The relation between polar coordinates #(r,theta)# and Cartesian coordinates #(x,y)# is #x=rcostheta#, #y=rsintheta# and #r^2=x^2+y^2#.

We can use this to convert equation in polar coordinates to an equation with Cartesian coordinates.

#r^2=3sin2theta#

#hArrr^2=3xx2sinthetacostheta#

or #r^2xxr^2=6xxrsinthetaxxrcostheta#

or #(x^2+y^2)^2=6xy#

Note that

**(a)** As #6xy# is a complete square, it is positive and hence curve can lie only in first and third quadrant.

**(b)** Further as maximum value of #r^2=3sin2theta#, maximum possible value for #r^2# is #3# and so #r# cannot be more than #sqrt3=1.732....#

**(c)** As replacing #x# and #y# with each other does not change the equation, it is symmetric along #x=y#.

Now we can put different values of #x# to get #y# (both less than #sqrt3#) and draw the graph.

The function appears as follows.

graph{((x^2+y^2)^2-6xy)(x-y)=0 [-5, 5, -2.5, 2.5]}