Parametrically let us consider that #(t^2,2t)# represents the coordinates of any point #P# on the given parabola,#y^2=4x#.

Differentiating the equation of the parabola w r to #x# we get

#2y(dy)/(dx)=4#

#=>(dy)/(dx)=2/y#

So slope of the tangent at #P# and also that of a normal at #Q# parallel to this tangent will be given by

#m=[(dy)/(dx)]_("("t^2,2t")")=2/(2t)=1/t#

Hence the **tangent** at the point #Q# will have slope #m'=-t#. So applying reverse logic we can say that the coordinates of #Q# will be #(1/t^2,-2/t)#

Now equation of the tangent at #Q# will be

#y-2t=m(x-t^2)#

#=>y-2t=1/t(x-t^2)#

#=>yt-2t^2=(x-t^2)#

#=>x-yt+t^2=0#

So length of the perpendicular drawn from #Q# on the tangent drawn at #P# will be

#=(1/t^2-(-2/t)t+t^2)/sqrt(t^2+1)#

#=(t+1/t)^2/sqrt(t^2+1)#

#=(t^2+1)^2/(t^2sqrt(t^2+1))#

By the condition of the problem

#(t^2+1)^2/(t^2sqrt(t^2+1))=2sqrt2#

#=>(t^2+1)^(3/2)/t^2=2sqrt2#

Let #t^2=a#

#=>(a+1)^(3/2)/a=2sqrt2#

#=>(a+1)^3=(2sqrt2)^2a^2#

#=>a^3+3a^2+3a+1-8a^2=0#

#=>a^3-5a^2+3a+1=0#

#=>a^3-a^2-4a^2+4a-a+1=0#

#=>a^2(a-1)-4a(a-1)-1(a-1)=0#

#=>(a-1)(a^2-4a-1)=0#

When #a=1#

#=>t^2=1#

#=>t=pm1#

When

#(a^2-4a-1)=0#

#=>a=(4pmsqrt((-4)^2-4*1*(-1)))/(2*1)#

#=>a=(4pmsqrt20)/(2*1)#

#=>a=(4pm2sqrt5)/(2*1)#

#=>t^2=(2pmsqrt5)#

Neglecting imaginary roots we get

#t=pmsqrt(sqrt5+2)#

Hence the possible values of gradient #m=1/t=pm1 and pm1/sqrt(sqrt5+2)=pmsqrt(sqrt5-2)#

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