The Student Room Group

Polar equations

I have the equation x - y = 3 and need to find its polar equivalent.
I got to the expression r = 3 / (cos theta - sin theta) but the mark scheme seems to divide r(cos theta - sin theta) = 3 by root 2 and then manipulate to get (3 over root 2) sec(theta plus pi over 4)

Would my answer still be correct, and how does their one work?
It looks as though the mark scheme has simply used cos@ - sin@ = Rcos(@ + a), and found the values of R and a.

Although your answer is correct, I don't know whether it would have gained full marks. The mark scheme answer is arguably more useful as it provides an easier route to finding the value(s) of @ corresponding to a given value of r.
(edited 2 months ago)

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