The given line is $y= \frac{3}{4}x+\frac{5}{4} = mx +c$
where $m=\frac{3}{4}, c $
$ = \frac{5}{4} $
$ y= mx+c $ touches $y^{2}= 4ax$ if $c= \frac{a}{m} $
$ \therefore$ the given line touches $y^{2}= 4ax$
if $\frac{5}{4}= \frac{a}{{3}/{4}} $
$ \Rightarrow a=\frac{5}{4}\times\frac{3}{4}$
$ = \frac{15}{16}$