In the function #r=a(1+sintheta)#, as #sintheta# can take a value between #[-1,1}#, #r# can take values between #[0,2a]#

Let us select a few points such as #theta={0,pi/3,pi/2,(2pi)/3, pi,(4pi)/3,(3pi)/2,(5pi)/3,2pi}#

Then putting these values of #theta# in #r=a(1+sintheta)#, we get values of #r# as #{a,(3a)/2,2a,(3a)/2,a,a/2,0,a/2,a}#. As #r=0# is one of the points, it passes trough origin.

Observe that values move in a cycle and hence the curve is a closed one. Further for each #theta# and corresponding #pi-theta# value is same, hence the curve is symmetric around #y#-axis.

Let us also convert them to rectangular coordinates using #x=rcostheta#, #y=rsintheta# and #r^2=x^2+y^2#

Hence #r=a(1+sintheta)hArr r^2=ar+arsintheta# or

#x^2+y^2=asqrt(x^2+y^2)+ay#

Again note that when #x=0#, we have #y^2=2ay# i.e. #y=0# or #y=2a# - (relating to #theta=pi/2# and #theta=(3pi)/2#.

If we assume #a=4# the graph appears as below. Such a curve is known as **cardioid**.

graph{x^2+y^2-4sqrt(x^2+y^2)-4y=0 [-9.71, 10.29, -2, 9]}