Two distinct chords of the parabola $y^2=4ax$  passing through (a, 2a) is bisected by the line  $x+y=1$.The length of the latus rectum of the parabola can be

Any point on x+y=1 is (t, 1-t)
Eq.of chord with this point as mid-point is  $y(1-t)-2a(x+t)={(1-t)}^2-4at$
It passes through (a, 2a) so  $t^2-t+2a^2-2a+1=0$
This should have two distinct real roots
$\begin{array}{l}D>0\Rightarrow a^2-a<0\Rightarrow 0<a<1\\ \Rightarrow 0<4a<4\end{array}$