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10.4: The Parabola

  • Page ID
    114099
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    Learning Objectives

    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.
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    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.

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    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.

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    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.

    A vertical upward opening parabola with Vertex (0, 0), Focus (0, p) and Directrix y = negative p. Lines of length d connect a point on the parabola (x, y) to the Focus and the Directrix. The line to the Directrix is perpendicular to it.

    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.

    STANDARD FORMS OF PARABOLAS WITH VERTEX (0, 0)

    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

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    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.

    This is a graph labeled y squared = 24 x, a horizontal parabola opening to the right with Vertex (0, 0), Focus (6, 0) and Directrix x = negative 6. Two lines extend to the parabola from the point (negative 6, 0) and are tangent to the parabola at (6, 12) and (6, negative 12).

    Figure 6

    HOW TO

    Given a standard form equation for a parabola centered at (0, 0), sketch the graph.

    1. Determine which of the standard forms applies to the given equation: y2=4pxy2=4px or x2=4py.x2=4py.
    2. 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.
      1. 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.
      2. 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)
    3. 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
    This is a horizontal parabola opening to the right with Vertex (0, 0), Focus (6, 0), and Directrix x = negative 6. The Latus Rectum is shown, a vertical line passing through the Focus and terminating on the parabola at (6, 12) and (6, negative 12).
    TRY IT #1

    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
    This is the graph labeled x squared = negative 6 y, a vertical parabola opening down with Vertex (0, 0), Focus (0, negative 3/2) and Directrix y = 3/2. The Latus Rectum is shown, a horizontal line passing through the Focus and terminating on the parabola at (negative 3, negative 3/2) and (3, negative 3/2).
    TRY IT #2

    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.

    HOW TO

    Given its focus and directrix, write the equation for a parabola in standard form.

    1. Determine whether the axis of symmetry is the x- or y-axis.
      1. 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.
      2. 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.
    2. Multiply 4p.4p.
    3. 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
    TRY IT #3

    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.

    STANDARD FORMS OF PARABOLAS WITH VERTEX (H, K)

    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

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    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.

    HOW TO

    Given a standard form equation for a parabola centered at (h, k), sketch the graph.

    1. 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).
    2. 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.
      1. 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)
      2. 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)
    3. 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
    This is the graph labeled (y minus 1) squared = negative 16(x + 3), a horizontal parabola opening to the left with Vertex (negative 3, 1), Focus (negative 7, 1), and Directrix x = 1. The Latus Rectum is shown, a vertical line passing through the Focus and terminating on the parabola at (negative 7, negative 7) and (negative 7, 9). The Axis of Symmetry, the horizontal line y = 1, is also shown, passing through the Vertex and the Focus.
    TRY IT #4

    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
    This is the graph labeled (x minus 4)squared = 28 times (y + 8), a vertical parabola opening upward with Vertex (4, negative 8), Focus (4, negative 1), and Directrix y = negative 15. The Latus Rectum is shown, a horizontal line passing through the Focus and terminating on the parabola at (negative 10, negative 1) and (18, negative 1). The Axis of Symmetry, the vertical line x = 4, is also shown, passing through the Vertex and the Focus.
    TRY IT #5

    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.

    A parabolic reflector is shown with its Focus labeled. Rays of sunlight parallel to the Axis of Symmetry all bounce off the reflector and pass through the Focus.

    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.
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    Figure 13 Cross-section of a travel-sized solar fire starter

    Answer
    TRY IT #6

    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.

    MEDIA

    Access these online resources for additional instruction and practice with parabolas.

    10.3 Section Exercises

    Verbal

    1.

    Define a parabola in terms of its focus and directrix.

    2.

    If the equation of a parabola is written in standard form and pp 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 pp is negative and the directrix is a horizontal line, then what can we conclude about its graph?

    4.

    What is the effect on the graph of a parabola if its equation in standard form has increasing values of p?p?

    5.

    As the graph of a parabola becomes wider, what will happen to the distance between the focus and directrix?

    Algebraic

    For the following exercises, determine whether the given equation is a parabola. If so, rewrite the equation in standard form.

    6.

    y2=4−x2y2=4−x2

    7.

    y=4x2y=4x2

    8.

    3x2−6y2=123x2−6y2=12

    9.

    (y−3)2=8(x−2)(y−3)2=8(x−2)

    10.

    y2+12x−6y−51=0y2+12x−6y−51=0

    For the following exercises, rewrite the given equation in standard form, and then determine the vertex (V),(V), focus (F),(F), and directrix (d)(d) of the parabola.

    11.

    x=8y2x=8y2

    12.

    y=14x2y=14x2

    13.

    y=−4x2y=−4x2

    14.

    x=18y2x=18y2

    15.

    x=36y2x=36y2

    16.

    x=136y2x=136y2

    17.

    (x−1)2=4(y−1)(x−1)2=4(y−1)

    18.

    (y−2)2=45(x+4)(y−2)2=45(x+4)

    19.

    (y−4)2=2(x+3)(y−4)2=2(x+3)

    20.

    (x+1)2=2(y+4)(x+1)2=2(y+4)

    21.

