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    • Thread Starter

    This is probably easy stuff for anyone here who's done FP1, but I'm having trouble getting my head round the basics:

    Find the coordinates of the focus and an equation of the directrix of the parabola with equation: 4y=x^2 - 16

    See I thought:
    Standard form is x^2 = 4ay
    For x^2 = 4y, a is (0,1) and directrix is y=-1
    x^2 = 4(y + 4) is a vertical positive translation of x^2 = 4y by 4
    so a is (0,5) and directrix is y=3

    But no - plz help!
    • Thread Starter

    Actually solved it myself, i was transofrming wrong:

    x^2 = 4(y + 4)
    y=0.25x^2 - 4

    so downward translation of 4 from x^2 = 4y [y = 0.25x^2]
    to give a is (0,-3), directrix is y + 5 = 0

    That's FP3.

    Off topic but it's scary how the grade 10/11 people in the US do this for Advanced Algebra.
    I had to teach that and coming from an English educational background and not done the further maths modules before, I was like wt...o_O

    Its the old FP2 actually. Does the new FP1 start next year?

    To be honest, I'm not surprised its taught at grade 10/11. Limited as my knowledge of American education system is, I have an inclinged that they emphasis alot more on geometry than us brits do. So yeah I can definitely see this being taught. Its not difficult, just different really.
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Updated: June 13, 2008

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