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fp1 focus,directrix and parabolas

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    • Thread Starter


    question 2 here

    I know that since the co-ordinates which will be equidistant away from the point and line. It'll form a parabola which is a U shape with the curved lines pointing left.

    But how am i supposed to work out the equation of that curved line?

    is it just y=-4ax? or do i need to show some proof of that?

    You can use y^2 =4ax you wouldn't have to show proof.
    • Thread Starter

    (Original post by NotNotBatman)
    You can use y^2 =4ax you wouldn't have to show proof.
    so y^2 =4ax

    and since it's on the other half of the graph the it's a minus and since you got a point -3 a=-3

    • Thread Starter

    tfw on question 5 when u did everything right and then u did z=x^-1 y instead of z=yx^-1 on the last part


    It always helps to be able to derive these in an emergency

    take a generic point (x,y)
    distance to point (-3,0) is sqrt((x+3)^2 + y^2)
    distance to line x = 3 is x-3

    these must be equal

    x-3 = sqrt((x+3)^2 + y^2)
    (x-3)^2 = (x+3)^2 + y^2
    x^2 - 6x + 9 = x^2 + 6x + 9 + y^2
    y^2 = -12x
    x = -y^2/12
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