Any line through origin is $y=mx$ .
Since it is a tangent to $y^2=4a(x-a)$
it will cut it in two coincident points.
Roots of $m^2x^2-4ax+4a^2$ are $B^2-4AC=0$
$16a^2-16a^2m^2=0$
(or) $m^2=1 \;(or)\; m=1,-1$
Hence it is right angle.
Hence A is the correct answer.