Question

`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?

Thanks!

Answer 1

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.