Ampersand curve in polar coordinates

Question

I got an assignment to write a program which draws the ampersand curve. The equation of ampersand curve looks like this: $(y^2-x^2)(x-1)(2x-3)=4(x^2+y^2-2x)^2$

I was given an advice to convert this equation into the polar form and so I did. With help of maple I got this little beastie(I wonder if there is a simpler form...): $$\frac{1}{4}\cdot \frac{11\cdot cos(\theta)+10\cdot cos(\theta)^3+\sqrt3\cdot \sqrt{-(21\cdot cos(\theta)^2-16)\cdot (2\cdot cos(\theta)^2-1)^2)}}{-cos(\theta)^2+2\cdot cos(\theta)^4+2}$$

But there is a problem. I get discontinuities in the curve(look at the loops). You can see the picture below. I guess something is wrong with the polar equation. Is there a way to get the polar form of ampersand curve without discontinuities?

Answer 1

restart:
xyexpr:=(y^2-x^2)*(x-1)*(2*x-3)=4*(x^2+y^2-2*x)^2:
plots:-implicitplot(xyexpr,x=-1..2,y=-2..2,
                    grid=[200,200],gridrefine=3);

#ee:=eval(xyexpr,[x=r*cos(t),y=r*sin(t)]):
#S:=[solve(simplify((rhs-lhs)(ee)),r)][2]:
#S:=subs(-252*cos(t)^6+444*cos(t)^4-255*cos(t)^2+48
#        =-(3*(21*cos(t)^2-16))*(2*cos(t)^2-1)^2,S);
S:=(1/4)*(10*cos(t)^3+11*cos(t)+(-3*(21*cos(t)^2-16)
   *(2*cos(t)^2-1)^2)^(1/2))/(2*cos(t)^4-cos(t)^2+2):
plot(S,t=0..2*Pi,gridlines=false,
     coords=polar,adaptive=true,numpoints=2000);

Answer 2

Substituting $x=r\cos\theta$ and $y=r\sin\theta$ and solving quadratically for $r$ yields

$$\small{r=\frac{\cos\theta(5\cos2\theta+16)\pm\sqrt{\cos^2\theta(5\cos2\theta+16)^2-8(16\cos^2\theta+3\cos2\theta)(\cos2\theta\cos^2\theta+2)}}{4(\cos2\theta\cos^2\theta+2)}} $$