1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Downward parabola: value of p = 1/a?

  1. Feb 29, 2016 #1
    1. The problem statement, all variables and given/known data
    I am repeating conics, and have stumbled upon this paragraph on purplemath.com
    Then a = –1/9. With a being the leading coefficient from the regular quadratic equation y = ax^2 + bx + c, I also know that the value of 1/a is the same as the value of 4p, so 1/(–1/9) = –9 = 4p, and thus p = –9/4. "

    I am not aware of the rule that states that 4p equals 1/a for a parabola that has a vertex other than
    (0, 0), which is the case here. The author refers to some general formula of parabola y = ax^2 + bx + c.
    If vertex is at (0, 0 ), then the formula is x^2 = 4py, hence y = 1/4p * x^2, where a = 1/4p, hence
    a = 1/4p. But in the given example, the vertex is not at (0, 0) - it's at (0, 25). Why a = 1/4p if the
    vertex is not (0, 0)?

    I would be grateful for explanation.

    Thank you!

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 29, 2016 #2
    This result can easily be shown for any parabola, not just one with vertex at the orgin e.g in this section of a Wikipedia page.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted