The given equation of the parabola

#y=4ax^2#

#=>x^2=4xx1/(16a)y#

Hence the parabola is symmetric about y-axis i,e its axis is Y-axis .

The co-ordinates of its vertex is #=>(0,0)#.

The co-ordinates of its focus ( F)#=>(0,1/(16a))#

The equation of its directrix,#=>y=-1/(16a)#

Let the co-ordinates of any point P on it in parametric form be #((2t)/(16a),t^2/(16a))#,where # t# is the parameter.

Now slope of the tangent at P

#((dy)/(dx))_(((2t)/(16a),t^2/(16a))) =4axx2xx(2t)/(16a)=t#

So slope of the normal at P #=-1/t#

So equation of the normal at P

#y-t^2/(16a)=-1/txx(x-(2t)/(16a))#

Putting ,#y=-1/(16a)# in the equation of the normal we get the X=coordinate of the point of intersection (Q) of the normal with the directrix

#-1/(16a)-t^2/(16a)=-1/txx(x-(2t)/(16a))#

#=>-1/(16a)-t^2/(16a)=-x/t+2/(16a)#

#=>x/t=t^2/(16a)+3/(16a)#

#=>x=t^3/(16a)+(3t)/(16a)#

So co-ordinates of Q is #(t^3/(16a)+(3t)/(16a),-1/(16a))#

So

#PQ^2=[(t^3/(16a)+(3t)/(16a)-(2t)/(16a))^2+(t^2/(16a)+1/(16a))^2]#

#=1/(16a)^2[(t^3+t)^2+(t^2+1 )^2]#

#=1/(16a)^2[t^2(t^2+t)^2+(t^2+1 )^2]#

#=1/(16a)^2[(t^2+t)^2(t^2+1 )]#

#=1/(16a)^2[(t^2+t)^3]#

No

#FP^2=((2t)/(16a))^2+(t^2/(16a)-1/(16a))^2=(t^2+1)^2/(16a)^2#

#=>FP^3=(t^2+1)^3/(16a)^3#

Hence

#(PQ^2)/(FP^3)=16a#

#=>PQ^2=16aFP^3#