Let S be the given focus and ZM be the given line.
Then, SZ=ea−ae =ea(1−e2)=aeb2=k (say)
as b2=a2(1−e2)
Now, take SC as the x-axis and LSL' as the y-axis. Let (x,y) be the coordinates of B with respect to these axes. Then, x=SC= ae and y=CB=b.
Hence, xy2=aeb2=SZ which is constant
Therefore, y2=kx is the required locus which is a parabola.