    (x+4)2=24(y+1)(x+4)2=24(y+1)

    22.

    (y+4)2=16(x+4)(y+4)2=16(x+4)

    23.

    y2+12x−6y+21=0y2+12x−6y+21=0

    24.

    x2−4x−24y+28=0x2−4x−24y+28=0

    25.

    5x2−50x−4y+113=05x2−50x−4y+113=0

    26.

    y2−24x+4y−68=0y2−24x+4y−68=0

    27.

    x2−4x+2y−6=0x2−4x+2y−6=0

    28.

    y2−6y+12x−3=0y2−6y+12x−3=0

    29.

    3y2−4x−6y+23=03y2−4x−6y+23=0

    30.

    x2+4x+8y−4=0x2+4x+8y−4=0

    Graphical

    For the following exercises, graph the parabola, labeling the focus and the directrix.

    31.

    x=18y2x=18y2

    32.

    y=36x2y=36x2

    33.

    y=136x2y=136x2

    34.

    y=−9x2y=−9x2

    35.

    (y−2)2=−43(x+2)(y−2)2=−43(x+2)

    36.

    −5(x+5)2=4(y+5)−5(x+5)2=4(y+5)

    37.

    −6(y+5)2=4(x−4)−6(y+5)2=4(x−4)

    38.

    y2−6y−8x+1=0y2−6y−8x+1=0

    39.

    x2+8x+4y+20=0x2+8x+4y+20=0

    40.

    3x2+30x−4y+95=03x2+30x−4y+95=0

    41.

    y2−8x+10y+9=0y2−8x+10y+9=0

    42.

    x2+4x+2y+2=0x2+4x+2y+2=0

    43.

    y2+2y−12x+61=0y2+2y−12x+61=0

    44.

    −2x2+8x−4y−24=0−2x2+8x−4y−24=0

    For the following exercises, find the equation of the parabola given information about its graph.

    45.

    Vertex is (0,0);(0,0); directrix is y=4,y=4, focus is (0,−4).(0,−4).

    46.

    Vertex is (0,0);(0,0); directrix is x=4,x=4, focus is (−4,0).(−4,0).

    47.

    Vertex is (2,2);(2,2); directrix is x=2−2–√,x=2−2, focus is (2+2–√,2).(2+2,2).

    48.

    Vertex is (−2,3);(−2,3); directrix is x=−72,x=−72, focus is (−12,3).(−12,3).

    49.

    Vertex is (2–√,−3–√);(2,−3); directrix is x=22–√,x=22, focus is (0,−3–√).(0,−3).

    50.

    Vertex is (1,2); (1,2);  directrix is y=113,y=113, focus is (1,13).(1,13).

    For the following exercises, determine the equation for the parabola from its graph.

    51.

    This figure shows two thumbtacks stuck in a piece of paper with a slack piece of string between them. A pencil pulls the string taught and by moving around, draws an ellipse.

    52.

    This is a horizontal parabola in the x y plane, opening to the left, with Vertex (3, 2) and Focus (negative 1, 2). The Axis of Symmetry, a horizontal line, is shown, passing through the Vertex and the Focus.

    53.

    This is a horizontal parabola in the x y plane, opening to the right, with Vertex (negative 2, 2) and Focus (negative 31/16, 2). The Axis of Symmetry, a horizontal line, is shown, passing through the Vertex and the Focus.

    54.

    This is a vertical parabola in the x-y plane, opening down, with Vertex (negative 3, 5) and Focus (negative 3, 319/64). The Axis of Symmetry, a vertical line, is shown, passing through the Vertex and the Focus.

    55.

    This is a horizontal parabola in the x y plane, opening to the right, with Vertex (negative square root of 2, square root of 3) and Focus (negative square root of 2 + square root of 5, square root of 3). The Axis of Symmetry, a horizontal line, is shown, passing through the Vertex and the Focus.

    Extensions

    For the following exercises, the vertex and endpoints of the latus rectum of a parabola are given. Find the equation.

    56.

    V(0,0)V(0,0), Endpoints (2,1)(2,1), (−2,1)(−2,1)

    57.

    V(0,0)V(0,0), Endpoints (−2,4)(−2,4), (−2,−4)(−2,−4)

    58.

    V(1,2)V(1,2), Endpoints (−5,5)(−5,5), (7,5)(7,5)

    59.

    V(−3,−1)V(−3,−1), Endpoints (0,5)(0,5), (0,−7)(0,−7)

    60.

    V(4,−3)V(4,−3), Endpoints (5,−72)(5,−72), (3,−72)(3,−72)

    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 x2=4y.x2=4y. At what coordinates should you place the light bulb?

    62.

    If we want to construct the mirror from the previous exercise such that the focus is located at (0,0.25),(0,0.25), 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?

    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.

    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.

    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=−x2+96x,y=−x2+96x, where xx is the horizontal distance traveled in feet and yy is the height. Determine the maximum height the object reaches.

    70.

    For the object from the previous exercise, assume the path followed is given by y=−0.5x2+80x.y=−0.5x2+80x. Determine how far along the horizontal the object traveled to reach maximum height.


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