Question

What the is the polar form of `2 = x^2y-x/y^2 +xy^2`?

Answer 1

`r^4sin theta cos theta ( cos theta+sin theta )`
`- 2 r - cos theta csc^2theta= 0, r ne 0.`

Explanation:

I have realized that my previous answer was wrong. Thank you

Sonnhard. Now, I do it again.

Using

`0 <= sqrt ( x^2 + y^2 ) = r and ( x, y ) = r ( cos theta, sin theta )`,

`2 = x^2y -x/y^2 + xy^2` converts to

`r^4sin theta cos theta ( cos theta+sin theta )`

`- 2 r - cos theta csc^2theta= 0, r ne 0.`

Graph, from the given Cartesian equation:
graph{ x^2y -x/y^2 + xy^2 - 2 = 0}

As `x and y to 0, parabolic x/y^2 to - 2`.

In polars, as `r to 0, cos theta/sin^2theta to 0 rArr cos theta to 0`

`rArr abs theta to pi/2`