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#
