Let the equation of the parabola in standard form be #y^2=4ax#.So the coordinates of its focus S will be #(a,0)#.
Let the coordinates of one end point A of the given focal chord AB be #(at_1^2,2at_1)# and that of other end point B be #(at_2^2,2at_2)#
AS and SB line segments are on same straight line. So their slopes must be same.
Hence #(2at_1-0)/(at_1^2-a)=(2at_2-0)/(at_2^2-a)#
#=>t_1/(t_1^2-1)=t_2/(t_2^2-1)#
#=>t_1(t_2^2-1)=t_2(t_1^2-1)#
#=>t_1(t_2^2-1)-t_2(t_1^2-1)=0#
#=>t_1t_2^2-t_1-t_2t_1^2+t_2=0#
#=>t_1t_2(t_2-t_1)+(t_ 2-t_1)=0#
#=>(t_1t_2+1)(t_2-t_1)=0#
As A and B are different points then #t_1!=t_2#
So #t_2=-1/t_1#
Now it is given that #AS=2and SB=4#
So
#(SB)/(SA)=4/2=2#
#=>(SB)^2/(SA)^2=4#
#=>((at_2^2-a)^2+(2at_2-0)^2)/((at_1^2 -a)^2+(2at_1-0)^2)=4#
#=>((t_2^2-1)^2+4t_2^2)/((t_1^2 -1)^2+4t_1^2)=4#
#=>(t_2^2+1)^2/(t_1^2 +1)^2=4#
#=>(t_2^2+1)/(t_1^2 +1)=2#
Inserting #t_2=-1/t_1# we get
#=>(1/t_ 1^2+1)/(t_1^2 +1)=2#
#=>1/t_1^2=2#
#=>t_1^2=1/2#
Now
#SA^2=4#
#=>(at_1^2-a)^2+(2at_1-0)^2=4#
#=>a^2[t_1^2-1)^2+4t_1^2]=4#
#=>a^2[(t_1^2+1)^2]=4#
#=>a(t_1^2+1)=2#
Inserting #t_1^2=1/2# we get
#a(1/2+1)=2#
#=>a xx3/2=2#
#=>a=4/3#
So length of the Latus rectum will be
#=4a=4*4/3=16/3#
A shortcut
As #SA:SB=1:2# and Coordinates of #S# is #(a,0)#
So
#(2at_1*2+2at_2*1)/3=0#
#=>t_2=-2t_1#
#=>-1/t_ 1=-2t_1#
#=>t_1^2=1/2#
