.5625 feet
66. If the reflector in 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.
67. 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.
Solution (click to reveal)
\(x^{2} = -125\left( {y - 20} \right),\) height is 7.2 feet
68. 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.
69. An object is projected so as to follow a parabolic path given by \(y = - x^{2} + 96x,\) where \(x\) is the horizontal distance traveled in feet and \(y\) is the height. Determine the maximum height the object reaches.
Solution (click to reveal)
2304 feet
70. For the object from the previous exercise, assume the path followed is given by \(y = -0.5x^{2} + 80x.\) Determine how far along the horizontal the object traveled to reach maximum height.