Analytic Geometry – Problem 37: Find the minimum distance from the point to the curve , and give the point(s) on…

Let $C$ be the parabola $x^2=8y$. Find the minimum distance from the point $P(0,5)$ to the curve $C$, and give the point(s) on $C$ where it is attained.

(1) Parameterize $C: x^2=8y$ as $$(x,y)=(4t,\,2t^2),\quad t\in\mathbb R.$$ (2) Distance squared from $(4t,2t^2)$ to $P(0,5)$ is $$D^2(t)=(4t-0)^2+(2t^2-5)^2=16t^2+4t^4-20t^2+25=4t^4-4t^2+25.$$ Let $u=t^2\ge 0$. Then $$D^2=4u^2-4u+25=4\left(u-\frac12\right)^2+24\ge 24.$$ So $D_{\min}=\sqrt{24}=2\sqrt6$, attained when $u=\frac12\Rightarrow t=\pm\frac{1}{\sqrt2}$.

(3) The nearest points are $$(4t,2t^2)=\left(\pm 2\sqrt2,\,1\right).$$

The minimum distance is $2\sqrt6$, attained at $(2\sqrt2,1)$ and $(-2\sqrt2,1)$.