Which point of the graph of $y=\sqrt{x}$ is closest to the point $(1,0)$? This problem was assigned for an AP Calculus AB class and was not allowed a calculator:
Which point of the graph of $y=\sqrt{x}$ is closest to the point $(1,0)$?
We are not given answers and the teacher will be absent for $2$ weeks. I need to check my answer $(1/2,1/4)$ before she returns.
Here is my work:
$d=\sqrt{(x-1)^2+(y-0)^2}$
$d=\sqrt{(x-1)^2+x}$
$d^2=(x-1)^2+x$
$2dd'=2(x-1)+1$
$d'=(2(x-1)+1)/(2d)$
$0=2(x-1)+1$
$x=1/2$

An algebra-free approach:

In order to draw the tangent from a point $P$ on a parabola, it is sufficient to project $P$ on the axis, reflect this point with respect to the vertex and join the new point with $P$. Since the tangent drawn from $P=\left(\frac{1}{2},\frac{1}{\sqrt{2}}\right)$ is orthogonal to the line joining $P$ with $(1,0)$ (by Euclid's second theorem on right triangles, or just by computing slopes), $P$ is the wanted solution.
