Question 751296
Assuming you mean


{{{y=1/(2x)}}}



we can write


{{{
2yx=1
}}}

then 

{{{
2*(r*sin(theta))*(r*cos(theta))=1
}}}

then

{{{
r^2(2sin(theta)cos(theta))=1
}}}

{{{
r^2*sin(2*theta)=1
}}}


and finally

{{{
r(theta)=sqrt(csc(2*theta))
}}}



:)