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Aqa FP3 Polar Geometry(?)

I've been trying the paper below (Question 7). I've managed part a) but not part b).

The diagram looks as if the curve is symmetrical about the initial line and from there I would have done a bit of geometry to find where S was and so on. But I didn't know how to prove the curve was symmetrical so I decided to look at the answer. The first line says OS = 8 and it's only worth 1 mark so it must be obvious. But I don't know why.

Can anyone help please?

http://filestore.aqa.org.uk/subjects/AQA-MFP3-W-QP-JAN07.PDF

http://filestore.aqa.org.uk/subjects/AQA-MFP3-W-MS-JAN07.PDF
Original post by maggiehodgson
I've been trying the paper below (Question 7). I've managed part a) but not part b).

The diagram looks as if the curve is symmetrical about the initial line and from there I would have done a bit of geometry to find where S was and so on. But I didn't know how to prove the curve was symmetrical so I decided to look at the answer. The first line says OS = 8 and it's only worth 1 mark so it must be obvious. But I don't know why.

Can anyone help please?

http://filestore.aqa.org.uk/subjects/AQA-MFP3-W-QP-JAN07.PDF

http://filestore.aqa.org.uk/subjects/AQA-MFP3-W-MS-JAN07.PDF


What's the value of theta for OS? Look at the geometry.

To show symmetry about the intial line you want to show r(θ)=r(θ)r(\theta) = r(-\theta) treating r as a function of theta, which it is.
Original post by maggiehodgson
I've been trying the paper below (Question 7). I've managed part a) but not part b).

The diagram looks as if the curve is symmetrical about the initial line and from there I would have done a bit of geometry to find where S was and so on. But I didn't know how to prove the curve was symmetrical so I decided to look at the answer. The first line says OS = 8 and it's only worth 1 mark so it must be obvious. But I don't know why.

Can anyone help please?

http://filestore.aqa.org.uk/subjects/AQA-MFP3-W-QP-JAN07.PDF

http://filestore.aqa.org.uk/subjects/AQA-MFP3-W-MS-JAN07.PDF


cosθ=cos(θ)\cos\theta=\cos(-\theta) so curve must be symmetrical about the initial line
Original post by ghostwalker
What's the value of theta for OS? Look at the geometry.

To show symmetry about the intial line you want to show r(θ)=r(θ)r(\theta) = r(-\theta) treating r as a function of theta, which it is.


Thank you. Learn such a lot on TSR.
Original post by brianeverit
cosθ=cos(θ)\cos\theta=\cos(-\theta) so curve Thanks. Makes sense. Another trick to add to my notebook.

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