# 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!

Nov 20, 2017

The variable distance $A P = s$ is given by

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

$\implies {s}^{2} = {t}^{2} - 8 t + 16 + {t}^{4} / 4 - {t}^{2} + 1$

$\implies {s}^{2} = {t}^{4} / 4 - 8 t + 17$

Differentiating w r to t we get

$2 s \frac{\mathrm{ds}}{\mathrm{dt}} = {t}^{3} - 8$

Imposing the condition of minimum $\frac{\mathrm{ds}}{\mathrm{dt}} = 0$ we get ${t}^{3} = 8 \implies t = 2$

At this position the coordinates of $P$ becomes $\left(2 , {2}^{2} / 2\right) \mathmr{and} \left(2 , 2\right)$

At this stage the slope of $A P$ will be $\frac{2 - 1}{2 - 4} = - \frac{1}{2}$

Now the equation of the parabola ${x}^{2} = 2 y$

So $\frac{\mathrm{dy}}{\mathrm{dx}} = x$

Hence slope of the normal at $\left(2 , 2\right)$ will be

$= - \frac{1}{\frac{\mathrm{dy}}{\mathrm{dx}}} _ \left(2 , 2\right) = - \frac{1}{2}$ which is slope of $A P$.Hence $A P$ is the normal to the parabola.