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FP2 Cartesian to Polar

Determine the polar equation of (1/x^2)+(1/y^2)=1. I substituted x=rcosΘ and y=rsinΘ to obtain

(1/(r^2cos^2Θ) + (1/r^2sin^2Θ) = 1

It follows that

1 + (r^2cos^2Θ)/(r^2sin^2Θ) = r^2cos^2Θ

r^2sin^2Θ + r^2cos^2Θ=r^4cos^2Θsin^2Θ
Then factorising LHS and dividing by r^2 gives

Sin^2Θ+cos^2Θ=r^2cos^2Θsin^2Θ
And hence r^2=cosec^2Θ+sec^2Θ
But the answer in the book is r = cosec^2Θ+sec^2Θ
Not r^2 so have I done something algebraically incorrect or is this a mistake in the book ?
Original post by Mathematicus65

And hence r^2=cosec^2Θ+sec^2Θ
But the answer in the book is r = cosec^2Θ+sec^2Θ
Not r^2 so have I done something algebraically incorrect or is this a mistake in the book ?


Typo in the book.
Reply 2
Avatar for Rkai01
Rkai01
6
just a bit of suggestions needed. Im thinking of doing M2 edexcel. I did M1 last year (only got C with a weeks revision). If I take M2 is it worth taking M1 again if they overlap alot? I have to get like 70 in M2 for A but am aiming for A* (Need 172ums)
Original post by Rkai01
just a bit of suggestions needed. Im thinking of doing M2 edexcel. I did M1 last year (only got C with a weeks revision). If I take M2 is it worth taking M1 again if they overlap alot? I have to get like 70 in M2 for A but am aiming for A* (Need 172ums)

I would say do it again, it won't be too much extra work and will reassure you on some M2 topics
Reply 4
Avatar for Rkai01
Rkai01
6
Original post by Mathematicus65
I would say do it again, it won't be too much extra work and will reassure you on some M2 topics

Thats what my head is telling me. My heart is saying that it'll detract attention from FP2 and other modules so its annoying. I also hear M2 is easier than M1.
Original post by Rkai01
Thats what my head is telling me. My heart is saying that it'll detract attention from FP2 and other modules so its annoying. I also hear M2 is easier than M1.


I think some parts are easier, because a lot of it is just AS Physics (well it is for OCR anyway) but some parts are far harder I think as well

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