If the area of the triangle inscribed in the parabola 

$y^2=4ax$ with one vertex at the vertex of the parabola
and other two vertices at the extremities of
a focal chord is $5a^2/2$then the length of the focal
chord is

Let the vertices be $O(0,0),A\left(a{t}^2,2at\right)$

$B\left(\frac{a}{t^2},\frac{-2a}{t}\right)$ then 

${\left.\frac{1}{2}\left|\right|\begin{array}{ccc}0& 0& 1\\ 2t^2& 2at& 1\\ \frac{a}{t^2}& \frac{2a}{t}& 1\end{array}\right|}_1=\frac{5a^2}{2}\Rightarrow 2t^2-5t+2=0$

$\Rightarrow t=2\text{ or }1/2$so the vertices of a focal chord are

$(4a, 4a)$and $(a/4, – a)$ (Taking $t = 2$) and length
of this focal chord is $25 a/4$.