What the is the polar form of #y^2 = 1/y-1/x+x^2/y #?

1 Answer
Ananda Dasgupta
Feb 26, 2018

# r^3 sin^3 theta cos theta -r^2 cos^3 theta = cos theta -sin theta #

Explanation:

Using the relation #x = r cos theta , quad y= r sin theta# between Cartesian and polar coordinates, we have

#r^2 sin^2 theta = 1/{r sin theta}-1/{r cos theta}+{r^2 cos ^2 theta}/{r sin theta}#

Multiplying both sides by #r sin theta cos theta# lets us simplify this a bit to

# r^3 sin^3 theta cos theta = cos theta -sin theta +r^2 cos^3 theta#
or
# r^3 sin^3 theta cos theta -r^2 cos^3 theta = cos theta -sin theta #

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