In the function r=a(1+sinθ), as sinθ can take a value between [−1,1}, r can take values between [0,2a]
Let us select a few points such as θ={0,π3,π2,2π3,π,4π3,3π2,5π3,2π}
Then putting these values of θ in r=a(1+sinθ), we get values of r as {a,3a2,2a,3a2,a,a2,0,a2,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 θ and corresponding π−θ value is same, hence the curve is symmetric around y-axis.
Let us also convert them to rectangular coordinates using x=rcosθ, y=rsinθ and r2=x2+y2
Hence r=a(1+sinθ)⇔r2=ar+arsinθ or
x2+y2=a√x2+y2+ay
Again note that when x=0, we have y2=2ay i.e. y=0 or y=2a - (relating to θ=π2 and θ=3π2.
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]}
