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the dome over a town hall has a parabolic shape. the dome measures 48 m across and rises 12m at the centre. A vertical column needs to be attached to the dome at a point that is 4m away from its rim. How tall is the dome at this point? Also, what is the equation that models the shape of this dome?

  • math -

    Take the vertex of the parabola at x=0.
    Then the equation of the parabola (opening downwards) is y=12-ax^2 where a is a constant to be found.
    We know that y=0 at x=±24 (half span), so
    0=12-a(24^2)
    =>
    a=12/576=1/48
    The equation of the parabola is therefore
    y=12-x^2/48.

    I will let you finish the problem.

  • math -

    all these numbers kind of confuse me. Im not sure what ^ means. Right now the only equation i have learned is y= a(x-s)(x-t)

  • math -

    x^2 means x squared, or x².

    If you just started with parabolas, it won't hurt to say what you've learned so far.

    Let's start over:
    We first assume the vertex of the parabola is at x=0.
    We know that y=0 at x=±24, or
    y=a(x-24)(x+24).
    We also know that y=12 at x=0, i.e.
    12=a(0-24)(0+24)=-576
    so a=12/(-576)=-1/48
    The equation of the parabola is therefore:
    y=(x-24)(x+24)/48

    Can you then continue?

  • math -

    Correction to one of the above lines:

    12=a(0-24)(0+24)=-576a

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