the length of the focal chord of the parabola  y² = 4ax at a distance b from the vertex is c. Then 

1)  a² = bc

2)  a³ = b²c

3)  b² = ac

4)  b²c = 4a³

 The ans is b2c = 4a3.

the proof is :

Consider any focal chord passing through focus S(a,0) and meeting parabola at P(at²,2at) and Q(at₁²,2at₁).

Slope of QS = Slope of PS (the focal chord)

Equate them and u'll get t₁ = - 1 / t

So u can write point Q as ( a / t², -2a / t )

According to question QP = c ( length of focal chord )

So, (2a)² (t + 1/t)² + (a)² (t² - 1/t²)² = c²

Solve it to get a( t + 1/t )² = c ------- (1)

We are given that the focal chord is at a distance b form vertex.

So, equation of Focal chord comes out to be

(2t) x - (t² - 1) y - 2at = 0

So, length of focal chord = - 2at / (t² + 1)

So, - 2at / (t² + 1) = b

or, ( t + 1/t ) = -2a / b ------- (2)

From (1) and (2),

a ( - 2a / b )² = c

rearranging,

a ( 4a² / b² ) = c

4a³ = b²c..............!!!!