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Homework Help: Exercise on parabolic arch

  1. Dec 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Please, take a look at this exercise.
    A parabolic arch is constructed which is 6 feet wide at the base and 9 feet tall in the middle.
    Find the height of the arch exactly 1 foot in from the base of the arch.

    I have trouble solving it. Please, guide towards a correct understanding.

    Here is how I have started thinking over the task:
    I assume the base is located on OX axis, therefore x = 3 and y = 0 for a point at the right bottom edge.
    9 feet tall implies that the focus has y value equal 9, and x value = 0, if I assume the vertex is at (0;0).

    If so, I should use x^2 = 4py formula.
    The textbook suggests this solution and I have trouble arriving at it.
    "The arch can be modeled by x2 = -(y - 9) or y = 9 - x2. One foot in from the base of the
    arch corresponds to either x = +-2, so the height is y = 9 - (+-2)^2 = 5 feet."

    Thank you!
    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 28, 2015 #2


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    Why do you consider the focus is at (0,9)? The top of a parabolic archway is not the focus of the parabola.
  4. Dec 28, 2015 #3
    9 is the value on OY axis and is not at the top of archway, because it is said, if I am not mistaken, that the arch is "9 feet tall in the middle"
  5. Dec 28, 2015 #4


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    This image labels the key points of a parabola which opens upward.


    You'll have to imagine the figure is flipped upside down to match the archway in this problem.

    What you are calling the focus of the parabola is actually the vertex. Usually, the focal point cannot easily be identified by inspection like the vertex can be.

    Since you know two points on the x-axis where the value of the parabola's equation is zero, you can use these coordinates to re-construct the equation of the parabola, even if you don't know the location of the focal point.
  6. Jan 4, 2016 #5
    Thank you. Yes, I have not used a correct name. Of course, if I turn the parabola upside down, then vertex is not at (0,0) but at
    (0, 9), because the given height of the arch in the middle is 9, correct?
    Given its width of 6 feet at the base (on OX axis), I assume there is a point at (3, 0); therefore, x = 3, y = 0.
    Also, arch is an upside parabola, hence p is negative.
    Standard parabola formula is:
    (x - h)^2 = 4p(y - k)
    (3 - 0)^2 = -4p(0 - 9)
    9 = 4p*9
    4p = 1
    p = 1/4
    Is this correct?
    I don't see how the result from textbook was achieved, and how 1 feet should be applied. Should I use the condition
    of 1 feet in from the base as an indicator of another point (3, 1) and use this point for my computations?
    Or does 1 feet in means horizontal movement from parabola edges towards its middle?
    Thank you!
  7. Jan 4, 2016 #6


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    (bolding mine)
    You see the contradiction?

    The last: "1 feet in means horizontal movement from parabola edges towards its middle", so a point on the x-axis.
  8. Jan 4, 2016 #7


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    I don't think you can make this claim that p is negative.

    p is the distance from the vertex to the focus of the parabola, and p is also the distance from the vertex to the directrix. The focus and the directrix lie on opposite sides of the vertex, regardless of whether the parabola opens upward or downward.

    The calculation of p looks OK, but you should write out the equation of the parabola. I prefer y = something.

    The problem statement wants you to find the height of the arch at a location 1 foot in from the base, i.e. where the arch intersects the ground.

    You've already found the equation of the arch, so this task is not an additional condition to impose on the shape of the arch, merely an exercise to use your previous work and find out some additional information about the arch, given its equation.
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