The length of a focal chord of the parabola ${\mathrm{y}}^2=4\mathrm{ax}$ at a distance 'b' from the vertex is 'c' then 

$$\begin{gathered} Let one end of the focal chord be ( a{t}^2,2a t ) \\ , then the length of the that chord is c = a {( t + \frac{1}{t} )}^2\cdots ( 1 ) \\ Now equation of the focal chord will be y - 0 = ( \frac{2at-0}{a{t}^2-a} ) ( x - a ) \\ \Rightarrow y=\frac{2t}{t^2-1}(x-a) \\ As we know that vertex coordinates of the parabola is ( 0 , 0 ) \\ Hence perpendicular dis\tan ce from will be \\ b = \frac{\left|{\displaystyle \frac{-2at}{t^2-1}}\right|}{\sqrt{1+{\displaystyle \frac{4t^2}{{(t^2-1)}^2}}}}= \frac{| 2 a t |}{t^2+1} \cdots ( 2 ) \\ from ( 1 ) \times {( 2 )}^2{ }^{ } , \\ t will get eliminated and we get \\ \Rightarrow b^2 c= 4a^3 \end{gathered}$$