If the area of the triangle inscribed in the parabola y2=4ax with one vertex at the vertex of the parabolaand other two vertices at the extremities ofa focal chord is 5a2/2then the length of the focalchord is

Let the vertices be O(0,0),Aat2,2atBat2,−2at then 12||0012t22at1at22at11=5a22⇒2t2−5t+2=0⇒ t=2 or 1/2so the vertices of a focal chord are(4a, 4a) and  (a/4, – a) (Taking t = 2) and lengthof this focal chord is 25 a/4.