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...

The slope of $y = 1/x$ at $x$ is $-1/x^2$,
so the tangent at $(p, 1/p)$ is
$(y-1/p)/(x-p) = -1/p^2$
or $y = 1/p - (x-p)/p^2$.
At $y = 0$, $(x-p)/p^2 = 1/p$
or $x = 2p$;
at $x = 0$, $y = 1/p + p/p^2 = 2/p$.
The intercept points are thus
$(0, 2/p)$ and $(2p, 0)$.
I will minimize the distance by making
the line from $(a, b)$ to
$(p, 1/p)$ perpendicular to this tangent
at $(p, 1/p)$
and do this by making their
dot product zero.
The dot product of the vector
from $(a, b)$ to $(p, 1/p)$
with the tangent is
$\begin{align}
(p-a, 1/p-b)\cdot (2p, -2/p)
&= 2p(p-a) - 2/p^2+2b/p \\
&= (2p^3(p-a) - 2 + 2pb)/p^2 \\
&= (2p^4 - 2ap^3 + 2bp-2)/p^2 \\
&= 2(p^4 - ap^3 + bp-1)/p^2
\end{align}
$
and this is zero when
$f(p; a, b) = p^4 - ap^3 + bp-1 = 0$.
I will consider the special case $a = b$
to check this;
there should be a root at $p = 1$.
If $a = b$,
$\begin{align}
f(p; a, a) &= p^4 - ap^3 + ap-1 \\
&= (p^4-1) - a(p^3-1) \\
&= (p-1)(p^3+p^2+p^1 - a(p^2+p+1)) \\
&= (p-1)(p^3 + (1-a)(p^2+p+1))
\end{align}
$
which has a root at $p = 1$
(as expected) and,
if $a > 1$
(so $(a, a)$ is to the right and above the hyperbola),
 another positive real root
since $p^3 + (1-a)(p^2+p+1) < 0$
for small $p$
and $p^3 + (1-a)(p^2+p+1) > 0$
for large $p$.
If $a < 1$,
so $(a, a)$ is between the hyperbola and the origin,
there are no other positive real roots
since $p^3 + (1-a)(p^2+p+1) > 0$
for all $p \ge 0$.
I could argue further about
where the real roots
of $f(p; a, b)$
are using
$f(1/p; a, b) = (1/p)^4 - a(1/p)^3 + b/p-1
= (1 - ap + bp^3 - p^4)/p^4
= -f(p; b, a)/p^4
$
so the roots of
$f(1/p, a, b)$ are the same as the roots of
$f(p; b, a)$,
but that's enough for now for me.
In general, I find it much easier
algebraically to make two vectors
orthogonal by making their dot product zero
than minimizing their distance
using the Pythagorean theorem.
