P is a point on the parabola x=t, y=t^2/2. A(4,1) is a fixed point. As P varies, find the minimum distance of P from A and prove that for this position of P, AP is normal to the parabola?

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1 Answer
Nov 20, 2017

The variable distance AP=s is given by

s=sqrt((t-4)^2+(t^2/2-1)^2

=>s^2=t^2-8t+16+t^4/4-t^2+1

=>s^2=t^4/4-8t+17

Differentiating w r to t we get

2s(ds)/(dt)=t^3-8

Imposing the condition of minimum (ds)/(dt)=0 we get t^3=8=>t=2

At this position the coordinates of P becomes (2,2^2/2) or (2,2)

At this stage the slope of AP will be (2-1)/(2-4)=-1/2

Now the equation of the parabola x^2=2y

So (dy)/(dx)=x

Hence slope of the normal at (2,2) will be

=-1/[(dy)/(dx)]_(2,2)=-1/2 which is slope of AP.Hence AP is the normal to the parabola.