Question

Sol. [Key Concept: Same midpoint-chord condition as Q1 ($T = S_{1}$ with midpoints on $x + y = 1$); the discriminant for two distinct chords gives the admissible latus-rectum values.]

A chord of $y^{2} = 4ax$ with midpoint $(h,k)$ is $ky - 2a(x + h) = k^{2} - 4ah$;

it passes through $(a,2a)$ with the midpoint on $x + y = 1\ (k = 1 - h)$.

This yields $h^{2} - 2h + (1 - 2a + 2a^{2}) = 0$;

two distinct chords need discriminant $> 0:\ \ 2a(1 - a) > 0 \Rightarrow \ 0 < a < 1$.

So the latus rectum $4a \in (0,4)$ — it can be $1,\ 2,\ 3$ but not $4$ → (a), (c), (d).