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P3 question help !! watch

    • Thread Starter

    Jan 02 paper P3, last question, q8


    Help would be appreciated, mostly on c onwards but all would be great as I am not sure about rest

    Guarenteed rep if you are bothered about it

    when x = 0, y = 20/5sin alpha

    when y = 0, x = 0/4cos alpha

    0.5 x base x height=

    0.5 x 20/4cosalpha x 20/5sinalpha

    = 20/sin2alpha

    thats a quarter of the total area

    total area = 4 x 20/sin2alpha - 20pi = 80/sin 2alpha - 20pi

    d) 80/sin 2alpha - 20pi = 20 pi

    80/sin 2alpha= 40 pi

    work out alpha.......0.345054
    • Thread Starter

    Can anyone do part b pls?

    (Original post by lgs98jonee)
    Can anyone do part b pls?
    The top half of the ellipse can be modelled by taking only positive values of x.

    So the equation y = rt(16(1-(x^2)/25)) will be for the top half. You also know the ellipse crosses the x axis at 5 and -5. So integrating y between limits 5 and -5, will bring half the area.

    Int[rt(16(1-(x^2)/25))] (-5 to 5) = 10Pi.

    So area of entire ellipse = 20Pi

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Updated: January 10, 2005

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