$y^2=4ax$-----(i)
Vertex $=(0,0)$
$m= \tan \theta$
Hence equation of the chord is $y=(\tan \theta)x$
Put in (i) we have $\tan ^2 \theta (x^2)=4ax$
=> $x= \large\frac{4a}{\tan ^2 \theta}$
$y= \large\frac{4a}{\tan \theta}$
Length of the chord $=\sqrt {\bigg(\large\frac{4a}{\tan 2t} \bigg)^2+\bigg(\large\frac{4a}{\tan \theta}\bigg)^2}$
$\qquad= \large\frac{4a}{\tan ^2 \theta} $$ \sqrt {1+\tan ^2 \theta}$
$\qquad= \large\frac{4a \cos ^2 \theta}{\sin ^2t } \times \large\frac{1}{\cot \theta}$
$\qquad= \large\frac{4a \cos \theta}{\sin ^2 \theta}$
Hence A is the correct answer.