FP1 - Parabola question

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  1. ian.slater's Avatar
    • Exalted and Worshipped Member
    Re: FP1 - Parabola question
    (Original post by Aeyuin)
    Sorry, could you explain this again? I understand how parametric equations work but I don't see how you've formed this one.

    x = at^2 ; y = 2at ; y^2 = 4ax

    Basically, where did these values come from?

    And why is mc, a? I've never even seen a question where the intercept x the gradient was significant...
    The first bit is standard bookwork. The parabola y^2 = 4ax (focus at (a,0), directrix x = -a) can be represented as (at^2, 2at). So I quoted this without justification.

    The question defined the general tangent as y = mx +c.

    The general tangent came out as y = (1/t)x + at

    So m = 1/t , c = at, and hence mc = a which is what the question wants - a constant. Importantly this result has no 't' in it. This is just a property of parabolas.
  2. kitkat19's Avatar
    • Benevolent Member
    • Posts: 774
    Re: FP1 - Parabola question
    What exam board is this question from. I do AQA and we never get questions like this....
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