closest point to on $y=1/x$ to a given point I feel like I'm missing something basic - given a point $(a,b)$ how do I find the closest point to it on the curve $y=1/x$? I tried the direct approach of pluggin in $y=1/x$ into the distance formula but it leads to an order-4 polynomial...

Well. We have by the distance formula that the point on $xy = 1$ closest to $(a,b)$ will be the solution to:
$$\frac{d}{dx}\sqrt{(a-x)^2+(b-\frac{1}{x})^2} = 0$$
Which has the same soutions as 
$$\frac{d}{dx}\left( (a-x)^2+(b-\frac{1}{x})^2\right)= 0\\
-2(a-x)+\frac{2(b-\frac{1}{x})}{x^2} = 0\\
x^4-ax^3+bx-1=0$$
Which has as the first real solution:
$$x==\frac{a}{4}-\frac{1}{2} \surd \left(\frac{a^2}{4}+\frac{2^{1/3} (-4+a b)}{\left(-27 a^2+27 b^2+\sqrt{-4 (-12+3 a b)^3+\left(-27 a^2+27 b^2\right)^2}\right)^{1/3}}+\frac{\left(-27 a^2+27 b^2+\sqrt{-4 (-12+3 a b)^3+\left(-27 a^2+27 b^2\right)^2}\right)^{1/3}}{3\ 2^{1/3}}\right)-\frac{1}{2} \surd \left(\frac{a^2}{2}-\frac{2^{1/3} (-4+a b)}{\left(-27 a^2+27 b^2+\sqrt{-4 (-12+3 a b)^3+\left(-27 a^2+27 b^2\right)^2}\right)^{1/3}}-\frac{\left(-27 a^2+27 b^2+\sqrt{-4 (-12+3 a b)^3+\left(-27 a^2+27 b^2\right)^2}\right)^{1/3}}{3\ 2^{1/3}}-\frac{a^3-8 b}{4 \sqrt{\frac{a^2}{4}+\frac{2^{1/3} (-4+a b)}{\left(-27 a^2+27 b^2+\sqrt{-4 (-12+3 a b)^3+\left(-27 a^2+27 b^2\right)^2}\right)^{1/3}}+\frac{\left(-27 a^2+27 b^2+\sqrt{-4 (-12+3 a b)^3+\left(-27 a^2+27 b^2\right)^2}\right)^{1/3}}{3\ 2^{1/3}}}}\right)$$
And the others are just as complicated. This was produced with Mathematica and simplifying under the assumption that the root was real ,$b>-a$ and, $x>0$.
This is such a simply posed problem, but it seems as if the solution is incredibly complicated. 
