Answer
(a) $r^2=sin~2θ$
(b) See graph below
Work Step by Step
(a)
$x=r~cos~θ$
$y=r~sin~θ$
$x^2+y^2=r^2$
$(x^2+y^2)^2=2xy$
$r^4=2r^2cos~θ~sin~θ$
$r^2(r^2-2~cos~θ~sin~θ)=0~~$ Since $r\ne0$
$r^2-2~cos~θ~sin~θ=0$
$r^2=2~cos~θ~sin~θ$
$r^2=sin~2θ$
$r=±\sqrt {sin~2θ}$
$r=\sqrt {sin~2θ}$ and $r=-\sqrt {sin~2θ}$ provide the same graph.
(b) Use:
$r=\sqrt {sin~2θ}$
