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P3 question help !! watch

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1. Jan 02 paper P3, last question, q8

http://www.mathsexams.ukteachers.com...3_2002_Jan.pdf

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
2. 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
3. Can anyone do part b pls?
4. (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

Euclid

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