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Thread: [RESOLVED] Parabolas & Directrix/Focus

  1. #1

    Thread Starter
    WiggleWiggle dclamp's Avatar
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    Resolved [RESOLVED] Parabolas & Directrix/Focus

    Using the formula y=ax2 i can use a to find c, the distance to the focus and the directrix from the vertex, using the forumula a=1/4c

    however i am trying to do this in code, so i need to isolate C on one side, and when i work it out i get c=1/4a, and that doesn't seem right...

    Can some one enlighten me on this?

  2. #2

    Thread Starter
    WiggleWiggle dclamp's Avatar
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    Re: Parabolas & Directrix/Focus

    Resolved...

    I worked it out, and they came out the same.

    Assuming a=1/8:

    using a=1/4c
    1/8=1/4c
    8=4c (reciprocal)
    8/4 = 4c/4 (divide by 4, each side)
    2 = c

    using c=1/4a:
    c=1/(4(1/8))
    c=1/(1/2)
    c=1 x 2 (multiply by reciprocal)
    c=2

    c = c


  3. #3
    Only Slightly Obsessive jemidiah's Avatar
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    Re: [RESOLVED] Parabolas & Directrix/Focus

    Yup, hyperbolas like you've given (equations of the form x=q/y) are their own inverses. Not many functions have that property, so it's understandable that your result seemed strange.
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